Brilliant numbers: Difference between revisions
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main :: IO () |
main :: IO () |
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main = do |
main = do |
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let indexedBrilliants = |
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| ⚫ | |||
zip |
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[1 ..] |
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| ⚫ | |||
putStrLn $ |
putStrLn $ |
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table " " $ |
table " " $ |
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chunksOf 10 $ |
chunksOf 10 $ |
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show . snd |
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<$> take 100 indexedBrilliants |
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putStrLn "(index, brilliant)" |
putStrLn "(index, brilliant)" |
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Revision as of 13:30, 18 June 2022
You are encouraged to solve this task according to the task description, using any language you may know.
Brilliant numbers are a subset of semiprime numbers. Specifically, they are numbers that are the product of exactly two prime numbers that both have the same number of digits when expressed in base 10.
Brilliant numbers are useful in cryptography and when testing prime factoring algorithms.
- E.G.
- 3 × 3 (9) is a brilliant number.
- 2 × 7 (14) is a brilliant number.
- 113 × 691 (78083) is a brilliant number.
- 2 × 31 (62) is semiprime, but is not a brilliant number (different number of digits in the two factors).
- Task
- Find and display the first 100 brilliant numbers.
- For the orders of magnitude 1 through 6, find and show the first brilliant number greater than or equal to the order of magnitude, and, its position in the series (or the count of brilliant numbers up to that point).
- Stretch
- Continue for larger orders of magnitude.
- See also
ALGOL 68
<lang algol68>BEGIN # Find Brilliant numbers - semi-primes whose two prime factors have #
# the same number of digits #
PR read "primes.incl.a68" PR # include prime utilities #
INT max prime = 1 010; # maximum prime we will consider #
# should be enough to find the first brilliant number > 10^6 #
[]BOOL prime = PRIMESIEVE max prime; # sieve of primes to max prime #
# construct a table of brilliant numbers #
[ 1 : max prime * max prime ]BOOL brilliant;
FOR n FROM LWB brilliant TO UPB brilliant DO brilliant[ n ] := FALSE OD;
# brilliant numbers where one of the fators is 2 #
brilliant[ 4 ] := TRUE;
FOR p FROM 3 BY 2 TO 7 DO brilliant[ 2 * p ] := TRUE OD;
# brilliant numbers where both factors are odd #
INT p start := 1, p end := 9;
WHILE pstart < max prime DO
FOR p FROM p start BY 2 TO p end DO
IF prime[ p ] THEN
brilliant[ p * p ] := TRUE;
FOR q FROM p + 2 BY 2 TO p end DO
IF prime[ q ] THEN
brilliant[ p * q ] := TRUE
FI
OD
FI
OD;
p start := p end + 2;
p end := ( ( p start - 1 ) * 10 ) - 1;
IF p end > max prime THEN p end := max prime FI
OD;
# show the first 100 brilliant numbers #
INT b count := 0;
FOR n TO UPB brilliant WHILE b count < 100 DO
IF brilliant[ n ] THEN
print( ( whole( n, -6 ) ) );
IF ( b count +:= 1 ) MOD 10 = 0 THEN print( ( newline ) ) FI
FI
OD;
# first brilliant number >= 10^n, n = 1, 2, ..., 6 #
b count := 0;
INT power of 10 := 10;
FOR n TO UPB brilliant DO
IF brilliant[ n ] THEN
b count +:= 1;
IF n >= power of 10 THEN
print( ( "First brilliant number >= ", whole( power of 10, -8 )
, ": " , whole( n, -8 )
, " at position " , whole( b count, -6 )
, newline
)
);
power of 10 *:= 10
FI
FI
OD
END</lang>
- Output:
4 6 9 10 14 15 21 25 35 49 121 143 169 187 209 221 247 253 289 299 319 323 341 361 377 391 403 407 437 451 473 481 493 517 527 529 533 551 559 583 589 611 629 649 667 671 689 697 703 713 731 737 767 779 781 793 799 803 817 841 851 869 871 893 899 901 913 923 943 949 961 979 989 1003 1007 1027 1037 1067 1073 1079 1081 1121 1139 1147 1157 1159 1189 1207 1219 1241 1247 1261 1271 1273 1333 1343 1349 1357 1363 1369 First brilliant number >= 10: 10 at position 4 First brilliant number >= 100: 121 at position 11 First brilliant number >= 1000: 1003 at position 74 First brilliant number >= 10000: 10201 at position 242 First brilliant number >= 100000: 100013 at position 2505 First brilliant number >= 1000000: 1018081 at position 10538
Arturo
<lang rebol>brilliant?: function [x][
pf: factors.prime x
and? -> 2 = size pf
-> equal? size digits first pf
size digits last pf
]
brilliants: new [] i: 2 while [100 > size brilliants][
if brilliant? i -> 'brilliants ++ i i: i + 1
]
print "First 100 brilliant numbers:" loop split.every: 10 brilliants 'row [
print map to [:string] row 'item -> pad item 4
] print ""
i: 4 nth: 0 order: 1 while [order =< 6] [
if brilliant? i [
nth: nth + 1
if i >= 10^order [
print ["First brilliant number >= 10 ^" order "is" i "at position" nth]
order: order + 1
]
]
i: i + 1
]</lang>
- Output:
First 100 brilliant numbers: 4 6 9 10 14 15 21 25 35 49 121 143 169 187 209 221 247 253 289 299 319 323 341 361 377 391 403 407 437 451 473 481 493 517 527 529 533 551 559 583 589 611 629 649 667 671 689 697 703 713 731 737 767 779 781 793 799 803 817 841 851 869 871 893 899 901 913 923 943 949 961 979 989 1003 1007 1027 1037 1067 1073 1079 1081 1121 1139 1147 1157 1159 1189 1207 1219 1241 1247 1261 1271 1273 1333 1343 1349 1357 1363 1369 First brilliant number >= 10 ^ 1 is 10 at position 4 First brilliant number >= 10 ^ 2 is 121 at position 11 First brilliant number >= 10 ^ 3 is 1003 at position 74 First brilliant number >= 10 ^ 4 is 10201 at position 242 First brilliant number >= 10 ^ 5 is 100013 at position 2505 First brilliant number >= 10 ^ 6 is 1018081 at position 10538
C++
<lang cpp>#include <algorithm>
- include <chrono>
- include <iomanip>
- include <iostream>
- include <locale>
- include <vector>
- include <primesieve.hpp>
auto get_primes_by_digits(uint64_t limit) {
primesieve::iterator pi;
