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Line 30: <br> =={{header\|11l}}== {{trans\|C++}} <syntaxhighlight lang="11l"> F is_curzon(n, k) V m = k * n + 1 R pow(Int64(k), n, m) + 1 == m L(k) [2, 4, 6, 8, 10] V n = 1 [Int] curzons L curzons.len < 1000 I is_curzon(n, k) curzons.append(n) n++ print(‘Curzon numbers with k = ’k‘:’) L(c) curzons[0.<50] V i = L.index print(f:‘{commatize(c):6}’, end' I (i + 1) % 25 == 0 {"\n"} E ‘’) print(‘ Thousandth Curzon with k = ’k‘: ’curzons[999]".\n") </syntaxhighlight> {{out}} <pre> Curzon numbers with k = 2: 1 2 5 6 9 14 18 21 26 29 30 33 41 50 53 54 65 69 74 78 81 86 89 90 98 105 113 114 125 134 138 141 146 153 158 165 173 174 186 189 194 198 209 210 221 230 233 245 249 254 Thousandth Curzon with k = 2: 8646. Curzon numbers with k = 4: 1 3 7 9 13 15 25 27 37 39 43 45 49 57 67 69 73 79 87 93 97 99 105 115 127 135 139 153 163 165 169 175 177 183 189 193 199 205 207 213 219 235 249 253 255 265 267 273 277 279 Thousandth Curzon with k = 4: 9375. Curzon numbers with k = 6: 1 6 30 58 70 73 90 101 105 121 125 146 153 166 170 181 182 185 210 233 241 242 266 282 290 322 373 381 385 390 397 441 445 446 450 453 530 557 562 585 593 601 602 605 606 621 646 653 670 685 Thousandth Curzon with k = 6: 20717. Curzon numbers with k = 8: 1 14 35 44 72 74 77 129 131 137 144 149 150 185 200 219 236 266 284 285 299 309 336 357 381 386 390 392 402 414 420 441 455 459 470 479 500 519 527 536 557 582 600 602 617 639 654 674 696 735 Thousandth Curzon with k = 8: 22176. Curzon numbers with k = 10: 1 9 10 25 106 145 190 193 238 253 306 318 349 385 402 462 486 526 610 649 658 678 733 762 810 990 994 1,033 1,077 1,125 1,126 1,141 1,149 1,230 1,405 1,422 1,441 1,485 1,509 1,510 1,513 1,606 1,614 1,630 1,665 1,681 1,690 1,702 1,785 1,837 Thousandth Curzon with k = 10: 46845. </pre> =={{header\|ALGOL 68}}== {{Trans\|C++}} <syntaxhighlight lang="algol68"> BEGIN # find some generalised Curzon numbers - translation of the C++ sample # PROC modpow = ( LONG INT rqd base, rqd exp, mod )LONG INT: IF mod = 1 THEN 0 ELSE LONG INT result := 1; LONG INT base := rqd base MOD mod; LONG INT exp := rqd exp; WHILE exp > 0 DO IF ODD exp THEN result TIMESAB base MODAB mod FI; base TIMESAB base MODAB mod; exp OVERAB 2 OD; result FI # modpow # ; PROC is curzon = ( LONG INT n, k )BOOL: BEGIN LONG INT m = k * n + 1; modpow( k, n, m ) + 1 = m END # is curon # ; FOR k FROM 2 BY 2 TO 10 DO print( ( "Curzon numbers with base ", whole( k, 0 ), ":", newline ) ); INT count := 0, n := 0; WHILE n +:= 1; count < 50 DO IF is curzon( n, k ) THEN print( ( whole( n, -4 ) , IF ( count +:= 1 ) MOD 10 = 0 THEN newline ELSE " " FI ) ) FI OD; WHILE IF is curzon( n, k ) THEN count +:= 1 FI; count < 1000 DO n +:= 1 OD; print( ( "1000th Curzon number with base ", whole( k, 0 ), ": ", whole( n, 0 ) ) ); print( ( newline, newline ) ) OD END </syntaxhighlight> {{out}} <pre> Curzon numbers with base 2: 1 2 5 6 9 14 18 21 26 29 30 33 41 50 53 54 65 69 74 78 81 86 89 90 98 105 113 114 125 134 138 141 146 153 158 165 173 174 186 189 194 198 209 210 221 230 233 245 249 254 1000th Curzon number with base 2: 8646 Curzon numbers with base 4: 1 3 7 9 13 15 25 27 37 39 43 45 49 57 67 69 73 79 87 93 97 99 105 115 127 135 139 153 163 165 169 175 177 183 189 193 199 205 207 213 219 235 249 253 255 265 267 273 277 279 1000th Curzon number with base 4: 9375 Curzon numbers with base 6: 1 6 30 58 70 73 90 101 105 121 125 146 153 166 170 181 182 185 210 233 241 242 266 282 290 322 373 381 385 390 397 441 445 446 450 453 530 557 562 585 593 601 602 605 606 621 646 653 670 685 1000th Curzon number with base 6: 20717 Curzon numbers with base 8: 1 14 35 44 72 74 77 129 131 137 144 149 150 185 200 219 236 266 284 285 299 309 336 357 381 386 390 392 402 414 420 441 455 459 470 479 500 519 527 536 557 582 600 602 617 639 654 674 696 735 1000th Curzon number with base 8: 22176 Curzon numbers with base 10: 1 9 10 25 106 145 190 193 238 253 306 318 349 385 402 462 486 526 610 649 658 678 733 762 810 990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 1000th Curzon number with base 10: 46845 </pre> =={{header\|Arturo}}== <~~lang~~syntaxhighlight lang="rebol">curzon?: function [n,base]-> zero? (inc base^n) % inc basen Line 63 ⟶ 207: print ["\n1000th Curzon with base" withBase "=" oneThousandth withBase] print "" ]</~~lang~~syntaxhighlight> {{out}} Line 111 ⟶ 255: 1000th Curzon with base 10 = 46845</pre> =={{header\|C}}== {{trans\|C++}} <syntaxhighlight lang="c">#include <stdio.h> #include <stdbool.h> #include <stdint.h> #include <locale.h> uint64_t modPow(uint64_t base, uint64_t exp, uint64_t mod) { if (mod == 1) return 0; uint64_t result = 1; base %= mod; for (; exp > 0; exp >>= 1) { if ((exp & 1) == 1) result = (result base) % mod; base = (base * base) % mod; } return result; } bool isCurzon(uint64_t n, uint64_t k) { const uint64_t r = k * n; return modPow(k, n, r+1) == r; } int main() { uint64_t k, n, count; setlocale(LC_NUMERIC, ""); for (k = 2; k <= 10; k += 2) { printf("Curzon numbers with base %ld:\n", k); for (n = 1, count = 0; count < 50; ++n) { if (isCurzon(n, k)) { printf("%4ld ", n); if (++count % 10 == 0) printf("\n"); } } for (;;) { if (isCurzon(n, k)) ++count; if (count == 1000) break; ++n; } printf("1,000th Curzon number with base %ld: %'ld\n\n", k, n); } return 0; }</syntaxhighlight> {{out}} <pre> Curzon numbers with base 2: 1 2 5 6 9 14 18 21 26 29 30 33 41 50 53 54 65 69 74 78 81 86 89 90 98 105 113 114 125 134 138 141 146 153 158 165 173 174 186 189 194 198 209 210 221 230 233 245 249 254 1,000th Curzon number with base 2: 8,646 Curzon numbers with base 4: 1 3 7 9 13 15 25 27 37 39 43 45 49 57 67 69 73 79 87 93 97 99 105 115 127 135 139 153 163 165 169 175 177 183 189 193 199 205 207 213 219 235 249 253 255 265 267 273 277 279 1,000th Curzon number with base 4: 9,375 Curzon numbers with base 6: 1 6 30 58 70 73 90 101 105 121 125 146 153 166 170 181 182 185 210 233 241 242 266 282 290 322 373 381 385 390 397 441 445 446 450 453 530 557 562 585 593 601 602 605 606 621 646 653 670 685 1,000th Curzon number with base 6: 20,717 Curzon numbers with base 8: 1 14 35 44 72 74 77 129 131 137 144 149 150 185 200 219 236 266 284 285 299 309 336 357 381 386 390 392 402 414 420 441 455 459 470 479 500 519 527 536 557 582 600 602 617 639 654 674 696 735 1,000th Curzon number with base 8: 22,176 Curzon numbers with base 10: 1 9 10 25 106 145 190 193 238 253 306 318 349 385 402 462 486 526 610 649 658 678 