Curzon numbers: Difference between revisions
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→{{header|EasyLang}}
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</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}}==
Line 1,256 ⟶ 1,296:
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 1,433 ⟶ 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,b*n+1)== b*n);
/* 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}}==
Line 2,015 ⟶ 2,207:
{{libheader|Wren-gmp}}
{{libheader|Wren-fmt}}
<syntaxhighlight lang="
import "./gmp" for Mpz
Line 2,098 ⟶ 2,290:
{{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="
import "./fmt" for Fmt
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