std::vector<std::vector<uint64_t>> primes_by_digits;
std::vector<uint64_t> primes;
for (uint64_t p = 10; p <= limit;) {
uint64_t prime = pi.next_prime();
if (prime > p) {
primes_by_digits.push_back(std::move(primes));
p *= 10;
}
primes.push_back(prime);
}
return primes_by_digits;
}
int main() {
std::cout.imbue(std::locale(""));
auto start = std::chrono::high_resolution_clock::now();
auto primes_by_digits = get_primes_by_digits(1000000000);
std::cout << "First 100 brilliant numbers:\n";
std::vector<uint64_t> brilliant_numbers;
for (const auto& primes : primes_by_digits) {
for (auto i = primes.begin(); i != primes.end(); ++i)
for (auto j = i; j != primes.end(); ++j)
brilliant_numbers.push_back(*i * *j);
if (brilliant_numbers.size() >= 100)
break;
}
std::sort(brilliant_numbers.begin(), brilliant_numbers.end());
for (size_t i = 0; i < 100; ++i) {
std::cout << std::setw(5) << brilliant_numbers[i]
<< ((i + 1) % 10 == 0 ? '\n' : ' ');
}
std::cout << '\n';
uint64_t power = 10;
size_t count = 0;
for (size_t p = 1; p < 2 * primes_by_digits.size(); ++p) {
const auto& primes = primes_by_digits[p / 2];
size_t position = count + 1;
uint64_t min_product = 0;
for (auto i = primes.begin(); i != primes.end(); ++i) {
uint64_t p1 = *i;
auto j = std::lower_bound(i, primes.end(), (power + p1 - 1) / p1);
if (j != primes.end()) {
uint64_t p2 = *j;
uint64_t product = p1 * p2;
if (min_product == 0 || product < min_product)
min_product = product;
position += std::distance(i, j);
if (p1 >= p2)
break;
}
}
std::cout << "First brilliant number >= 10^" << p << " is "
<< min_product << " at position " << position << '\n';
power *= 10;
if (p % 2 == 1) {
size_t size = primes.size();
count += size * (size + 1) / 2;
}
}
auto end = std::chrono::high_resolution_clock::now(); std::chrono::duration<double> duration(end - start); std::cout << "\nElapsed time: " << duration.count() << " seconds\n";
}</lang>
- Output:
First 100 brilliant numbers:
4 6 9 10 14 15 21 25 35 49
121 143 169 187 209 221 247 253 289 299
319 323 341 361 377 391 403 407 437 451
473 481 493 517 527 529 533 551 559 583
589 611 629 649 667 671 689 697 703 713
731 737 767 779 781 793 799 803 817 841
851 869 871 893 899 901 913 923 943 949
961 979 989 1,003 1,007 1,027 1,037 1,067 1,073 1,079
1,081 1,121 1,139 1,147 1,157 1,159 1,189 1,207 1,219 1,241
1,247 1,261 1,271 1,273 1,333 1,343 1,349 1,357 1,363 1,369
First brilliant number >= 10^1 is 10 at position 4
First brilliant number >= 10^2 is 121 at position 11
First brilliant number >= 10^3 is 1,003 at position 74
First brilliant number >= 10^4 is 10,201 at position 242
First brilliant number >= 10^5 is 100,013 at position 2,505
First brilliant number >= 10^6 is 1,018,081 at position 10,538
First brilliant number >= 10^7 is 10,000,043 at position 124,364
First brilliant number >= 10^8 is 100,140,049 at position 573,929
First brilliant number >= 10^9 is 1,000,000,081 at position 7,407,841
First brilliant number >= 10^10 is 10,000,600,009 at position 35,547,995
First brilliant number >= 10^11 is 100,000,000,147 at position 491,316,167
First brilliant number >= 10^12 is 1,000,006,000,009 at position 2,409,600,866
First brilliant number >= 10^13 is 10,000,000,000,073 at position 34,896,253,010
First brilliant number >= 10^14 is 100,000,380,000,361 at position 174,155,363,187
First brilliant number >= 10^15 is 1,000,000,000,000,003 at position 2,601,913,448,897
First brilliant number >= 10^16 is 10,000,001,400,000,049 at position 13,163,230,391,313
First brilliant number >= 10^17 is 100,000,000,000,000,831 at position 201,431,415,980,419
Elapsed time: 1.50048 seconds
Factor
<lang factor>USING: assocs formatting grouping io kernel lists lists.lazy math math.functions math.primes.factors prettyprint project-euler.common sequences ;
MEMO: brilliant? ( n -- ? )
factors [ length 2 = ] keep [ number-length ] map all-eq? and ;
- lbrilliant ( -- list )
2 lfrom [ brilliant? ] lfilter 1 lfrom lzip ;
- first> ( m -- n )
lbrilliant swap '[ first _ >= ] lfilter car ;
- .first> ( n -- )
dup first> first2 "First brilliant number >= %7d: %7d at position %5d\n" printf ;
100 lbrilliant ltake list>array keys 10 group simple-table. nl { 1 2 3 4 5 6 } [ 10^ .first> ] each</lang>
- Output:
4 6 9 10 14 15 21 25 35 49 121 143 169 187 209 221 247 253 289 299 319 323 341 361 377 391 403 407 437 451 473 481 493 517 527 529 533 551 559 583 589 611 629 649 667 671 689 697 703 713 731 737 767 779 781 793 799 803 817 841 851 869 871 893 899 901 913 923 943 949 961 979 989 1003 1007 1027 1037 1067 1073 1079 1081 1121 1139 1147 1157 1159 1189 1207 1219 1241 1247 1261 1271 1273 1333 1343 1349 1357 1363 1369 First brilliant number >= 10: 10 at position 4 First brilliant number >= 100: 121 at position 11 First brilliant number >= 1000: 1003 at position 74 First brilliant number >= 10000: 10201 at position 242 First brilliant number >= 100000: 100013 at position 2505 First brilliant number >= 1000000: 1018081 at position 10538
Go
<lang go>package main
import (
"fmt" "math" "rcu" "sort"
)
var primes = rcu.Primes(1e8 - 1)
type res struct {
bc interface{}
next int
}
func getBrilliant(digits, limit int, countOnly bool) res {
var brilliant []int
count := 0
pow := 1
next := math.MaxInt
for k := 1; k <= digits; k++ {
var s []int
for _, p := range primes {
if p >= pow*10 {
break
}
if p > pow {
s = append(s, p)
}
}
for i := 0; i < len(s); i++ {
for j := i; j < len(s); j++ {
prod := s[i] * s[j]
if prod < limit {
if countOnly {
count++
} else {
brilliant = append(brilliant, prod)
}
} else {
if next > prod {
next = prod
}
break
}
}
}
pow *= 10
}
if countOnly {
return res{count, next}
}
return res{brilliant, next}
}
func main() {
fmt.Println("First 100 brilliant numbers:")
brilliant := getBrilliant(2, 10000, false).bc.([]int)
sort.Ints(brilliant)
brilliant = brilliant[0:100]
for i := 0; i < len(brilliant); i++ {
fmt.Printf("%4d ", brilliant[i])
if (i+1)%10 == 0 {
fmt.Println()
}
}
fmt.Println()
for k := 1; k <= 13; k++ {
limit := int(math.Pow(10, float64(k)))
