November 11th, 2009, 8:49 pm
Prove that p^2 divides binomial sum a(n) = 16^n * Sum( Binomial(2k,k) / 16^k, {k, 0, n} ) for n = (p - 3)/2 and prime p of the form p = 6k + 1, such as {7,13,19,31,37,43,61,...}.Binomial(2k,k) = (2k)!/(k!)! are the central binomial coefficients.For example, for p = 7 and n = (p - 3)/2 = 2a(2) == 16^2 * ( C(4,2)/16^2 + C(2,1)/16^1 + C(0,0)/16^0 ) = = C(4,2) + 16 * C(2,1) + 16^2 * C(0,0) == 4!/(2!)^2 + 16 * 2!/(1!)^2 + 16^2 * 0!/(0!)^2 == 6 + 16 * 2 + 16^2 * 1 == 6 + 32 + 256 == 294 == 2 * 3 * 7^2 Thus p^2 = 7^2 divides a(2).
Last edited by
Chukchi on November 10th, 2009, 11:00 pm, edited 1 time in total.