733 762 810 990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 1,000th Curzon number with base 10: 46,845 </pre> =={{header\|C++}}== <syntaxhighlight lang="cpp">#include <cstdint> ~~{{libheader\|GMP}}~~ ~~<lang cpp>~~#include <iomanip> #include <iostream> #include <vector> uint64_t modpow(uint64_t base, uint64_t exp, uint64_t mod) { ~~#include <gmpxx.h>~~ if (mod == 1) return 0; uint64_t result = 1; base %= mod; for (; exp > 0; exp >>= 1) { if ((exp & 1) == 1) result = (result * base) % mod; base = (base * base) % mod; } return result; } bool is_curzon(~~int~~uint64_t n, ~~int~~uint64_t k) { const uint64_t r = k * n; ~~mpz_class p;~~ return modpow(k, n, r + 1) == r; ~~mpz_ui_pow_ui(p.get_mpz_t(), k, n);~~ ~~return (p + 1) % (k * n + 1) == 0;~~ } int main() { for (~~int~~uint64_t k = 2; k <= 10; k += 2) { std::cout << "Curzon numbers with base " << k << ":\n"; ~~int~~uint64_t count = 0, n = 1; for (; count < 50; ++n) { if (is_curzon(n, k)) { Line 147 ⟶ 388: } return 0; }</~~lang~~syntaxhighlight> {{out}} Line 192 ⟶ 433: </pre> =={{header\|EasyLang}}== {{trans\|FreeBASIC}} <syntaxhighlight> func pow_mod b power modulus . x = 1 while power > 0 if power mod 2 = 1 x = x * b mod modulus . b = b * b mod modulus power = power div 2 . return x . for k = 2 step 2 to 10 numfmt 0 0 print "First 50 Curzon numbers using a base of " & k & ":" numfmt 0 4 n = 1 count = 0 repeat m = k * n + 1 p = pow_mod k n m + 1 if p = m count += 1 if count <= 50 write " " & n if count mod 9 = 0 print "" . . . until count = 1000 n += 1 . print "" ; print "One thousandth: " & n print "" . </syntaxhighlight> =={{header\|Factor}}== {{works with\|Factor\|0.99 2022-04-03}} <~~lang~~syntaxhighlight lang="factor">USING: grouping interpolate io kernel make math math.functions prettyprint ranges sequences ; Line 209 ⟶ 490: curzon 10 group simple-table. ; 2 10 2 <range> [ curzon. nl ] each</~~lang~~syntaxhighlight> {{out}} <pre> Line 247 ⟶ 528: 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 </pre> =={{header\|FreeBASIC}}== ===Normal basic=== <syntaxhighlight lang="freebasic">' limit: k * n +1 must be smaller then 2^32-1 Function pow_mod(b As ULongInt, power As ULongInt, modulus As ULongInt) As ULongInt ' returns b ^ power mod modulus Dim As ULongInt x = 1 While power > 0 If (power And 1) = 1 Then x = (x * b) Mod modulus End If b = (b * b) Mod modulus power = power Shr 1 Wend Return x End Function For k As ULongInt= 2 To 10 Step 2 Print "The first 50 Curzon numbers using a base of "; k; ":" Dim As ULongInt count, n = 1, p, m Do m = k * n +1 p = pow_mod(k, n ,m) +1 If p = m Then count += 1 If count <= 50 Then Print Using "#####"; n; If (count Mod 10) = 0 Then Print ElseIf count = 1000 Then Print : Print "One thousandth: "; n Print : Print Exit Do End If End If n += 1 Loop Next Sleep</syntaxhighlight>{{out}}<pre>The first 50 Curzon numbers using a base of 2: 1 2 5 6 9 14 18 21 26 29 30 33 41 50 53 54 65 69 74 78 81 86 89 90 98 105 113 114 125 134 138 141 146 153 158 165 173 174 186 189 194 198 209 210 221 230 233 245 249 254 One thousandth: 8646 The first 50 Curzon numbers using a base of 4: 1 3 7 9 13 15 25 27 37 39 43 45 49 57 67 69 73 79 87 93 97 99 105 115 127 135 139 153 163 165 169 175 177 183 189 193 199 205 207 213 219 235 249 253 255 265 267 273 277 279 One thousandth: 9375 The first 50 Curzon numbers using a base of 6: 1 6 30 58 70 73 90 101 105 121 125 146 153 166 170 181 182 185 210 233 241 242 266 282 290 322 373 381 385 390 397 441 445 446 450 453 530 557 562 585 593 601 602 605 606 621 646 653 670 685 One thousandth: 20717 The first 50 Curzon numbers using a base of 8: 1 14 35 44 72 74 77 129 131 137 144 149 150 185 200 219 236 266 284 285 299 309 336 357 381 386 390 392 402 414 420 441 455 459 470 479 500 519 527 536 557 582 600 602 617 639 654 674 696 735 One thousandth: 22176 The first 50 Curzon numbers using a base of 10: 1 9 10 25 106 145 190 193 238 253 306 318 349 385 402 462 486 526 610 649 658 678 733 762 810 990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 One thousandth: 46845</pre> ===GMP version=== {{libheader\|GMP}} <syntaxhighlight lang="freebasic">#include once "gmp.bi" Dim As Longint t = Len(__mpz_struct) Dim As mpz_ptr pow = Allocate(t) Dim As mpz_ptr z = Allocate(t) mpz_init(pow): mpz_init(z) For k As Uinteger = 2 To 10 Step 2 Print "The first 50 Curzon numbers using a base of"; k; ":" Dim As Integer count = 0, n = 1 mpz_set_si(pow,k) Do mpz_add_ui(z,pow,1) Dim As Integer d = kn + 1 If mpz_divisible_ui_p(z,d) Then count += 1 If count <= 50 Then Print Using "#####"; n If (count Mod 25) = 0 Then Print Elseif count=1000 Then Print "One thousandth: "; n Print : Print Exit Do End If End If n += 1 mpz_mul_si(pow,pow,k) Loop Next k Sleep</syntaxhighlight> =={{header\|Go}}== {{trans\|Wren}} Based on Version 1. ~~<lang go>package main~~ <syntaxhighlight lang="go">package main import ( Line 296 ⟶ 703: fmt.Println() } }</~~lang~~syntaxhighlight> {{out}} Line 344 ⟶ 751: One thousandth: 46845 </pre> =={{header\|Haskell}}== <syntaxhighlight lang=Haskell> import Data.List.Split ( chunksOf ) isGeneralizedCurzon :: Integer -> Integer -> Bool isGeneralizedCurzon base n = mod ( base ^ n + 1 ) ( base n + 1 ) == 0 solution :: Integer -> [Integer] solution base = take 50 $ filter (\i -> isGeneralizedCurzon base i ) [1..] printChunk :: [Integer] -> String printChunk chunk = foldl1 (++) $ map (\i -> (take ( 4 - (length $ show i) ) $ repeat ' ' ) ++ show i ++ " ") chunk prettyPrint :: [Integer] -> [String] prettyPrint list = map printChunk $ chunksOf 10 list oneThousandth :: Integer -> Integer oneThousandth base = last $ take 950 $ filter (\i -> isGeneralizedCurzon base i ) [(last $ solution base) + 1 ..] printBlock :: Integer -> [String] printBlock base = ["first 50 Curzon numbers using a base of " ++ show base ++ " :"] ++ (prettyPrint $ solution base) ++ ["one thousandth at base " ++ show base ++ ": " ++ (show $ oneThousandth base)] ++ [take 50 $ repeat '-'] main :: IO ( ) main = do blocks <- return $ concat $ map (\i -> printBlock i ) [2 , 4 , 6 , 8 , 10] mapM_ putStrLn blocks</syntaxhighlight> {{out}} <pre>first 50 Curzon numbers using a base of 2 : 1 2 5 6 9 14 18 21 26 29 30 33 41 50 53 54 65 69 74 78 81 86 89 90 98 105 113 114 125 134 138 141 146 153 158 165 173 174 186 189 194 198 209 210 221 230 233 245 249 254 one thousandth at base 2: 8646 -------------------------------------------------- first 50 Curzon numbers using a base of 4 : 1 3 7 9 13 15 25 27 37 