r := getBrilliant(k, limit, true)
total := r.bc.(int)
next := r.next
climit := rcu.Commatize(limit)
ctotal := rcu.Commatize(total + 1)
cnext := rcu.Commatize(next)
fmt.Printf("First >= %18s is %14s in the series: %18s\n", climit, ctotal, cnext)
}
}</lang>
- Output:
First 100 brilliant numbers: 4 6 9 10 14 15 21 25 35 49 121 143 169 187 209 221 247 253 289 299 319 323 341 361 377 391 403 407 437 451 473 481 493 517 527 529 533 551 559 583 589 611 629 649 667 671 689 697 703 713 731 737 767 779 781 793 799 803 817 841 851 869 871 893 899 901 913 923 943 949 961 979 989 1003 1007 1027 1037 1067 1073 1079 1081 1121 1139 1147 1157 1159 1189 1207 1219 1241 1247 1261 1271 1273 1333 1343 1349 1357 1363 1369 First >= 10 is 4 in the series: 10 First >= 100 is 11 in the series: 121 First >= 1,000 is 74 in the series: 1,003 First >= 10,000 is 242 in the series: 10,201 First >= 100,000 is 2,505 in the series: 100,013 First >= 1,000,000 is 10,538 in the series: 1,018,081 First >= 10,000,000 is 124,364 in the series: 10,000,043 First >= 100,000,000 is 573,929 in the series: 100,140,049 First >= 1,000,000,000 is 7,407,841 in the series: 1,000,000,081 First >= 10,000,000,000 is 35,547,995 in the series: 10,000,600,009 First >= 100,000,000,000 is 491,316,167 in the series: 100,000,000,147 First >= 1,000,000,000,000 is 2,409,600,866 in the series: 1,000,006,000,009 First >= 10,000,000,000,000 is 34,896,253,010 in the series: 10,000,000,000,073
Haskell
<lang haskell>import Control.Monad (join) import Data.Bifunctor (bimap) import Data.List (intercalate, transpose) import Data.List.Split (chunksOf, splitWhen) import Data.Numbers.Primes (primeFactors) import Text.Printf (printf)
BRILLIANT NUMBERS -------------------
isBrilliant :: (Integral a, Show a) => a -> Bool isBrilliant n = case primeFactors n of
[a, b] -> length (show a) == length (show b) _ -> False
TEST -------------------------
main :: IO () main = do
let indexedBrilliants =
zip
[1 ..]
(filter isBrilliant [1 ..])
putStrLn $
table " " $
chunksOf 10 $
show . snd
<$> take 100 indexedBrilliants
putStrLn "(index, brilliant)"
mapM_ print $
take 6 $
fmap (fst . head) $
splitWhen
(uncurry (<) . join bimap (length . show . snd))
$ zip indexedBrilliants (tail indexedBrilliants)
DISPLAY ------------------------
table :: String -> String -> String table gap rows =
let ws = maximum . fmap length <$> transpose rows
pw = printf . flip intercalate ["%", "s"] . show
in unlines $ intercalate gap . zipWith pw ws <$> rows</lang>
- Output:
4 6 9 10 14 15 21 25 35 49 121 143 169 187 209 221 247 253 289 299 319 323 341 361 377 391 403 407 437 451 473 481 493 517 527 529 533 551 559 583 589 611 629 649 667 671 689 697 703 713 731 737 767 779 781 793 799 803 817 841 851 869 871 893 899 901 913 923 943 949 961 979 989 1003 1007 1027 1037 1067 1073 1079 1081 1121 1139 1147 1157 1159 1189 1207 1219 1241 1247 1261 1271 1273 1333 1343 1349 1357 1363 1369 (index, brilliant) (1,4) (4,10) (11,121) (74,1003) (242,10201) (2505,100013)
J
<lang J>oprimes=: {{ NB. all primes of order y
p:(+i.)/-/\ p:inv +/\1 9*10^y
}}
obrill=: {{ NB. all brilliant numbers of order y primes
~.,*/~oprimes y
}}
brillseq=: {{ NB. sequences of brilliant numbers up through order y-1 primes
/:~;obrill each i.y
}}</lang>
Task examples:
<lang J> 10 10 $brillseq 2
4 6 9 10 14 15 21 25 35 49 121 143 169 187 209 221 247 253 289 299 319 323 341 361 377 391 403 407 437 451 473 481 493 517 527 529 533 551 559 583 589 611 629 649 667 671 689 697 703 713 731 737 767 779 781 793 799 803 817 841 851 869 871 893 899 901 913 923 943 949 961 979 989 1003 1007 1027 1037 1067 1073 1079
1081 1121 1139 1147 1157 1159 1189 1207 1219 1241 1247 1261 1271 1273 1333 1343 1349 1357 1363 1369 NB. order, index, value
(brillseq 4) (],.(I. 10^]) ([,.{) [) 1 2 3 4 5 6
1 3 10 2 10 121 3 73 1003 4 241 10201 5 2504 100013 6 10537 1018081</lang>
Stretch goal (results are order, index, value): <lang J> (brillseq 4) (],.(I. 10^]) ([,.{) [) ,7 7 124363 10000043
(brillseq 5) (],.(I. 10^]) ([,.{) [) 8 9
8 573928 100140049 9 7407840 1000000081</lang>
Java
Uses the PrimeGenerator class from Extensible prime generator#Java. <lang java>import java.util.*;
public class BrilliantNumbers {
public static void main(String[] args) {
var primesByDigits = getPrimesByDigits(100000000);
System.out.println("First 100 brilliant numbers:");
List<Integer> brilliantNumbers = new ArrayList<>();
for (var primes : primesByDigits) {
int n = primes.size();
for (int i = 0; i < n; ++i) {
int prime1 = primes.get(i);
for (int j = i; j < n; ++j) {
int prime2 = primes.get(j);
brilliantNumbers.add(prime1 * prime2);
}
}
if (brilliantNumbers.size() >= 100)
break;
}
Collections.sort(brilliantNumbers);
for (int i = 0; i < 100; ++i) {
char c = (i + 1) % 10 == 0 ? '\n' : ' ';
System.out.printf("%,5d%c", brilliantNumbers.get(i), c);
}
System.out.println();
long power = 10;
long count = 0;
for (int p = 1; p < 2 * primesByDigits.size(); ++p) {
var primes = primesByDigits.get(p / 2);
long position = count + 1;
long minProduct = 0;
int n = primes.size();
for (int i = 0; i < n; ++i) {
long prime1 = primes.get(i);
var primes2 = primes.subList(i, n);
int q = (int)((power + prime1 - 1) / prime1);
int j = Collections.binarySearch(primes2, q);
if (j == n)
continue;
if (j < 0)
j = -(j + 1);
long prime2 = primes2.get(j);
long product = prime1 * prime2;
if (minProduct == 0 || product < minProduct)
minProduct = product;
position += j;
if (prime1 >= prime2)
break;
}
System.out.printf("First brilliant number >= 10^%d is %,d at position %,d\n",
p, minProduct, position);
power *= 10;
if (p % 2 == 1) {
long size = primes.size();
count += size * (size + 1) / 2;
}
}
}
private static List<List<Integer>> getPrimesByDigits(int limit) {
PrimeGenerator primeGen = new PrimeGenerator(100000, 100000);
List<List<Integer>> primesByDigits = new ArrayList<>();
List<Integer> primes = new ArrayList<>();
for (int p = 10; p <= limit; ) {
int prime = primeGen.nextPrime();
if (prime > p) {
