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}]
KEYWORD
sign
AUTHOR
Zhi-Wei Sun, Aug 29 2014
STATUS
approved