39 43 45 49 57 67 69 73 79 87 93 97 99 105 115 127 135 139 153 163 165 169 175 177 183 189 193 199 205 207 213 219 235 249 253 255 265 267 273 277 279 one thousandth at base 4: 9375 -------------------------------------------------- first 50 Curzon numbers using a base of 6 : 1 6 30 58 70 73 90 101 105 121 125 146 153 166 170 181 182 185 210 233 241 242 266 282 290 322 373 381 385 390 397 441 445 446 450 453 530 557 562 585 593 601 602 605 606 621 646 653 670 685 one thousandth at base 6: 20717 -------------------------------------------------- first 50 Curzon numbers using a base of 8 : 1 14 35 44 72 74 77 129 131 137 144 149 150 185 200 219 236 266 284 285 299 309 336 357 381 386 390 392 402 414 420 441 455 459 470 479 500 519 527 536 557 582 600 602 617 639 654 674 696 735 one thousandth at base 8: 22176 -------------------------------------------------- first 50 Curzon numbers using a base of 10 : 1 9 10 25 106 145 190 193 238 253 306 318 349 385 402 462 486 526 610 649 658 678 733 762 810 990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 one thousandth at base 10: 46845 -------------------------------------------------- </pre> =={{header\|J}}== A simple test for whether a number is a (generalized) Curzon number is: <syntaxhighlight lang="j">isCurzon =: 2&$: : (0 = * \|&:>: ^)</syntaxhighlight> Here, the monadic case uses 2 as the left argument for the dyadic case, which handles generalized Curzon numbers. However, this is inefficient for large numbers, due to it explicitly exponentiating. Since we test the remainder immediately after, we can use a special combination instead, <syntaxhighlight lang="j">modpow =: {{ m&\|@^ }} isCurzon =: {{ z =: >: x * y z = >: x z modpow y }}</syntaxhighlight> This avoids performing the large exponentiation, which is significantly faster. For generating the results: <syntaxhighlight lang="j">generateCurzons =: {{ found =. i. 0x current =. 0x while. 1000 > # found do. if. y isCurzon current do. found =. found , current end. current =. >: current end. y ; (5 10 $ found) ; {: found }} ('Base';'First 50';'1000th') , generateCurzons"0 +:>:i.5</syntaxhighlight> {{out}} <pre> +----+-------------------------------------------------+------+ \|Base\|First 50 \|1000th\| +----+-------------------------------------------------+------+ \|2 \| 1 2 5 6 9 14 18 21 26 29 \|8646 \| \| \| 30 33 41 50 53 54 65 69 74 78 \| \| \| \| 81 86 89 90 98 105 113 114 125 134 \| \| \| \|138 141 146 153 158 165 173 174 186 189 \| \| \| \|194 198 209 210 221 230 233 245 249 254 \| \| +----+-------------------------------------------------+------+ \|4 \| 1 3 7 9 13 15 25 27 37 39 \|9375 \| \| \| 43 45 49 57 67 69 73 79 87 93 \| \| \| \| 97 99 105 115 127 135 139 153 163 165 \| \| \| \|169 175 177 183 189 193 199 205 207 213 \| \| \| \|219 235 249 253 255 265 267 273 277 279 \| \| +----+-------------------------------------------------+------+ \|6 \| 1 6 30 58 70 73 90 101 105 121 \|20717 \| \| \|125 146 153 166 170 181 182 185 210 233 \| \| \| \|241 242 266 282 290 322 373 381 385 390 \| \| \| \|397 441 445 446 450 453 530 557 562 585 \| \| \| \|593 601 602 605 606 621 646 653 670 685 \| \| +----+-------------------------------------------------+------+ \|8 \| 1 14 35 44 72 74 77 129 131 137 \|22176 \| \| \|144 149 150 185 200 219 236 266 284 285 \| \| \| \|299 309 336 357 381 386 390 392 402 414 \| \| \| \|420 441 455 459 470 479 500 519 527 536 \| \| \| \|557 582 600 602 617 639 654 674 696 735 \| \| +----+-------------------------------------------------+------+ \|10 \| 1 9 10 25 106 145 190 193 238 253\|46845 \| \| \| 306 318 349 385 402 462 486 526 610 649\| \| \| \| 658 678 733 762 810 990 994 1033 1077 1125\| \| \| \|1126 1141 1149 1230 1405 1422 1441 1485 1509 1510\| \| \| \|1513 1606 1614 1630 1665 1681 1690 1702 1785 1837\| \| +----+-------------------------------------------------+------+ </pre> =={{header\|Java}}== <syntaxhighlight lang="java"> public final class CurzonNumbers { public static void main(String[] aArgs) { for ( int k = 2; k <= 10; k += 2 ) { System.out.println("Generalised Curzon numbers with base " + k + ":"); int n = 1; int count = 0; while ( count < 50 ) { if ( isGeneralisedCurzonNumber(k, n) ) { System.out.print(String.format("%4d%s", n, ( ++count % 10 == 0 ? "\n" : " " ))); } n += 1; } while ( count < 1_000 ) { if ( isGeneralisedCurzonNumber(k, n) ) { count += 1; } n += 1; } System.out.println("1,000th Generalised Curzon number with base " + k + ": " + ( n - 1 )); System.out.println(); } } private static boolean isGeneralisedCurzonNumber(int aK, int aN) { final long r = aK * aN; return modulusPower(aK, aN, r + 1) == r; } private static long modulusPower(long aBase, long aExponent, long aModulus) { if ( aModulus == 1 ) { return 0; } aBase %= aModulus; long result = 1; while ( aExponent > 0 ) { if ( ( aExponent & 1 ) == 1 ) { result = ( result * aBase ) % aModulus; } aBase = ( aBase * aBase ) % aModulus; aExponent >>= 1; } return result; } } </syntaxhighlight> {{ out }} <pre> Generalised Curzon numbers with base 2: 1 2 5 6 9 14 18 21 26 29 30 33 41 50 53 54 65 69 74 78 81 86 89 90 98 105 113 114 125 134 138 141 146 153 158 165 173 174 186 189 194 198 209 210 221 230 233 245 249 254 1,000th Generalised Curzon number with base 2: 8646 Generalised Curzon numbers with base 4: 1 3 7 9 13 15 25 27 37 39 43 45 49 57 67 69 73 79 87 93 97 99 105 115 127 135 139 153 163 165 169 175 177 183 189 193 199 205 207 213 219 235 249 253 255 265 267 273 277 279 1,000th Generalised Curzon number with base 4: 9375 Generalised Curzon numbers with base 6: 1 6 30 58 70 73 90 101 105 121 125 146 153 166 170 181 182 185 210 233 241 242 266 282 290 322 373 381 385 390 397 441 445 446 450 453 530 557 562 585 593 601 602 605 606 621 646 653 670 685 1,000th Generalised Curzon number with base 6: 20717 Generalised Curzon numbers with base 8: 1 14 35 44 72 74 77 129 131 137 144 149 150 185 200 219 236 266 284 285 299 309 336 357 381 386 390 392 402 414 420 441 455 459 470 479 500 519 527 536 557 582 600 602 617 639 654 674 696 735 1,000th Generalised Curzon number with base 8: 22176 Generalised Curzon numbers with base 10: 1 9 10 25 106 145 190 193 238 253 306 318 349 385 402 462 486 526 610 649 658 678 733 762 810 990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 1,000th Generalised Curzon number with base 10: 46845 </pre> Line 354 ⟶ 996: '''Preliminaries''' <~~lang~~syntaxhighlight lang="jq"># To take advantage of gojq's arbitrary-precision integer arithmetic: def power($b): . as $in \| reduce range(0;$b) as $i (1; . * $in); Line 364 ⟶ 1,006: n; def printRows($m): _nwise($m) \| map(lpad(5)) \| join("");</~~lang~~syntaxhighlight> '''The task''' <~~lang~~syntaxhighlight lang="jq">def isCurzon($n; $k): ($k \| power($n) + 1) % ($k * $n + 1) == 0; Line 384 ⟶ 1,026: ""; printcurzons(2; 10; 1000)</~~lang~~syntaxhighlight> {{out}} <pre> Line 415 ⟶ 1,057: =={{header\|Julia}}== <~~lang~~syntaxhighlight lang="julia">isCurzon(n, k) = (BigInt(k)^n + 1) % (k * n + 1) == 0 function printcurzons(klow, khigh) Line 431 ⟶ 1,073: printcurzons(2, 10) </~~lang~~syntaxhighlight>{{out}} <pre> Curzon numbers with k = 2: Line 453 ⟶ 1,095: 990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 Thousandth Curzon with k = 10: 46845 </pre> =={{header\|Lambdatalk}}== Using Javascript's BigInt. <syntaxhighlight lang="scheme"> {def is_curzon {lambda {:n :k} {= {BI.