primesByDigits.add(primes);
primes = new ArrayList<>();
p *= 10;
}
primes.add(prime);
}
return primesByDigits;
}
}</lang>
- Output:
First 100 brilliant numbers:
4 6 9 10 14 15 21 25 35 49
121 143 169 187 209 221 247 253 289 299
319 323 341 361 377 391 403 407 437 451
473 481 493 517 527 529 533 551 559 583
589 611 629 649 667 671 689 697 703 713
731 737 767 779 781 793 799 803 817 841
851 869 871 893 899 901 913 923 943 949
961 979 989 1,003 1,007 1,027 1,037 1,067 1,073 1,079
1,081 1,121 1,139 1,147 1,157 1,159 1,189 1,207 1,219 1,241
1,247 1,261 1,271 1,273 1,333 1,343 1,349 1,357 1,363 1,369
First brilliant number >= 10^1 is 10 at position 4
First brilliant number >= 10^2 is 121 at position 11
First brilliant number >= 10^3 is 1,003 at position 74
First brilliant number >= 10^4 is 10,201 at position 242
First brilliant number >= 10^5 is 100,013 at position 2,505
First brilliant number >= 10^6 is 1,018,081 at position 10,538
First brilliant number >= 10^7 is 10,000,043 at position 124,364
First brilliant number >= 10^8 is 100,140,049 at position 573,929
First brilliant number >= 10^9 is 1,000,000,081 at position 7,407,841
First brilliant number >= 10^10 is 10,000,600,009 at position 35,547,995
First brilliant number >= 10^11 is 100,000,000,147 at position 491,316,167
First brilliant number >= 10^12 is 1,000,006,000,009 at position 2,409,600,866
First brilliant number >= 10^13 is 10,000,000,000,073 at position 34,896,253,010
First brilliant number >= 10^14 is 100,000,380,000,361 at position 174,155,363,187
First brilliant number >= 10^15 is 1,000,000,000,000,003 at position 2,601,913,448,897
Julia
<lang julia> using Primes
function isbrilliant(n)
p = factor(n).pe
return (length(p) == 1 && p[1][2] == 2) ||
length(p) == 2 && ndigits(p[1][1]) == ndigits(p[2][1]) && p[1][2] == p[2][2] == 1
end
function testbrilliants()
println("First 100 brilliant numbers:")
foreach(p -> print(lpad(p[2], 5), p[1] % 20 == 0 ? "\n" : ""),
enumerate(filter(isbrilliant, 1:1370)))
bcount, results, positions = 0, zeros(Int, 9), zeros(Int, 9)
for n in 1:10^10
if isbrilliant(n)
bcount += 1
for i in 1:9
if n >= 10^i && results[i] == 0
results[i] = n
positions[i] = bcount
println("First >=", lpad(10^i, 12), " is", lpad(bcount, 8),
" in the series: $n")
end
end
end
end
return results, positions
end
testbrilliants()
</lang>
- Output:
First 100 brilliant numbers:
4 6 9 10 14 15 21 25 35 49 121 143 169 187 209 221 247 253 289 299
319 323 341 361 377 391 403 407 437 451 473 481 493 517 527 529 533 551 559 583
589 611 629 649 667 671 689 697 703 713 731 737 767 779 781 793 799 803 817 841
851 869 871 893 899 901 913 923 943 949 961 979 989 1003 1007 1027 1037 1067 1073 1079
1081 1121 1139 1147 1157 1159 1189 1207 1219 1241 1247 1261 1271 1273 1333 1343 1349 1357 1363 1369
First >= 10 is 4 in the series: 10
First >= 100 is 11 in the series: 121
First >= 1000 is 74 in the series: 1003
First >= 10000 is 242 in the series: 10201
First >= 100000 is 2505 in the series: 100013
First >= 1000000 is 10538 in the series: 1018081
First >= 10000000 is 124364 in the series: 10000043
First >= 100000000 is 573929 in the series: 100140049
First >= 1000000000 is 7407841 in the series: 1000000081
Mathematica/Wolfram Language
<lang Mathematica>ClearAll[PrimesDecade] PrimesDecade[n_Integer] := Module[{bounds},
bounds = {PrimePi[10^n] + 1, PrimePi[10^(n + 1) - 1]};
Prime[Range @@ bounds]
]
ds = Union @@ Table[Union[Times @@@ Tuples[PrimesDecade[d], 2]], {d, 0, 4}];
Multicolumn[Take[ds, 100], {Automatic, 8}, Appearance -> "Horizontal"]
sel = Min /@ GatherBy[Select[ds, GreaterEqualThan[10]], IntegerLength]; Grid[{#, FirstPosition[ds, #]1} & /@ sel]</lang>
- Output:
4 6 9 10 14 15 21 25 35 49 121 143 169 187 209 221 247 253 289 299 319 323 341 361 377 391 403 407 437 451 473 481 493 517 527 529 533 551 559 583 589 611 629 649 667 671 689 697 703 713 731 737 767 779 781 793 799 803 817 841 851 869 871 893 899 901 913 923 943 949 961 979 989 1003 1007 1027 1037 1067 1073 1079 1081 1121 1139 1147 1157 1159 1189 1207 1219 1241 1247 1261 1271 1273 1333 1343 1349 1357 1363 1369 10 4 121 11 1003 74 10201 242 100013 2505 1018081 10538 10000043 124364 100140049 573929 1000000081 7407841
Perl
<lang perl>use strict; use warnings; use feature 'say'; use List::AllUtils <max head firstidx uniqint>; use ntheory <primes is_semiprime forsetproduct>;
sub table { my $t = shift() * (my $c = 1 + length max @_); ( sprintf( ('%'.$c.'d')x@_, @_) ) =~ s/.{1,$t}\K/\n/gr } sub comma { reverse ((reverse shift) =~ s/(.{3})/$1,/gr) =~ s/^,//r }
my(@B,@Br); for my $oom (1..5) {
my @P = grep { $oom == length } @{primes(10**$oom)};
forsetproduct { is_semiprime($_[0] * $_[1]) and push @B, $_[0] * $_[1] } \@P, \@P;
@Br = uniqint sort { $a <=> $b } @Br, @B;
}
say "First 100 brilliant numbers:\n" . table 10, head 100, @Br;
for my $oom (1..9) {
my $key = firstidx { $_ > 10**$oom } @Br;
printf "First >= %13s is position %9s in the series: %13s\n", comma(10**$oom), comma($key), comma $Br[$key];
}</lang>
- Output:
First 100 brilliant numbers:
4 6 9 10 14 15 21 25 35 49
121 143 169 187 209 221 247 253 289 299
319 323 341 361 377 391 403 407 437 451
473 481 493 517 527 529 533 551 559 583
589 611 629 649 667 671 689 697 703 713
731 737 767 779 781 793 799 803 817 841
851 869 871 893 899 901 913 923 943 949
961 979 989 1003 1007 1027 1037 1067 1073 1079
1081 1121 1139 1147 1157 1159 1189 1207 1219 1241
1247 1261 1271 1273 1333 1343 1349 1357 1363 1369
First >= 10 is position 4 in the series: 14
First >= 100 is position 10 in the series: 121
First >= 1,000 is position 73 in the series: 1,003
First >= 10,000 is position 241 in the series: 10,201
First >= 100,000 is position 2,504 in the series: 100,013
First >= 1,000,000 is position 10,537 in the series: 1,018,081
First >= 10,000,000 is position 124364 in the series: 10,000,043
First >= 100,000,000 is position 573929 in the series: 100,140,049
First >= 1,000,000,000 is position 7407841 in the series: 1,000,000,081
Faster approach
