A122485 -id:A122485 - cp's OEIS frontend

A167912 a(n) = (1/(3^n)^2) * Sum_{k=0..(3^n-1)} binomial(2k,k).

1, 217, 913083596083, 18744974860247264575032720770000376335095039

Offset: 1

Alexander Adamchuk, Nov 15 2009

nonn

Note that a(n) mod 27 = a(n) mod 9 = a(n) mod 3 = 1.

The Maple program yields the first seven terms; easily adjustable for obtaining more terms. However, a(4) has 44 digits, a(5) has 140 digits, a(6) has 432 digits and a(7) has 1308 digits. - Emeric Deutsch, Nov 22 2009

Eric Weisstein's World of Mathematics, Central Binomial Coefficient.
Eric Weisstein's World of Mathematics, Binomial Sums.

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

a := proc (n) options operator, arrow: (sum(binomial(2*k, k), k = 0 .. 3^n-1))/3^(2*n) end proc: seq(a(n), n = 1 .. 7); # Emeric Deutsch, Nov 22 2009

Table[(1/3^n)^2 * Sum[Binomial[2 k, k], {k, 0, 3^n - 1}], {n, 1, 5}] (* G. C. Greubel, Jul 01 2016 *)

a(4) from Emeric Deutsch, Nov 22 2009

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

Original entry on oeis.org

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

Offset: 2

Alexander Adamchuk, Apr 04 2012

nonn

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.

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

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

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

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

cp's OEIS Frontend

A167912 a(n) = (1/(3^n)^2) * Sum_{k=0..(3^n-1)} binomial(2k,k).

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