def non_coprime_sum_pattern(m, x_max, k_max=4):
'''Run this once to get an idea of what the sum of Fnx is for a non-co-prime factor m of 15!
for n=10^k, k=0,1,2,3,... .'''
for n in 10 ** np.arange(k_max):
fnx_bf = Fnx_bf(n, m, x_max)
s_bf = sum_mod(fnx_bf, m)
print 'n %6d s %5d' % (n, s_bf), 'BF fnx', fnx_bf # Turn on debugging printouts to observe pattern
return s_bf, m # Assuming constant s_bf for large k
F = Fib(r=m)
f = np.array([F(i) for i in xrange(n + 1)])
return map(long, (polyval_mod(f, x, m) for x in xrange(x_max + 1)))
def non_coprime_sum_pattern(m, x_max, k_max=4):
'''Run this once to get an idea of what the sum of Fnx is for a non-co-prime factor m of 15!
for n=10^k, k=0,1,2,3,... .'''
for n in 10 ** np.arange(k_max):
fnx_bf = Fnx_bf(n, m, x_max)
s_bf = sum_mod(fnx_bf, m)
print 'n %6d s %5d' % (n, s_bf), 'BF fnx', fnx_bf # Turn on debugging printouts to observe pattern
return s_bf, m # Assuming constant s_bf for large k
def crt(a):
'''Returns the basic solution to the system x = a[i][0] (mod a[i][1]), i=0..len(a)-1 using the Chinese
Remainder Theorem. All a[i][1] must be co-prime (although a weaker condition involving the a[i][0]''s is
also sufficient for a solution, this implementation is not guaranteed to work for it).'''
a, n = zip(*a)
N = np.prod(map(long, n))
return sum(a[i] * (N / n[i]) * inv_mod(N / n[i], n[i]) for i in xrange(len(a))) % N
'''Decompose 15! into factors with co-prime denominators (using division of Fn terms)
and factors with non-co-prime denominators (using patterns observed for n=10^k)
Solve for sum modulo 15! using the Chinese remainder theorem.'''
sum_fnx_mod_fifteen_factorial = lambda n, x_max: crt([(sum_mod((Fnx(n, m, x) for x in xrange(x_max + 1)), m), m)
for m in [2 ** 11, 3 ** 6, 7 ** 2, 13]] + [non_coprime_sum_pattern(m, x_max) for m in [5 ** 3, 11]])
if __name__ == "__main__":
n = 10 ** 15
print sum_fnx_mod_fifteen_factorial(10 ** 15, 100)