<lang perl>use 5.020; use strict; use warnings;
use ntheory qw(:all); use experimental qw(signatures);
sub is_briliant_number ($n) {
is_semiprime($n) || return; my @f = factor($n); length($f[0]) == length($f[1]);
}
sub next_brilliant_number ($n) {
++$n while not is_briliant_number($n); $n;
}
sub brilliant_numbers_count ($n) {
use integer;
my $count = 0; my $len = length(sqrtint($n));
foreach my $k (1 .. $len - 1) {
my $pi = prime_count(10**($k - 1), 10**$k - 1);
$count += binomial($pi, 2) + $pi;
}
my $min = 10**($len - 1); my $max = 10**$len - 1;
my $pi_min = prime_count($min); my $pi_max = prime_count($max);
my $j = -1;
forprimes {
if ($_*$_ <= $n) {
$count += (($max <= $n/$_) ? $pi_max : prime_count($n/$_)) - $pi_min - ++$j;
}
else {
lastfor;
}
} $min, $max;
return $count;
}
say "First 100 brilliant numbers:";
my @nums;
for (my $k = 1 ; scalar(@nums) < 100 ; ++$k) {
push(@nums, $k) if is_briliant_number($k);
}
while (@nums) {
my @slice = splice(@nums, 0, 10);
say join ' ', map { sprintf("%4s", $_) } @slice;
}
say ;
foreach my $n (1 .. 13) {
my $v = next_brilliant_number(vecprod((10) x $n));
printf("First brilliant number >= 10^%d is %s", $n, $v);
printf(" at position %s\n", brilliant_numbers_count($v));
}</lang>
- Output:
First 100 brilliant numbers: 4 6 9 10 14 15 21 25 35 49 121 143 169 187 209 221 247 253 289 299 319 323 341 361 377 391 403 407 437 451 473 481 493 517 527 529 533 551 559 583 589 611 629 649 667 671 689 697 703 713 731 737 767 779 781 793 799 803 817 841 851 869 871 893 899 901 913 923 943 949 961 979 989 1003 1007 1027 1037 1067 1073 1079 1081 1121 1139 1147 1157 1159 1189 1207 1219 1241 1247 1261 1271 1273 1333 1343 1349 1357 1363 1369 First brilliant number >= 10^1 is 10 at position 4 First brilliant number >= 10^2 is 121 at position 11 First brilliant number >= 10^3 is 1003 at position 74 First brilliant number >= 10^4 is 10201 at position 242 First brilliant number >= 10^5 is 100013 at position 2505 First brilliant number >= 10^6 is 1018081 at position 10538 First brilliant number >= 10^7 is 10000043 at position 124364 First brilliant number >= 10^8 is 100140049 at position 573929 First brilliant number >= 10^9 is 1000000081 at position 7407841 First brilliant number >= 10^10 is 10000600009 at position 35547995 First brilliant number >= 10^11 is 100000000147 at position 491316167 First brilliant number >= 10^12 is 1000006000009 at position 2409600866 First brilliant number >= 10^13 is 10000000000073 at position 34896253010
Phix
Replaced with C++ translation; much faster and now goes comfortably to 1e15 even on 32 bit. You can run this online here.
-- -- demo\rosetta\BrilliantNumbers.exw -- ================================= -- with javascript_semantics requires("1.0.2") -- (for in) atom t0 = time() function get_primes_by_digits(integer limit) sequence primes = get_primes_le(power(10,limit)), primes_by_digits = {} integer p = 10 while length(primes) do integer pi = abs(binary_search(p,primes))-1 primes_by_digits &= {primes[1..pi]} primes = primes[pi+1..$] p*= 10 end while return primes_by_digits end function sequence primes_by_digits = get_primes_by_digits(8) procedure first100() sequence brilliant_numbers = {} for primes in primes_by_digits do for i=1 to length(primes) do --see talk page -- for j=i to length(primes) do for j=1 to i do brilliant_numbers &= primes[i]*primes[j] end for end for if length(brilliant_numbers)>=100 then exit end if end for brilliant_numbers = sort(brilliant_numbers)[1..100] sequence j100 = join_by(brilliant_numbers,1,10," ","\n","%,5d") printf(1,"First 100 brilliant numbers:\n%s\n\n",{j100}) end procedure first100() atom pwr = 10, count = 0 for p=1 to 2*length(primes_by_digits)-1 do sequence primes = primes_by_digits[floor(p/2)+1] atom pos = count+1, min_product = 0 for i=1 to length(primes) do integer p1 = primes[i], j = abs(binary_search(floor((pwr+p1-1)/p1),primes,i)) if j<=length(primes) then -- (always is, I think) integer p2 = primes[j] atom prod = p1*p2 if min_product=0 or prod<min_product then min_product = prod end if pos += j-i if p1>=p2 then exit end if end if end for printf(1,"First brilliant number >= 10^%d is %,d at position %,d\n", {p, min_product, pos}) pwr *= 10; if odd(p) then integer size = length(primes) count += size * (size + 1) / 2; end if end for ?elapsed(time()-t0) {} = wait_key()
- Output:
First 100 brilliant numbers:
4 6 9 10 14 15 21 25 35 49
121 143 169 187 209 221 247 253 289 299
319 323 341 361 377 391 403 407 437 451
473 481 493 517 527 529 533 551 559 583
589 611 629 649 667 671 689 697 703 713
731 737 767 779 781 793 799 803 817 841
851 869 871 893 899 901 913 923 943 949
961 979 989 1,003 1,007 1,027 1,037 1,067 1,073 1,079
1,081 1,121 1,139 1,147 1,157 1,159 1,189 1,207 1,219 1,241
1,247 1,261 1,271 1,273 1,333 1,343 1,349 1,357 1,363 1,369
First brilliant number >= 10^1 is 10 at position 4
First brilliant number >= 10^2 is 121 at position 11
First brilliant number >= 10^3 is 1,003 at position 74
First brilliant number >= 10^4 is 10,201 at position 242
First brilliant number >= 10^5 is 100,013 at position 2,505
First brilliant number >= 10^6 is 1,018,081 at position 10,538
First brilliant number >= 10^7 is 10,000,043 at position 124,364
First brilliant number >= 10^8 is 100,140,049 at position 573,929
First brilliant number >= 10^9 is 1,000,000,081 at position 7,407,841
First brilliant number >= 10^10 is 10,000,600,009 at position 35,547,995
First brilliant number >= 10^11 is 100,000,000,147 at position 491,316,167
First brilliant number >= 10^12 is 1,000,006,000,009 at position 2,409,600,866
First brilliant number >= 10^13 is 10,000,000,000,073 at position 34,896,253,010
First brilliant number >= 10^14 is 100,000,380,000,361 at position 174,155,363,187
First brilliant number >= 10^15 is 1,000,000,000,000,003 at position 2,601,913,448,897
"3.3s"
Python
Using primesieve and numpy modules. If program is run to above 1018 it overflows 64 bit integers (that's what primesieve module backend uses internally).