% {BI.+ {BI.** :k :n} 1} {BI.+ {BI.* :k :n} 1} } 0}}} -> is_curzon {def curzon {lambda {:length :base} {S.replace \s by space in {S.brmap {{lambda {:length :base :count :i} {if {< {A.get 0 :count} :length} then {if {is_curzon :i :base} then {{lambda {:a :i} :i} {A.set! 0 {+ {A.get 0 :count} 1} :count} :i} else} else _break_} } :length :base {A.new 0}} 1 100000 1 }}}} -> curzon {S.map {lambda {:i} {div {b First 50 Curzon numbers using a base of :i:} {div {curzon 50 :i}}}} {S.serie 2 10 2}} -> First 50 Curzon numbers using a base of 2: 1 2 5 6 9 14 18 21 26 29 30 33 41 50 53 54 65 69 74 78 81 86 89 90 98 105 113 114 125 134 138 141 146 153 158 165 173 174 186 189 194 198 209 210 221 230 233 245 249 254 First 50 Curzon numbers using a base of 4: 1 3 7 9 13 15 25 27 37 39 43 45 49 57 67 69 73 79 87 93 97 99 105 115 127 135 139 153 163 165 169 175 177 183 189 193 199 205 207 213 219 235 249 253 255 265 267 273 277 279 First 50 Curzon numbers using a base of 6: 1 6 30 58 70 73 90 101 105 121 125 146 153 166 170 181 182 185 210 233 241 242 266 282 290 322 373 381 385 390 397 441 445 446 450 453 530 557 562 585 593 601 602 605 606 621 646 653 670 685 First 50 Curzon numbers using a base of 8: 1 14 35 44 72 74 77 129 131 137 144 149 150 185 200 219 236 266 284 285 299 309 336 357 381 386 390 392 402 414 420 441 455 459 470 479 500 519 527 536 557 582 600 602 617 639 654 674 696 735 First 50 Curzon numbers using a base of 10: 1 9 10 25 106 145 190 193 238 253 306 318 349 385 402 462 486 526 610 649 658 678 733 762 810 990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 {S.last {curzon 1000 6}} -> 20717 in 272193ms = 4.53655 minutes </syntaxhighlight> =={{header\|Maxima}}== <syntaxhighlight lang="maxima"> /* Predicate function that checks wether a positive integer is generalized k-Curzon number or not / g_curzonp(n,k):=if mod((k^n)+1,kn+1)=0 then true$ /* Function that returns a list of the first len generalized k-Curzon numbers / g_curzon_count(len,k):=block( [i:1,count:0,result:[]], while count<len do (if g_curzonp(i,k) then (result:endcons(i,result),count:count+1),i:i+1), result)$ / Test cases / g_curzon_count(50,2); g_curzon_count(50,4); g_curzon_count(50,6); g_curzon_count(50,8); g_curzon_count(50,10); </syntaxhighlight> {{out}} <pre> [1,2,5,6,9,14,18,21,26,29,30,33,41,50,53,54,65,69,74,78,81,86,89,90,98,105,113,114,125,134,138,141,146,153,158,165,173,174,186,189,194,198,209,210,221,230,233,245,249,254] [1,3,7,9,13,15,25,27,37,39,43,45,49,57,67,69,73,79,87,93,97,99,105,115,127,135,139,153,163,165,169,175,177,183,189,193,199,205,207,213,219,235,249,253,255,265,267,273,277,279] [1,6,30,58,70,73,90,101,105,121,125,146,153,166,170,181,182,185,210,233,241,242,266,282,290,322,373,381,385,390,397,441,445,446,450,453,530,557,562,585,593,601,602,605,606,621,646,653,670,685] [1,14,35,44,72,74,77,129,131,137,144,149,150,185,200,219,236,266,284,285,299,309,336,357,381,386,390,392,402,414,420,441,455,459,470,479,500,519,527,536,557,582,600,602,617,639,654,674,696,735] [1,9,10,25,106,145,190,193,238,253,306,318,349,385,402,462,486,526,610,649,658,678,733,762,810,990,994,1033,1077,1125,1126,1141,1149,1230,1405,1422,1441,1485,1509,1510,1513,1606,1614,1630,1665,1681,1690,1702,1785,1837] </pre> =={{header\|Nim}}== Nim standard library doesn’t provide a modular power, so we have to define our own function. <syntaxhighlight lang="Nim">import std/strformat func pow(a, n: Natural; m: Positive): Natural = var a = a mod m var n = n if a > 0: result = 1 while n > 0: if (n and 1) != 0: result = (result a) mod m n = n shr 1 a = (a * a) mod m iterator curzonNumbers(k: Positive): Natural = assert (k and 1) == 0, "base must be even." var n = 1 while true: let m = k * n + 1 if pow(k, n, m) + 1 == m: yield n inc n ### Task ### for k in countup(2, 10, 2): echo &"Curzon numbers for k = {k}:" var count = 0 for n in curzonNumbers(k): inc count stdout.write &"{n:>4}" stdout.write if count mod 10 == 0: '\n' else: ' ' if count == 50: break echo() ### Stretch task ### for k in countup(2, 10, 2): var count = 0 for n in curzonNumbers(k): inc count if count == 1000: echo &"1000th Curzon number for k = {k:>2}: {n:>5}" break </syntaxhighlight> {{out}} <pre>Curzon numbers for k = 2: 1 2 5 6 9 14 18 21 26 29 30 33 41 50 53 54 65 69 74 78 81 86 89 90 98 105 113 114 125 134 138 141 146 153 158 165 173 174 186 189 194 198 209 210 221 230 233 245 249 254 Curzon numbers for k = 4: 1 3 7 9 13 15 25 27 37 39 43 45 49 57 67 69 73 79 87 93 97 99 105 115 127 135 139 153 163 165 169 175 177 183 189 193 199 205 207 213 219 235 249 253 255 265 267 273 277 279 Curzon numbers for k = 6: 1 6 30 58 70 73 90 101 105 121 125 146 153 166 170 181 182 185 210 233 241 242 266 282 290 322 373 381 385 390 397 441 445 446 450 453 530 557 562 585 593 601 602 605 606 621 646 653 670 685 Curzon numbers for k = 8: 1 14 35 44 72 74 77 129 131 137 144 149 150 185 200 219 236 266 284 285 299 309 336 357 381 386 390 392 402 414 420 441 455 459 470 479 500 519 527 536 557 582 600 602 617 639 654 674 696 735 Curzon numbers for k = 10: 1 9 10 25 106 145 190 193 238 253 306 318 349 385 402 462 486 526 610 649 658 678 733 762 810 990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 1000th Curzon number for k = 2: 8646 1000th Curzon number for k = 4: 9375 1000th Curzon number for k = 6: 20717 1000th Curzon number for k = 8: 22176 1000th Curzon number for k = 10: 46845 </pre> =={{header\|OCaml}}== <syntaxhighlight lang="ocaml">let modpow m = let rec loop p b e = if e land 1 = 0 then if e = 0 then p else loop p (b * b mod m) (e lsr 1) else loop (p * b mod m) (b * b mod m) (e lsr 1) in loop 1 let is_curzon k n = let r = k * n in r = modpow (succ r) k n let () = List.iter (fun x -> Seq.