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#!/usr/bin/ruby
# An efficient algorithm for computing Fibonacci(n) (mod m) for arbitrary large integers.
# Algorithm from:
# https://metacpan.org/source/KRYDE/Math-NumSeq-72/lib/Math/NumSeq/Fibonacci.pm
# See also:
# https://en.wikipedia.org/wiki/Fibonacci_number
func fibmod(n, m) {
var (f, g, a) = (0, 1, -2)
for k in (n.ilog2 ^.. 0) {
g = (g*g)%m
f = (f*f)%m
var t = (g<<2 - f + a)
f += g
if (n.getbit(k)) {
(f, g, a) = (t-f, t, -2)
}
else {
(g, a) = (t-f, 2)
}
}
return (g % m)
}
func fibonacci_factorization(n, B=10000) {
var k = consecutive_lcm(B) # lcm(1..B)
var F = fibmod(k, n) # Fibonacci(k) (mod n)
return gcd(F, n)
}
say fibonacci_factorization(257221 * 470783, 700); #=> 470783 (p+1 is 700-smooth)
say fibonacci_factorization(333732865481 * 1632480277613, 3000); #=> 333732865481 (p-1 is 3000-smooth)