<lang Python>from primesieve.numpy import primes from math import isqrt import numpy as np
max_order = 9 blocks = [primes(10**n, 10**(n + 1)) for n in range(max_order)]
def smallest_brilliant(lb):
pos = 1 root = isqrt(lb)
for blk in blocks:
n = len(blk)
if blk[-1]*blk[-1] < lb:
pos += n*(n + 1)//2
continue
i = np.searchsorted(blk, root, 'left')
i += blk[i]*blk[i] < lb
if not i:
return blk[0]*blk[0], pos
p = blk[:i + 1]
q = (lb - 1)//p
idx = np.searchsorted(blk, q, 'right')
sel = idx < n
p, idx = p[sel], idx[sel]
q = blk[idx]
sel = q >= p
p, q, idx = p[sel], q[sel], idx[sel]
pos += np.sum(idx - np.arange(len(idx)))
return np.min(p*q), pos
res = [] p = 0 for i in range(100):
p, _ = smallest_brilliant(p + 1) res.append(p)
print(f'first 100 are {res}')
for i in range(max_order*2):
thresh = 10**i
p, pos = smallest_brilliant(thresh)
print(f'Above 10^{i:2d}: {p:20d} at #{pos}')</lang>
- Output:
first 100 are [4, 6, 9, 10, 14, 15, 21, 25, 35, 49, 121, 143, 169, 187, 209, 221, 247, 253, 289, 299, 319, 323, 341, 361, 377, 391, 403, 407, 437, 451, 473, 481, 493, 517, 527, 529, 533, 551, 559, 583, 589, 611, 629, 649, 667, 671, 689, 697, 703, 713, 731, 737, 767, 779, 781, 793, 799, 803, 817, 841, 851, 869, 871, 893, 899, 901, 913, 923, 943, 949, 961, 979, 989, 1003, 1007, 1027, 1037, 1067, 1073, 1079, 1081, 1121, 1139, 1147, 1157, 1159, 1189, 1207, 1219, 1241, 1247, 1261, 1271, 1273, 1333, 1343, 1349, 1357, 1363, 1369] Above 10^ 0: 4 at #1 Above 10^ 1: 10 at #4 Above 10^ 2: 121 at #11 Above 10^ 3: 1003 at #74 Above 10^ 4: 10201 at #242 Above 10^ 5: 100013 at #2505 Above 10^ 6: 1018081 at #10538 Above 10^ 7: 10000043 at #124364 Above 10^ 8: 100140049 at #573929 Above 10^ 9: 1000000081 at #7407841 Above 10^10: 10000600009 at #35547995 Above 10^11: 100000000147 at #491316167 Above 10^12: 1000006000009 at #2409600866 Above 10^13: 10000000000073 at #34896253010 Above 10^14: 100000380000361 at #174155363187 Above 10^15: 1000000000000003 at #2601913448897 Above 10^16: 10000001400000049 at #13163230391313 Above 10^17: 100000000000000831 at #201431415980419
Raku
1 through 7 are fast. 8 and 9 take a bit longer. <lang perl6>use Lingua::EN::Numbers;
- Find an abundance of primes to use to generate brilliants
my %primes = (2..100000).grep( &is-prime ).categorize: { .chars };
- Generate brilliant numbers
my @brilliant = lazy flat (1..*).map: -> $digits {
sort flat (^%primes{$digits}).race.map: { %primes{$digits}[$_] X× (flat %primes{$digits}[$_ .. *]) }
};
- The task
put "First 100 brilliant numbers:\n" ~ @brilliant[^100].batch(10)».fmt("%4d").join("\n") ~ "\n" ;
for 1 .. 7 -> $oom {
my $threshold = exp $oom, 10; my $key = @brilliant.first: :k, * >= $threshold; printf "First >= %13s is %9s in the series: %13s\n", comma($threshold), ordinal-digit(1 + $key, :u), comma @brilliant[$key];
}</lang>
- Output:
First 100 brilliant numbers: 4 6 9 10 14 15 21 25 35 49 121 143 169 187 209 221 247 253 289 299 319 323 341 361 377 391 403 407 437 451 473 481 493 517 527 529 533 551 559 583 589 611 629 649 667 671 689 697 703 713 731 737 767 779 781 793 799 803 817 841 851 869 871 893 899 901 913 923 943 949 961 979 989 1003 1007 1027 1037 1067 1073 1079 1081 1121 1139 1147 1157 1159 1189 1207 1219 1241 1247 1261 1271 1273 1333 1343 1349 1357 1363 1369 First >= 10 is 4ᵗʰ in the series: 10 First >= 100 is 11ᵗʰ in the series: 121 First >= 1,000 is 74ᵗʰ in the series: 1,003 First >= 10,000 is 242ⁿᵈ in the series: 10,201 First >= 100,000 is 2505ᵗʰ in the series: 100,013 First >= 1,000,000 is 10538ᵗʰ in the series: 1,018,081 First >= 10,000,000 is 124364ᵗʰ in the series: 10,000,043 First >= 100,000,000 is 573929ᵗʰ in the series: 100,140,049 First >= 1,000,000,000 is 7407841ˢᵗ in the series: 1,000,000,081
Rust
<lang rust>// [dependencies] // primal = "0.3" // indexing = "0.4.1"
fn get_primes_by_digits(limit: usize) -> Vec<Vec<usize>> {
let mut primes_by_digits = Vec::new();
let mut power = 10;
let mut primes = Vec::new();
for prime in primal::Primes::all().take_while(|p| *p < limit) {
if prime > power {
primes_by_digits.push(primes);
primes = Vec::new();
power *= 10;
}
primes.push(prime);
}
primes_by_digits.push(primes);
primes_by_digits
}
fn main() {
use indexing::algorithms::lower_bound; use std::time::Instant;
let start = Instant::now();
let primes_by_digits = get_primes_by_digits(1000000000);
println!("First 100 brilliant numbers:");
let mut brilliant_numbers = Vec::new();
for primes in &primes_by_digits {
for i in 0..primes.len() {
let p1 = primes[i];
for j in i..primes.len() {
let p2 = primes[j];
brilliant_numbers.push(p1 * p2);
}
}
if brilliant_numbers.len() >= 100 {
break;
}
}
brilliant_numbers.sort();
for i in 0..100 {
let n = brilliant_numbers[i];
print!("{:4}{}", n, if (i + 1) % 10 == 0 { '\n' } else { ' ' });
}
println!();
let mut power = 10;
let mut count = 0;
for p in 1..2 * primes_by_digits.len() {
let primes = &primes_by_digits[p / 2];
let mut position = count + 1;
let mut min_product = 0;
for i in 0..primes.len() {
let p1 = primes[i];
let n = (power + p1 - 1) / p1;
let j = lower_bound(&primes[i..], &n);
let p2 = primes[i + j];
let product = p1 * p2;
if min_product == 0 || product < min_product {