(ints 0 \|> filter (is_curzon x) \|> take 50 \|> map string_of_int) \|> List.of_seq \|> String.concat " " \|> Printf.printf "base %u:\n%s\n" x) [2; 4; 6; 8; 10] let () = List.iter (fun x -> Seq.(ints 0 \|> filter (is_curzon x) \|> drop 999 \|> take 1 \|> iter (Printf.printf "base %u (1000th): %u\n" x))) [2; 4; 6; 8; 10]</syntaxhighlight> <pre> base 2: 1 2 5 6 9 14 18 21 26 29 30 33 41 50 53 54 65 69 74 78 81 86 89 90 98 105 113 114 125 134 138 141 146 153 158 165 173 174 186 189 194 198 209 210 221 230 233 245 249 254 base 4: 1 3 7 9 13 15 25 27 37 39 43 45 49 57 67 69 73 79 87 93 97 99 105 115 127 135 139 153 163 165 169 175 177 183 189 193 199 205 207 213 219 235 249 253 255 265 267 273 277 279 base 6: 1 6 30 58 70 73 90 101 105 121 125 146 153 166 170 181 182 185 210 233 241 242 266 282 290 322 373 381 385 390 397 441 445 446 450 453 530 557 562 585 593 601 602 605 606 621 646 653 670 685 base 8: 1 14 35 44 72 74 77 129 131 137 144 149 150 185 200 219 236 266 284 285 299 309 336 357 381 386 390 392 402 414 420 441 455 459 470 479 500 519 527 536 557 582 600 602 617 639 654 674 696 735 base 10: 1 9 10 25 106 145 190 193 238 253 306 318 349 385 402 462 486 526 610 649 658 678 733 762 810 990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 base 2 (1000th): 8646 base 4 (1000th): 9375 base 6 (1000th): 20717 base 8 (1000th): 22176 base 10 (1000th): 46845 </pre> =={{header\|Odin}}== <syntaxhighlight lang="Go"> package curzon_numbers /* imports / import "core:c/libc" import "core:fmt" / main block / main :: proc() { for k: int = 2; k <= 10; k += 2 { fmt.println("\nCurzon numbers with base ", k) count := 0 n: int = 1 for ; count < 50; n += 1 { if is_curzon(n, k) { count += 1 libc.printf("%d ", 4, n) if (count) % 10 == 0 { fmt.printf("\n") } } } for { if is_curzon(n, k) { count += 1} if count == 1000 { break} n += 1 } libc.printf("1000th Curzon number with base %d: %d \n", k, n) } } /* definitions / modpow :: proc(base, exp, mod: int) -> int { if mod == 1 { return 0} result: int = 1 base := base exp := exp base %= mod for ; exp > 0; exp >>= 1 { if ((exp & 1) == 1) { result = (result base) % mod} base = (base * base) % mod } return result } is_curzon :: proc(n: int, k: int) -> bool { r := k * n //const? return modpow(k, n, r + 1) == r } </syntaxhighlight> {{out}} <pre> Curzon numbers with base 2 1 2 5 6 9 14 18 21 26 29 30 33 41 50 53 54 65 69 74 78 81 86 89 90 98 105 113 114 125 134 138 141 146 153 158 165 173 174 186 189 194 198 209 210 221 230 233 245 249 254 1000th Curzon number with base 2: 8646 Curzon numbers with base 4 1 3 7 9 13 15 25 27 37 39 43 45 49 57 67 69 73 79 87 93 97 99 105 115 127 135 139 153 163 165 169 175 177 183 189 193 199 205 207 213 219 235 249 253 255 265 267 273 277 279 1000th Curzon number with base 4: 9375 Curzon numbers with base 6 1 6 30 58 70 73 90 101 105 121 125 146 153 166 170 181 182 185 210 233 241 242 266 282 290 322 373 381 385 390 397 441 445 446 450 453 530 557 562 585 593 601 602 605 606 621 646 653 670 685 1000th Curzon number with base 6: 20717 Curzon numbers with base 8 1 14 35 44 72 74 77 129 131 137 144 149 150 185 200 219 236 266 284 285 299 309 336 357 381 386 390 392 402 414 420 441 455 459 470 479 500 519 527 536 557 582 600 602 617 639 654 674 696 735 1000th Curzon number with base 8: 22176 Curzon numbers with base 10 1 9 10 25 106 145 190 193 238 253 306 318 349 385 402 462 486 526 610 649 658 678 733 762 810 990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 1000th Curzon number with base 10: 46845 </pre> Line 459 ⟶ 1,397: It isn't clear why the task description says "generalized Curzon numbers only exist for even base integers." If k >= 3 is an odd base, then, besides the trivial solution n = 1, it can be checked that n = k^(k-1) is a Curzon number according to the given definition. It seems from the output below that Curzon numbers with an odd base are much scarcer than those with an even base. <~~lang~~syntaxhighlight lang="pascal"> program CurzonNumbers; uses SysUtils; Line 528 ⟶ 1,466: ListCurzonNumbers(11); end. </syntaxhighlight> ~~</lang>~~ {{out}} <pre> Line 589 ⟶ 1,527: =={{header\|Mathematica}}/{{header\|Wolfram Language}}== <~~lang~~syntaxhighlight ~~Mathematica~~lang="mathematica">ClearAll[CurzonNumberQ] CurzonNumberQ[b_Integer][n_Integer]:=PowerMod[b,n,b n+1]==b n val=Select[Range[100000],CurzonNumberQ[2]]; Line 609 ⟶ 1,547: val=Select[Range[100000],CurzonNumberQ[10]]; Take[val,50] val[[1000]]</~~lang~~syntaxhighlight> {{out}} <pre>{1, 2, 5, 6, 9, 14, 18, 21, 26, 29, 30, 33, 41, 50, 53, 54, 65, 69, 74, 78, 81, 86, 89, 90, 98, 105, 113, 114, 125, 134, 138, 141, 146, 153, 158, 165, 173, 174, 186, 189, 194, 198, 209, 210, 221, 230, 233, 245, 249, 254} Line 630 ⟶ 1,568: 46845</pre> =={{header\|PARI/GP}}== {{trans\|Mathematica/Wolfram_Language}} <syntaxhighlight lang="PARI/GP"> /* Define the CurzonNumberQ function for base b / powermod(a,k,n)=lift(Mod(a,n)^k) CurzonNumberQ(b, n) = (powermod(b,n,bn+1)== bn); / Define a function to find Curzon numbers within a range for base b / FindCurzonNumbers(b, maxrange) = { local(val, res, i); val = vector(maxrange); res = []; for(i = 1, maxrange, if(CurzonNumberQ(b, i), res = concat(res, i)); ); return(Vec(res)); } / Select and display the first 50 Curzon numbers and the 1000th for base 2 / val = FindCurzonNumbers(2, 100000); print(vector(50, i, val[i])); / First 50 / print(val[1000]); / 1000th Curzon number / / Select and display for base 4 / val = FindCurzonNumbers(4, 100000); print(vector(50, i, val[i])); / First 50 / print(val[1000]); / 1000th Curzon number / / Select and display for base 6 / val = FindCurzonNumbers(6, 100000); print(vector(50, i, val[i])); / First 50 / print(val[1000]); / 1000th Curzon number / / Select and display for base 8 / val = FindCurzonNumbers(8, 100000); print(vector(50, i, val[i])); / First 50 / print(val[1000]); / 1000th Curzon number / / Select and display for base 10 / val = FindCurzonNumbers(10, 100000); print(vector(50, i, val[i])); / First 50 / print(val[1000]); / 1000th Curzon number / </syntaxhighlight> {{out}} <pre> [1, 2, 5, 6, 9, 14, 18, 21, 26, 29, 30, 33, 41, 50, 53, 54, 65, 69, 74, 78, 81, 86, 89, 90, 98, 105, 113, 114, 125, 134, 138, 141, 146, 153, 158, 165, 173, 174, 186, 189, 194, 198, 209, 210, 221, 230, 233, 245, 249, 254] 8646 [1, 3, 7, 9, 13, 15, 