OFFSET
1,1
COMMENTS
On Jan 22 2011, Zhi-Wei Sun conjectured that a(n) is a positive integer for every n=1,2,3,... Let p > 3 be a prime. He also conjectured that a(p) == 5 + 6*(p/3)*(2-3^(p-1)) (mod p^2). Another conjecture of his states that Sum_{k=0..p-1} (-1)^k*binomial(2k,k)^2*T(k) is congruent to b(p) modulo p^2, where b(p)=0 if (p/15)=-1, b(p) = 4x^2-2p if p == 1,4 (mod 15) and p = x^2+15y^2 with x,y integers, and b(p) = 20x^2-2p if p == 2,8 (mod 15) and p=5x^2+3y^2 with x,y integers.
LINKS
D. S. McNeil, Table of n, a(n) for n = 1..900
Zhi-Wei Sun, Open Conjectures on Congruences, preprint, arXiv:0911.5665 [math.NT], 2009-2011.
Zhi-Wei Sun, On sums related to central binomial and trinomial coefficients, preprint, arXiv:1101.0600 [math.NT], 2011-2014.
EXAMPLE
For n=2 we have a(2) = (44*1^2*T(0)(-1) + (105+44)*2^2*T(1))/(2*2*binomial(4,2)) = 23.
MAPLE
T:= n-> `if`(n=0, 1, coeff ((x^2+x+1)^n, x, n)):
a:= n-> add ((105*k+44) *binomial (2*k, k)^2 *T(k)*(-1)^(n-1-k),
k=0..n-1)/ (2*n*binomial (2*n, n)):
seq (a(n), n=1..30);
MATHEMATICA
T[k_]:=If[k>0, Coefficient[(x^2+x+1)^k, x^k], 1]
A[n_]:=Sum[(105k+44)Binomial[2k, k]^2*T[k](-1)^(n-1-k), {k, 0, n-1}]/(2n*Binomial[2n, n])
Table[A[n], {n, 1, 50}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 24 2011
STATUS
approved