A181990 a(n) = Sum_{0 <= k <= m < p} (binomial(m, k)^(p-1))/p, where p is the n-th prime.

3, 399, 12708885, 124515078454872901983423, 39212583445587381894247266262023061, 43487633454143579523135045521112077473364484383507327790688372131, 157851796824901989964381293031623545741924564754192453966085327785455257503133278729

Offset: 2

Author

Alexander Adamchuk, Apr 04 2012

Keywords

• nonn

Comments

a(n) is a sum of all elements in the first p rows of Pascal's triangle each raised to the (p-1) power and divided by p, where p is the n-th prime.

For p = 3 and 7 (and their powers like 3, 9, 27, ... and 7, 49, ...) the sums of all elements in n = p^k top rows of Pascal's triangle each raised to the (n-1) = (p^k-1) power are divisible by n^2 = p^(2k) for all k > 0.

Links

• Eric Weisstein's World of Mathematics, Pascal's Triangle
• Eric Weisstein's World of Mathematics, Binomial Sums

Crossrefs

Cf. A007318, A006134, A083096, A066796, A083097, A081601, A010060, A122485, A167912.

Programs

Mathematica

Table[(Sum[Binomial[m, k]^(Prime[n] - 1), {m, 0, Prime[n] - 1}, {k, 0, m}])/Prime[n], {n, 2, 10}]

PARI

a(n) = my(p=prime(n)); sum(m=0, p-1, sum(k=0, m, binomial(m,k)^(p-1))/p); \\ Michel Marcus, Dec 03 2018