25, 27, 37, 39, 43, 45, 49, 57, 67, 69, 73, 79, 87, 93, 97, 99, 105, 115, 127, 135, 139, 153, 163, 165, 169, 175, 177, 183, 189, 193, 199, 205, 207, 213, 219, 235, 249, 253, 255, 265, 267, 273, 277, 279] 9375 [1, 6, 30, 58, 70, 73, 90, 101, 105, 121, 125, 146, 153, 166, 170, 181, 182, 185, 210, 233, 241, 242, 266, 282, 290, 322, 373, 381, 385, 390, 397, 441, 445, 446, 450, 453, 530, 557, 562, 585, 593, 601, 602, 605, 606, 621, 646, 653, 670, 685] 20717 [1, 14, 35, 44, 72, 74, 77, 129, 131, 137, 144, 149, 150, 185, 200, 219, 236, 266, 284, 285, 299, 309, 336, 357, 381, 386, 390, 392, 402, 414, 420, 441, 455, 459, 470, 479, 500, 519, 527, 536, 557, 582, 600, 602, 617, 639, 654, 674, 696, 735] 22176 [1, 9, 10, 25, 106, 145, 190, 193, 238, 253, 306, 318, 349, 385, 402, 462, 486, 526, 610, 649, 658, 678, 733, 762, 810, 990, 994, 1033, 1077, 1125, 1126, 1141, 1149, 1230, 1405, 1422, 1441, 1485, 1509, 1510, 1513, 1606, 1614, 1630, 1665, 1681, 1690, 1702, 1785, 1837] 46845 </pre> =={{header\|Perl}}== {{libheader\|ntheory}} ~~<lang perl>use strict;~~ <syntaxhighlight lang="perl">use strict; use warnings; use ~~Math::AnyNum~~ntheory '~~ipow~~powmod'; sub curzon { Line 640 ⟶ 1,636: my($n,@C) = 0; while (++$n) { ~~push @C,~~my $~~n if 0~~r == ~~(ipow($base,$n) + 1) % (~~$base $n ~~+ 1)~~; push @C, $n if powmod($base, $n, $r + 1) == $r; return @C if $cnt == @C; } Line 647 ⟶ 1,644: my $upto = 50; for my $k (<2 4 6 8 10>) { my @C = curzon( $k, 1000); print "First $upto Curzon numbers using a base of $k:\n" . (sprintf "@{['%5d' x $upto]}", @C[0..$upto-1]) =~ s/(.{~~125~~100})/$1&\n/gr; ~~print~~printf "%50s\n\n", "Thousandth: $C[-1]~~\n\n~~"; }</~~lang~~syntaxhighlight> {{out}} <pre> ~~<pre>First 50 Curzon numbers using a base of 2:~~ First 50 Curzon numbers using a base of 2: ~~1 2 5 6 9 14 18 21 26 29 30 33 41 50 53 54 65 69 74 78 81 86 89 90 98~~ ~~105~~ ~~113~~ ~~114~~1 ~~125~~ ~~134~~2 ~~138~~ ~~141~~5 ~~146~~ ~~153~~6 ~~158~~ ~~165~~9 ~~173~~ 14 ~~174~~ ~~186~~18 ~~189~~ 21 ~~194~~ ~~198~~26 ~~209~~ 29 ~~210~~ ~~221~~30 ~~230~~ 33 ~~233~~ ~~245~~41 ~~249~~ 50 ~~254~~ 53 54 65 69 74 78 81 86 89 90 98 105 113 114 125 134 138 141 146 153 158 165 173 174 186 189 ~~Thousandth: 8646~~ 194 198 209 210 221 230 233 245 249 254 Thousandth: 8646 First 50 Curzon numbers using a base of 4: 1 3 7 9 13 15 25 27 37 39 43 45 49 57 67 69 73 79 87 93 ~~97 99 105 115 127~~ 97 99 105 115 127 135 139 153 163 165 169 175 177 183 189 193 199 205 207 213 ~~219 235 249 253 255 265 267 273 277 279~~ 219 235 249 253 255 265 267 273 277 279 Thousandth: 9375 ~~Thousandth: 9375~~ First 50 Curzon numbers using a base of 6: 1 6 30 58 70 73 90 101 105 121 125 146 153 166 170 181 182 185 210 233 ~~241 242 266 282 290~~ 241 242 266 282 290 322 373 381 385 390 397 441 445 446 450 453 530 557 562 585 ~~593 601 602 605 606 621 646 653 670 685~~ 593 601 602 605 606 621 646 653 670 685 Thousandth: 20717 ~~Thousandth: 20717~~ First 50 Curzon numbers using a base of 8: 1 14 35 44 72 74 77 129 131 137 144 149 150 185 200 219 236 266 284 285 ~~299 309 336 357 381~~ 299 309 336 357 381 386 390 392 402 414 420 441 455 459 470 479 500 519 527 536 ~~557 582 600 602 617 639 654 674 696 735~~ 557 582 600 602 617 639 654 674 696 735 Thousandth: 22176 ~~Thousandth: 22176~~ First 50 Curzon numbers using a base of 10: 1 9 10 25 106 145 190 193 238 253 306 318 349 385 402 462 486 526 610 649 ~~658 678 733 762 810~~ 658 678 733 762 810 990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 ~~1513 1606 1614 1630 1665 1681 1690 1702 1785 1837~~ 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 Thousandth: 46845 ~~Thousandth: 46845</pre>~~ </pre> =={{header\|Phix}}== <!--<~~lang~~syntaxhighlight ~~Phix~~lang="phix">(phixonline)--> <span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span> <span style="color: #008080;">include</span> <span style="color: #004080;">mpfr</span><span style="color: #0000FF;">.</span><span style="color: #000000;">e</span> Line 703 ⟶ 1,702: <span style="color: #008080;">end</span> <span style="color: #008080;">while</span> <span style="color: #008080;">end</span> <span style="color: #008080;">for</span> <!--</~~lang~~syntaxhighlight>--> {{out}} <pre> Line 730 ⟶ 1,729: 990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 One thousandth: 46845 </pre> =={{header\|Python}}== <syntaxhighlight lang="python">def is_Curzon(n, k): r = k * n return pow(k, n, r + 1) == r for k in [2, 4, 6, 8, 10]: n, curzons = 1, [] while len(curzons) < 1000: if is_Curzon(n, k): curzons.append(n) n += 1 print(f'Curzon numbers with k = {k}:') for i, c in enumerate(curzons[:50]): print(f'{c: 5,}', end='\n' if (i + 1) % 25 == 0 else '') print(f' Thousandth Curzon with k = {k}: {curzons[999]}.\n')</syntaxhighlight> {{out}} <pre> Curzon numbers with k = 2: 1 2 5 6 9 14 18 21 26 29 30 33 41 50 53 54 65 69 74 78 81 86 89 90 98 105 113 114 125 134 138 141 146 153 158 165 173 174 186 189 194 198 209 210 221 230 233 245 249 254 Thousandth Curzon with k = 2: 8646. Curzon numbers with k = 4: 1 3 7 9 13 15 25 27 37 39 43 45 49 57 67 69 73 79 87 93 97 99 105 115 127 135 139 153 163 165 169 175 177 183 189 193 199 205 207 213 219 235 249 253 255 265 267 273 277 279 Thousandth Curzon with k = 4: 9375. Curzon numbers with k = 6: 1 6 30 58 70 73 90 101 105 121 125 146 153 166 170 181 182 185 210 233 241 242 266 282 290 322 373 381 385 390 397 441 445 446 450 453 530 557 562 585 593 601 602 605 606 621 646 653 670 685 Thousandth Curzon with k = 6: 20717. Curzon numbers with k = 8: 1 14 35 44 72 74 77 129 131 137 144 149 150 185 200 219 236 266 284 285 299 309 336 357 381 386 390 392 402 414 420 441 455 459 470 479 500 519 527 536 557 582 600 602 617 639 654 674 696 735 Thousandth Curzon with k = 8: 22176. Curzon numbers with k = 10: 1 9 10 25 106 145 190 193 238 253 306 318 349 385 402 462 486 526 610 649 658 678 733 762 810 990 994 1,033 1,077 1,125 1,126 1,141 1,149 1,230 1,405 1,422 1,441 1,485 1,509 1,510 1,513 1,606 1,614 1,630 1,665 1,681 1,690 1,702 1,785 1,837 Thousandth Curzon with k = 10: 46845. </pre> =={{header\|Quackery}}== <~~lang~~syntaxhighlight ~~Quackery~~lang="quackery"> [ number$ space 4 of swap join -5 split nip echo$ ] is rjust ( n --> ) Line 766 ⟶ 1,808: say " ... " -1 peek echo cr cr ] </syntaxhighlight> ~~</lang>~~ {{out}} Line 812 ⟶ 1,854: =={{header\|Raku}}== <syntaxhighlight lang="raku" ~~perl6~~line>sub curzon ($base) { lazy (1..∞).hyper.map: { $_ if (exp($_, $base) + 1) %% ($base × $_ + 1) } }; for <2 4 6 8 10> { my $curzon = .