def npartitions(n, verbose=False): """ Calculate the partition function P(n), i.e. the number of ways that n can be written as a sum of positive integers. P(n) is computed using the Hardy-Ramanujan-Rademacher formula, described e.g. at http://mathworld.wolfram.com/PartitionFunctionP.html The correctness of this implementation has been tested for 10**n up to n = 8. """ n = int(n) if n < 0: return 0 if n <= 5: return [1, 1, 2, 3, 5, 7][n] # Estimate number of bits in p(n). This formula could be tidied pbits = int((math.pi*(2*n/3.)**0.5-math.log(4*n))/math.log(10)+1)*\ math.log(10,2) prec = p = int(pbits*1.1 + 100) s = fzero M = max(6, int(0.24*n**0.5+4)) sq23pi = mpf_mul(mpf_sqrt(from_rational(2,3,p), p), mpf_pi(p), p) sqrt8 = mpf_sqrt(from_int(8), p) for q in xrange(1, M): a = A(n,q,p) d = D(n,q,p, sq23pi, sqrt8) s = mpf_add(s, mpf_mul(a, d), prec) if verbose: print "step", q, "of", M, to_str(a, 10), to_str(d, 10) # On average, the terms decrease rapidly in magnitude. Dynamically # reducing the precision greatly improves performance. p = bitcount(abs(to_int(d))) + 50 np = to_int(mpf_add(s, fhalf, prec)) return int(np)
def D(n, j, prec, sq23pi, sqrt8): """ Compute the sinh term in the outer sum of the HRR formula. The constants sqrt(2/3*pi) and sqrt(8) must be precomputed. """ j = from_int(j) pi = mpf_pi(prec) a = mpf_div(sq23pi, j, prec) b = mpf_sub(from_int(n), from_rational(1,24,prec), prec) c = mpf_sqrt(b, prec) ch, sh = cosh_sinh(mpf_mul(a,c), prec) D = mpf_div(mpf_sqrt(j,prec), mpf_mul(mpf_mul(sqrt8,b),pi), prec) E = mpf_sub(mpf_mul(a,ch), mpf_div(sh,c,prec), prec) return mpf_mul(D, E)