min_product = product;
}
position += j;
if p1 >= p2 {
break;
}
}
println!("First brilliant number >= 10^{p} is {min_product} at position {position}");
power *= 10;
if p % 2 == 1 {
let size = primes.len();
count += size * (size + 1) / 2;
}
}
let time = start.elapsed();
println!("\nElapsed time: {} milliseconds", time.as_millis());
}</lang>
- Output:
First 100 brilliant numbers: 4 6 9 10 14 15 21 25 35 49 121 143 169 187 209 221 247 253 289 299 319 323 341 361 377 391 403 407 437 451 473 481 493 517 527 529 533 551 559 583 589 611 629 649 667 671 689 697 703 713 731 737 767 779 781 793 799 803 817 841 851 869 871 893 899 901 913 923 943 949 961 979 989 1003 1007 1027 1037 1067 1073 1079 1081 1121 1139 1147 1157 1159 1189 1207 1219 1241 1247 1261 1271 1273 1333 1343 1349 1357 1363 1369 First brilliant number >= 10^1 is 10 at position 4 First brilliant number >= 10^2 is 121 at position 11 First brilliant number >= 10^3 is 1003 at position 74 First brilliant number >= 10^4 is 10201 at position 242 First brilliant number >= 10^5 is 100013 at position 2505 First brilliant number >= 10^6 is 1018081 at position 10538 First brilliant number >= 10^7 is 10000043 at position 124364 First brilliant number >= 10^8 is 100140049 at position 573929 First brilliant number >= 10^9 is 1000000081 at position 7407841 First brilliant number >= 10^10 is 10000600009 at position 35547995 First brilliant number >= 10^11 is 100000000147 at position 491316167 First brilliant number >= 10^12 is 1000006000009 at position 2409600866 First brilliant number >= 10^13 is 10000000000073 at position 34896253010 First brilliant number >= 10^14 is 100000380000361 at position 174155363187 First brilliant number >= 10^15 is 1000000000000003 at position 2601913448897 First brilliant number >= 10^16 is 10000001400000049 at position 13163230391313 First brilliant number >= 10^17 is 100000000000000831 at position 201431415980419 Elapsed time: 1515 milliseconds
Sidef
<lang ruby>func is_briliant_number(n) {
n.is_semiprime && (n.factor.map{.len}.uniq.len == 1)
}
func brilliant_numbers_count(n) {
var count = 0 var len = n.isqrt.len
for k in (1 .. len-1) {
var pi = prime_count(10**(k-1), 10**k - 1)
count += binomial(pi, 2)+pi
}
var min = (10**(len - 1)) var max = (10**len - 1)
each_prime(min, max, {|p|
count += prime_count(p, max `min` idiv(n, p))
})
return count
}
say "First 100 brilliant numbers:"
100.by(is_briliant_number).each_slice(10, {|*a|
say a.map { '%4s' % _}.join(' ')
})
say
for n in (1 .. 12) {
var v = (10**n .. Inf -> first_by(is_briliant_number))
printf("First brilliant number >= 10^%d is %s", n, v)
printf(" at position %s\n", brilliant_numbers_count(v))
}</lang>
- Output:
First 100 brilliant numbers: 4 6 9 10 14 15 21 25 35 49 121 143 169 187 209 221 247 253 289 299 319 323 341 361 377 391 403 407 437 451 473 481 493 517 527 529 533 551 559 583 589 611 629 649 667 671 689 697 703 713 731 737 767 779 781 793 799 803 817 841 851 869 871 893 899 901 913 923 943 949 961 979 989 1003 1007 1027 1037 1067 1073 1079 1081 1121 1139 1147 1157 1159 1189 1207 1219 1241 1247 1261 1271 1273 1333 1343 1349 1357 1363 1369 First brilliant number >= 10^1 is 10 at position 4 First brilliant number >= 10^2 is 121 at position 11 First brilliant number >= 10^3 is 1003 at position 74 First brilliant number >= 10^4 is 10201 at position 242 First brilliant number >= 10^5 is 100013 at position 2505 First brilliant number >= 10^6 is 1018081 at position 10538 First brilliant number >= 10^7 is 10000043 at position 124364 First brilliant number >= 10^8 is 100140049 at position 573929 First brilliant number >= 10^9 is 1000000081 at position 7407841 First brilliant number >= 10^10 is 10000600009 at position 35547995 First brilliant number >= 10^11 is 100000000147 at position 491316167 First brilliant number >= 10^12 is 1000006000009 at position 2409600866
Swift
Magnitudes of 1 to 3 is decent, 4 and beyond becomes slow. <lang rebol>// Refs: // https://www.geeksforgeeks.org/sieve-of-eratosthenes/?ref=leftbar-rightbar // https://developer.apple.com/documentation/swift/array/init(repeating:count:)-5zvh4 // https://www.geeksforgeeks.org/brilliant-numbers/#:~:text=Brilliant%20Number%20is%20a%20number,25%2C%2035%2C%2049%E2%80%A6.
// Using Sieve of Eratosthenes func primeArray(n: Int) -> [Bool] {
var primeArr = [Bool](repeating: true, count: n + 1)
primeArr[0] = false // setting zero to be not prime
primeArr[1] = false // setting one to be not prime
// finding all primes which are divisible by p and are greater than or equal to the square of it
var p = 2
while (p * p) <= n {
if primeArr[p] == true {
for j in stride(from: p * 2, through: n, by: p) {
primeArr[j] = false
}
}
p += 1
}
return primeArr
}
func digitsCount(n: Int) -> Int {
// count number of digits for a number
// increase the count if n divide by 10 is not equal to zero
var num = n
var count = 0;
while num != 0 {
num = num/10
count += 1
}
return count
}
func isBrilliant(n: Int) -> Bool {
// Set the prime array
var isPrime = [Bool]()
isPrime = primeArray(n: n)
// Check if the number is the product of two prime numbers
// Also check if the digit counts of those prime numbers are the same.