&curzon; say "\nFirst 50 Curzon numbers using a base of $_:\n" ~ $curzon[^50].batch(2510)».fmt("%4s").join("\n") ~ "\nOne thousandth: " ~ $curzon[999] }</~~lang~~syntaxhighlight> {{out}} <pre>First 50 Curzon numbers using a base of 2: 1 2 5 6 9 14 18 21 26 29 ~~30 33 41 50 53 54 65 69 74 78 81 86 89 90 98~~ ~~105~~ ~~113~~30 ~~114~~ 33 ~~125~~ ~~134~~41 ~~138~~ 50 ~~141~~ ~~146~~53 ~~153~~ 54 ~~158~~ ~~165~~65 ~~173~~ 69 ~~174~~ ~~186~~74 ~~189~~ ~~194 198 209 210 221 230 233 245 249 254~~78 81 86 89 90 98 105 113 114 125 134 138 141 146 153 158 165 173 174 186 189 194 198 209 210 221 230 233 245 249 254 One thousandth: 8646 First 50 Curzon numbers using a base of 4: 1 3 7 9 13 15 25 27 37 39 ~~43 45 49 57 67 69 73 79 87 93 97 99 105 115 127~~ ~~135~~ ~~139~~43 ~~153~~ 45 ~~163~~ ~~165~~49 ~~169~~ 57 ~~175~~ ~~177~~67 ~~183~~ 69 ~~189~~ ~~193~~73 ~~199~~ 79 ~~205~~ ~~207~~87 ~~213~~ ~~219 235 249 253 255 265 267 273 277 279~~93 97 99 105 115 127 135 139 153 163 165 169 175 177 183 189 193 199 205 207 213 219 235 249 253 255 265 267 273 277 279 One thousandth: 9375 First 50 Curzon numbers using a base of 6: 1 6 30 58 70 73 90 101 105 121 ~~125 146 153 166 170 181 182 185 210 233 241 242 266 282 290~~ 125 146 153 166 170 181 182 185 210 233 ~~322 373 381 385 390 397 441 445 446 450 453 530 557 562 585 593 601 602 605 606 621 646 653 670 685~~ 241 242 266 282 290 322 373 381 385 390 397 441 445 446 450 453 530 557 562 585 593 601 602 605 606 621 646 653 670 685 One thousandth: 20717 First 50 Curzon numbers using a base of 8: 1 14 35 44 72 74 77 129 131 137 ~~144 149 150 185 200 219 236 266 284 285 299 309 336 357 381~~ 144 149 150 185 200 219 236 266 284 285 ~~386 390 392 402 414 420 441 455 459 470 479 500 519 527 536 557 582 600 602 617 639 654 674 696 735~~ 299 309 336 357 381 386 390 392 402 414 420 441 455 459 470 479 500 519 527 536 557 582 600 602 617 639 654 674 696 735 One thousandth: 22176 First 50 Curzon numbers using a base of 10: 1 9 10 25 106 145 190 193 238 253 ~~306 318 349 385 402 462 486 526 610 649 658 678 733 762 810~~ 306 318 349 385 402 462 486 526 610 649 ~~990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837~~ 658 678 733 762 810 990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 One thousandth: 46845</pre> =={{header\|RPL}}== {{works with\|Halcyon Calc\|4.2.7}} {\| class="wikitable" ! Code ! Comments \|- \| ≪ → x n m ≪ '''IF''' OVER '''THEN''' 1 1 n '''START''' x * m MOD '''NEXT''' '''ELSE''' 1 '''END''' ≫ ≫ 'XnMOD' STO ≪ → k ≪ {} 1 '''WHILE''' OVER SIZE 50 < '''REPEAT''' k OVER * 1 + DUP2 k ROT ROT XnMOD 1 + SWAP MOD '''IF''' NOT '''THEN''' SWAP OVER + SWAP '''END''' 1 + '''END''' DROP ≫ ≫ 'QRZ50' STO \| ''( x n m -- (x^n mod m) )'' If n<>0... then x^n mod m = (((x mod m) * x mod m) ... * x mod m) else x^0 mod m = 1 ''( k -- { 50 Curzon numbers } )'' Initialize Get kn + 1 Get k^n+1 mod kn+1 If zero then add it to sequence Increment n \|} The following line of code delivers what is required: 2 QRZ50 10 QRZ50 {{out}} <pre> 2: { 1 2 5 6 9 14 18 21 26 29 30 33 41 50 53 54 65 69 74 78 81 86 89 90 98 105 113 114 125 134 138 141 146 153 158 165 173 174 186 189 194 198 209 210 221 230 233 245 249 254 } 1: { 1 9 10 25 106 145 190 193 238 253 306 318 349 385 402 462 486 526 610 649 658 678 733 762 810 990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 } </pre> =={{header\|Ruby}}== <syntaxhighlight lang="ruby">def curzons(k) Enumerator.new do \|y\| (1..).each do \|n\| r = k * n y << n if k.pow(n, r + 1) == r end end end [2,4,6,8,10].each do \|base\| puts "Curzon numbers with k = #{base}:" puts curzons(base).take(50).join(", ") puts "Thousandth Curzon with k = #{base}: #{curzons(base).find.each.with_index(1){\|_,i\| i == 1000} }","" end</syntaxhighlight> {{out}}<pre>Curzon numbers with k = 2: 1, 2, 5, 6, 9, 14, 18, 21, 26, 29, 30, 33, 41, 50, 53, 54, 65, 69, 74, 78, 81, 86, 89, 90, 98, 105, 113, 114, 125, 134, 138, 141, 146, 153, 158, 165, 173, 174, 186, 189, 194, 198, 209, 210, 221, 230, 233, 245, 249, 254 Thousandth Curzon with k = 2: 8646. Curzon numbers with k = 4: 1, 3, 7, 9, 13, 15, 25, 27, 37, 39, 43, 45, 49, 57, 67, 69, 73, 79, 87, 93, 97, 99, 105, 115, 127, 135, 139, 153, 163, 165, 169, 175, 177, 183, 189, 193, 199, 205, 207, 213, 219, 235, 249, 253, 255, 265, 267, 273, 277, 279 Thousandth Curzon with k = 4: 9375. Curzon numbers with k = 6: 1, 6, 30, 58, 70, 73, 90, 101, 105, 121, 125, 146, 153, 166, 170, 181, 182, 185, 210, 233, 241, 242, 266, 282, 290, 322, 373, 381, 385, 390, 397, 441, 445, 446, 450, 453, 530, 557, 562, 585, 593, 601, 602, 605, 606, 621, 646, 653, 670, 685 Thousandth Curzon with k = 6: 20717. Curzon numbers with k = 8: 1, 14, 35, 44, 72, 74, 77, 129, 131, 137, 144, 149, 150, 185, 200, 219, 236, 266, 284, 285, 299, 309, 336, 357, 381, 386, 390, 392, 402, 414, 420, 441, 455, 459, 470, 479, 500, 519, 527, 536, 557, 582, 600, 602, 617, 639, 654, 674, 696, 735 Thousandth Curzon with k = 8: 22176. Curzon numbers with k = 10: 1, 9, 10, 25, 106, 145, 190, 193, 238, 253, 306, 318, 349, 385, 402, 462, 486, 526, 610, 649, 658, 678, 733, 762, 810, 990, 994, 1033, 1077, 1125, 1126, 1141, 1149, 1230, 1405, 1422, 1441, 1485, 1509, 1510, 1513, 1606, 1614, 1630, 1665, 1681, 1690, 1702, 1785, 1837 Thousandth Curzon with k = 10: 46845. </pre> =={{header\|Rust}}== <syntaxhighlight lang="rust">fn modpow(mut base: usize, mut exp: usize, n: usize) -> usize { ~~<lang rust>// [dependencies]~~ // ~~rug~~ if n == "1~~.15.0"~~ { return 0; } let mut result = 1; base %= n; while exp > 0 { if (exp & 1) == 1 { result = (result * base) % n; } base = (base * base) % n; exp >>= 1; } result } fn is_curzon(n: usize, k: usize) -> bool { let m = k * n + 1; modpow(k, n, m) + 1 == m ~~fn is_curzon(n: u32, k: u32) -> bool {~~ ~~use rug::{Complete, Integer};~~ ~~(Integer::u_pow_u(k, n).complete() + 1) % (k * n + 1) == 0~~ } Line 878 ⟶ 2,031: println!