perfection
#!/usr/bin/python from sys import argv from math import sqrt def f(x): s = 1 for i in xrange(2, int(sqrt(x))+1): if x % i == 0: s += i + x/i return s == x def ff(p): if p <= 6: return True x = p-1 if x % 9 == 0: x = x / 9 else: return False j = 0 while True: t = 32*j**2 + 20*j + 3 j += 1 if t == x: return True elif t > x: return False def p(n): t1 = 2 j = 0 while True: t2 = t1 t1 = t1 << 1 p = (t1-1) * t2 if ff(p) and f(p): j += 1 if j == n: return p if __name__ == '__main__': if len(argv) <= 1: print p(2) else: print p(int(argv[1])) If you don't like Python, here's the C version (uses long long, which is technically avialable in C99 only, but gcc supports it as an extension anyway (for C89)):
#include #include #include int f(long long n); int ff(long long n); int main(int argc, char *argv[]) { long long j, n, p, t1=2, t2; if (argc <= 1) { n = 2; } else { n = strtol(argv[1], 0, 0); } for (j=0; j < n; ) { t2 = t1; t1 = t1 << 1; p = (t1-1)*t2; if (ff(p) && f(p)) { ++j; } } printf("%lld\n", p); return EXIT_SUCCESS; } int f(long long n) { long long int y = sqrt(n), s = 1, i; for (i=2; i <= y; ++i) { if (n%i == 0) { s += i + n/i; } } return s == n; } int ff(long long n) { long long i = n-1; long long j = 0; if (n <= 6) { return 1; } if (i % 9 == 0) { i /= 9; } else { return 0; } while(1) { long long t = 32*j*j + 20*j + 3; ++j; if (t == i) { return 1; } else if (t > i) { return 0; } } }