for i in stride(from: 2, through: n, by: 1) { // i=2, n=50
let x = n / i // i=2, n=50, x=25
if (isPrime[i] && isPrime[x] && x * i == n) { // i=2, x=50, false
if (digitsCount(n: i) == digitsCount(n: x)) {
return true
}
}
}
return false
}
func print100Brilliants() {
// Print the first 100 brilliant numbers
var brilNums = [Int]()
var count = 4
while brilNums.count != 100 {
if isBrilliant(n: count) {
brilNums.append(count)
}
count += 1
}
print("First 100 brilliant numbers:\n", brilNums)
}
func printBrilliantsOfMagnitude() {
// Print the brilliant numbers of base 10 up to magnitude of 6
// Including their positions in the array.
var basePower = 10.0
var brilNums: [Double] = [0.0]
var count = 1.0
while basePower != pow(basePower, 6) {
if isBrilliant(n: Int(count)) {
brilNums.append(count)
if count >= basePower {
print("First brilliant number >= \(Int(basePower)): \(Int(count)) at position \(brilNums.firstIndex(of: count)!)")
basePower *= 10
}
}
count += 1
}
}
print100Brilliants()
printBrilliantsOfMagnitude()</lang>
- Output:
First 100 brilliant numbers: 4 6 9 10 14 15 21 25 35 49 121 143 169 187 209 221 247 253 289 299 319 323 341 361 377 391 403 407 437 451 473 481 493 517 527 529 533 551 559 583 589 611 629 649 667 671 689 697 703 713 731 737 767 779 781 793 799 803 817 841 851 869 871 893 899 901 913 923 943 949 961 979 989 1003 1007 1027 1037 1067 1073 1079 1081 1121 1139 1147 1157 1159 1189 1207 1219 1241 1247 1261 1271 1273 1333 1343 1349 1357 1363 1369 First brilliant number >= 10: 10 at position 4 First brilliant number >= 100: 121 at position 11 First brilliant number >= 1000: 1003 at position 74 First brilliant number >= 10000: 10201 at position 242 First brilliant number >= 100000: 100013 at position 2505 First brilliant number >= 1000000: 1018081 at position 10538
Wren
<lang ecmascript>import "./math" for Int import "./seq" for Lst import "./fmt" for Fmt
var primes = Int.primeSieve(1e7-1)
var getBrilliant = Fn.new { |digits, limit, countOnly|
var brilliant = []
var count = 0
var pow = 1
var next = Num.maxSafeInteger
for (k in 1..digits) {
var s = primes.where { |p| p > pow && p < pow * 10 }.toList
for (i in 0...s.count) {
for (j in i...s.count) {
var prod = s[i] * s[j]
if (prod < limit) {
if (countOnly) {
count = count + 1
} else {
brilliant.add(prod)
}
} else {
next = next.min(prod)
break
}
}
}
pow = pow * 10
}
return countOnly ? [count, next] : [brilliant, next]
}
System.print("First 100 brilliant numbers:") var brilliant = getBrilliant.call(2, 10000, false)[0] brilliant.sort() brilliant = brilliant[0..99] for (chunk in Lst.chunks(brilliant, 10)) Fmt.print("$4d", chunk) System.print() for (k in 1..12) {
var limit = 10.pow(k)
var res = getBrilliant.call(k, limit, true)
var total = res[0]
var next = res[1]
Fmt.print("First >= $,17d is $,15r in the series: $,17d", limit, total + 1, next)
}</lang>
- Output:
First 100 brilliant numbers: 4 6 9 10 14 15 21 25 35 49 121 143 169 187 209 221 247 253 289 299 319 323 341 361 377 391 403 407 437 451 473 481 493 517 527 529 533 551 559 583 589 611 629 649 667 671 689 697 703 713 731 737 767 779 781 793 799 803 817 841 851 869 871 893 899 901 913 923 943 949 961 979 989 1003 1007 1027 1037 1067 1073 1079 1081 1121 1139 1147 1157 1159 1189 1207 1219 1241 1247 1261 1271 1273 1333 1343 1349 1357 1363 1369 First >= 10 is 4th in the series: 10 First >= 100 is 11th in the series: 121 First >= 1,000 is 74th in the series: 1,003 First >= 10,000 is 242nd in the series: 10,201 First >= 100,000 is 2,505th in the series: 100,013 First >= 1,000,000 is 10,538th in the series: 1,018,081 First >= 10,000,000 is 124,364th in the series: 10,000,043 First >= 100,000,000 is 573,929th in the series: 100,140,049 First >= 1,000,000,000 is 7,407,841st in the series: 1,000,000,081 First >= 10,000,000,000 is 35,547,995th in the series: 10,000,600,009 First >= 100,000,000,000 is 491,316,167th in the series: 100,000,000,147 First >= 1,000,000,000,000 is 2,409,600,866th in the series: 1,000,006,000,009
XPL0
<lang XPL0> func NumDigits(N); \Return number of digits in N int N, Cnt; [Cnt:= 0; repeat N:= N/10;
Cnt:= Cnt+1;
until N = 0; return Cnt; ];
func Brilliant(N); \Return 'true' if N is a brilliant number int N, Limit, Cnt, F; int A(3); [Limit:= sqrt(N); Cnt:= 0; F:= 2; loop [if rem(N/F) = 0 then
[A(Cnt):= F;
Cnt:= Cnt+1;
if Cnt > 2 then quit;
N:= N/F;
]
else F:= F+1;
if F > N then quit;
if F > Limit then
[A(Cnt):= N;
Cnt:= Cnt+1;
quit;
];
];
if Cnt # 2 then return false; return NumDigits(A(0)) = NumDigits(A(1)); ];
int Cnt, N, Mag; [Format(5, 0); Cnt:= 0; N:= 4; loop [if Brilliant(N) then
[RlOut(0, float(N));
Cnt:= Cnt+1;
if Cnt >= 100 then quit;
if rem(Cnt/10) = 0 then CrLf(0);
];
N:= N+1;
];
CrLf(0); CrLf(0); Format(7, 0); Cnt:= 0; N:= 4; Mag:= 10; loop [if Brilliant(N) then
[Cnt:= Cnt+1;
if N >= Mag then
[Text(0, "First >= ");
RlOut(0, float(Mag));
Text(0, " is ");
RlOut(0, float(Cnt));
Text(0, " in series: ");
RlOut(0, float(N));
CrLf(0);
if Mag >= 1_000_000 then quit;
Mag:= Mag*10;
];
];
N:= N+1;
];
]</lang>
- Output:
4 6 9 10 14 15 21 25 35 49 121 143 169 187 209 221 247 253 289 299 319 323 341 361 377 391 403 407 437 451 473 481 493 517 527 529 533 551 559 583 589 611 629 649 667 671 689 697 703 713 731 737 767 779 781 793 799 803 817 841 851 869 871 893 899 901 913 923 943 949 961 979 989 1003 1007 1027 1037 1067 1073 1079 1081 1121 1139 1147 1157 1159 1189 1207 1219 1241 1247 1261 1271 1273 1333 1343 1349 1357 1363 1369 First >= 10 is 4 in series: 10 First >= 100 is 11 in series: 121 First >= 1000 is 74 in series: 1003 First >= 10000 is 242 in series: 10201 First >= 100000 is 2505 in series: 100013 First >= 1000000 is 10538 in series: 1018081