("1000th Curzon number with base {k}: {n}\n"); } }</~~lang~~syntaxhighlight> {{out}} Line 925 ⟶ 2,078: =={{header\|Sidef}}== <~~lang~~syntaxhighlight lang="ruby">func is_curzon(n, k) { powmod(k, n, kn + 1).is_congruent(-1, kn + 1) && (n > 0) } Line 933 ⟶ 2,086: say 50.by {\|n\| is_curzon(n, k) }.join(' ') say ("1000th term: ", 1000.th {\|n\| is_curzon(n,k) }) }</~~lang~~syntaxhighlight> {{out}} Line 956 ⟶ 2,109: 1 9 10 25 106 145 190 193 238 253 306 318 349 385 402 462 486 526 610 649 658 678 733 762 810 990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 1000th term: 46845 </pre> =={{header\|V (Vlang)}}== {{trans\|go}} <syntaxhighlight lang="v (vlang)">import math.big fn main() { zero := big.zero_int one := big.one_int for k := i64(2); k <= 10; k += 2 { bk := big.integer_from_i64(k) println("The first 50 Curzon numbers using a base of $k:") mut count := 0 mut n := i64(1) mut pow := big.integer_from_i64(k) mut curzon50 := []i64{} for { z := pow + one d := kn + 1 bd := big.integer_from_i64(d) if z%bd == zero { if count < 50 { curzon50 << n } count++ if count == 50 { for i in 0..curzon50.len { print("${curzon50[i]:4} ") if (i+1)%10 == 0 { println('') } } print("\nOne thousandth: ") } if count == 1000 { println(n) break } } n++ pow = bk } println('') } }</syntaxhighlight> {{out}} <pre> The first 50 Curzon numbers using a base of 2 : 1 2 5 6 9 14 18 21 26 29 30 33 41 50 53 54 65 69 74 78 81 86 89 90 98 105 113 114 125 134 138 141 146 153 158 165 173 174 186 189 194 198 209 210 221 230 233 245 249 254 One thousandth: 8646 The first 50 Curzon numbers using a base of 4 : 1 3 7 9 13 15 25 27 37 39 43 45 49 57 67 69 73 79 87 93 97 99 105 115 127 135 139 153 163 165 169 175 177 183 189 193 199 205 207 213 219 235 249 253 255 265 267 273 277 279 One thousandth: 9375 The first 50 Curzon numbers using a base of 6 : 1 6 30 58 70 73 90 101 105 121 125 146 153 166 170 181 182 185 210 233 241 242 266 282 290 322 373 381 385 390 397 441 445 446 450 453 530 557 562 585 593 601 602 605 606 621 646 653 670 685 One thousandth: 20717 The first 50 Curzon numbers using a base of 8 : 1 14 35 44 72 74 77 129 131 137 144 149 150 185 200 219 236 266 284 285 299 309 336 357 381 386 390 392 402 414 420 441 455 459 470 479 500 519 527 536 557 582 600 602 617 639 654 674 696 735 One thousandth: 22176 The first 50 Curzon numbers using a base of 10 : 1 9 10 25 106 145 190 193 238 253 306 318 349 385 402 462 486 526 610 649 658 678 733 762 810 990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 One thousandth: 46845 </pre> =={{header\|Wren}}== ===Version 1 (Embedded using GMP)=== {{libheader\|Wren-gmp}} ~~{{libheader\|Wren-seq}}~~ {{libheader\|Wren-fmt}} <~~lang~~syntaxhighlight ~~ecmascript~~lang="wren">/* ~~curzon_numbers~~Curzon_numbers.wren / import "./gmp" for Mpz ~~import "./seq" for Lst~~ import "./fmt" for Fmt Line 981 ⟶ 2,225: count = count + 1 if (count == 50) { ~~for (chunk in Lst.chunks(curzon50, 10))~~ Fmt.~~print~~tprint("$4d", ~~chunk~~curzon50, 10) System.write("\nOne thousandth: ") } if (count == 1000) { ~~System~~Fmt.print("$,d", n) break } Line 993 ⟶ 2,237: } System.print() }</~~lang~~syntaxhighlight> {{out}} <pre> The first 50 Curzon numbers using a base of 2: 1 2 5 6 9 14 18 21 26 29 30 33 41 50 53 54 65 69 74 78 81 86 89 90 98 105 113 114 125 134 138 141 146 153 158 165 173 174 186 189 194 198 209 210 221 230 233 245 249 254 One thousandth: ~~8646~~8,646 The first 50 Curzon numbers using a base of 4: 1 3 7 9 13 15 25 27 37 39 43 45 49 57 67 69 73 79 87 93 97 99 105 115 127 135 139 153 163 165 169 175 177 183 189 193 199 205 207 213 219 235 249 253 255 265 267 273 277 279 One thousandth: ~~9375~~9,375 The first 50 Curzon numbers using a base of 6: 1 6 30 58 70 73 90 101 105 121 125 146 153 166 170 181 182 185 210 233 241 242 266 282 290 322 373 381 385 390 397 441 445 446 450 453 530 557 562 585 593 601 602 605 606 621 646 653 670 685 One thousandth: ~~20717~~20,717 The first 50 Curzon numbers using a base of 8: 1 14 35 44 72 74 77 129 131 137 144 149 150 185 200 219 236 266 284 285 299 309 336 357 381 386 390 392 402 414 420 441 455 459 470 479 500 519 527 536 557 582 600 602 617 639 654 674 696 735 One thousandth: ~~22176~~22,176 The first 50 Curzon numbers using a base of 10: 1 9 10 25 106 145 190 193 238 253 306 318 349 385 402 462 486 526 610 649 658 678 733 762 810 990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 One thousandth: ~~46845~~46,845 </pre> ===Version 2 (Wren-cli)=== {{trans\|C++}} {{libheader\|Wren-math}} Alternatively, using ''Int.modPow'' rather than ''GMP'' so it will run on Wren-cli, albeit about 6 times more slowly. <syntaxhighlight lang="wren">import "./math" for Int import "./fmt" for Fmt var isCurzon = Fn.new { \|n, k\| var r = k n return Int.modPow(k, n, r+1) == r } var k = 2 while (k <= 10) { System.print("Curzon numbers with base %(k):") var n = 1 var count = 0 while (count < 50) { if (isCurzon.call(n, k)) { Fmt.write("$4d ", n) count = count + 1 if (count % 10 == 0) System.print() } n = n + 1 } while (true) { if (isCurzon.call(n, k)) count = count + 1 if (count == 1000) break n = n + 1 } Fmt.print("1,000th Curzon number with base $d: $,d\n", k, n) k = k + 2 }</syntaxhighlight> {{out}} <pre> As Version 1. </pre> =={{header\|XPL0}}== {{trans\|C}} <syntaxhighlight lang "XPL0">func ModPow(Base, Exp, Mod); int Base, Exp, Mod, Result; [if Mod = 1 then return 0; Result:= 1; Base:= rem(Base/Mod); while Exp > 0 do [if (Exp&1) = 1 then Result:= rem((ResultBase)/Mod); Base:= rem((BaseBase) / Mod); Exp:= Exp >> 1; ]; return Result; ]; func IsCurzon(N, K); int N, K, R; [R:= K * N; return ModPow(K, N, R+1) = R; ]; int K, N, Count; [K:= 2; Format(5, 0); while K <= 10 do [Text(0, "Curzon numbers with base "); IntOut(0, K); CrLf(0); N:= 1; Count:= 0; while Count < 50 do [if IsCurzon(N, K) then [RlOut(0, float(N)); Count:= Count+1; if rem(Count/10) = 0 then CrLf(0); ]; N:= N+1; ]; K:= K+2; ]; ]</syntaxhighlight> {{out}} <pre> Curzon numbers with base 2 1 2 5 6 9 14 18 21 26 29 30 33 41 50 53 54 65 69 74 78 81 86 89 90 98 105 113 114 125 134 138 141 146 153 158 165 173 174 186 189 194 198 209 210 221 230 233 245 249 254 Curzon numbers with base 4 1 3 7 9 13 15 25 27 37 39 43 45 49 57 67 69 73 79 87 93 97 99 105 115 127 135 139 153 163 165 169 175 177 183 189 193 199 205 207 213 219 235 249 253 255 265 267 273 277 279 Curzon numbers with base 6 1 6 30 58 70 73 90 101 105 121 125 146 153 166 170 181 182 185 210 233 241 242 266 282 290 322 373 381 385 390 397 441 445 446 450 453 530 557 562 585 593 601 602 605 606 621 646 653 670 685 Curzon numbers with base 8 1 14 35 44 72 74 77 129 131 137 144 149 150 185 200 219 236 266 284 285 299 309 336 357 381 386 390 392 402 414 420 441 455 459 470 479 500 519 527 536 557 582 600 602 617 639 654 674 696 735 Curzon numbers with base 10 1 9 10 25 106 145 190 193 238 253 306 318 349 385 402 462 486 526 610 649 658 678 733 762 810 990 994 1033 1077 1125 1126 1141 1149 1230 1405 1422 1441 1485 1509 1510 1513 1606 1614 1630 1665 1681 1690 1702 1785 1837 </pre>