A246543 a(n) = (2/n^3)*( Sum_{k=0..n-1} (-1)^k*(3*k^2+3*k+1)*binomial(n-1,k)^3*binomial(n+k,k)^3 ).

2, -47, 1142, 3793, -4094806, 371557891, -13021558306, -1374157073639, 281067953420114, -22220280272696387, -51611579093593498, 257837341935815261683, -35155217354672369625958, 1761633462267526777842223, 202464167122130621896038062

Offset

1,1

Comments

Conjecture: Let n be any positive integer. For m = 0, 2, 4, ..., we have Sum_{k=0..n-1} (3k^2+3k+1)*(binomial(n-1,k)*binomial(n+k,k))^m == 0 (mod n^3); for m = 1, 3, 5, ... we have 2*Sum_{k=0..n-1} (-1)^k*(3k^2+3k+1)*(binomial(n-1,k)*binomial(n+k,k))^m == 0 (mod n^3).

The Zeilberger algorithm could yield a complicated fifth-order recurrence for a(n).

The author proved the conjecture in the latest version of arXiv:1408.5381. - Zhi-Wei Sun, Sep 14 2014

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..70

Zhi-Wei Sun, Two new kinds of numbers and related divisibility results, arXiv:1408.5381 [math.NT], 2014-2018.

Example

a(2) = -47 since (2/2^3)*( Sum_{k=0..1} (-1)^k*(3k^2+3k+1)*binomial(1,k)^3*binomial(2+k,k)^3 ) = (1/4)*(1-7*3^3) = -47.

Mathematica

a[n_] := Sum[(3 k^2 + 3 k + 1) (-1)^k (Binomial[n - 1, k] Binomial[n + k, k])^3, {k, 0, n - 1}] 2/n^3
Table[a[n], {n, 1, 14}]

Crossrefs

Cf. A246512, A246542.

Sequence in context: A120050 A112783 A277655 * A119776 A087265 A079307

Adjacent sequences: A246540 A246541 A246542 * A246544 A246545 A246546

Keyword

sign

Author

Zhi-Wei Sun, Aug 29 2014

Status

approved