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A179535
a(n) = Sum_{k=0..n} binomial(n,k)^2 * binomial(n-k,k)^2 * 81^k.
1
1, 1, 325, 2917, 247861, 5937301, 265793401, 10705726585, 378746444917, 18588932910901, 657940881863305, 32580334626782185, 1257522211980656425, 59212895251349313865, 2490039488311462939645, 112553667120196462181437
OFFSET
0,3
COMMENTS
On Jul 17 2010, Zhi-Wei Sun introduced this sequence and made the following conjecture: If p is a prime with p=1,9,11,19 (mod 40) and p = x^2+10y^2 with x,y integers, then Sum_{k=0..p-1} a(k) == 4x^2-2p (mod p^2); if p is a prime with p == 7,13,23,37 (mod 40) and 2p = x^2 + 10y^2 with x,y integers, then Sum_{k=0..p-1} a(k) == 2p - 2x^2 (mod p^2); if p is an odd prime with (-10/p)=-1, then Sum_{k=0..p-1} a(k) == 0 (mod p^2). He also conjectured that Sum_{k=0..n-1} (10k+9)*a(k) == 0 (mod n) for all n=1,2,3,... and that Sum_{k=0..p-1} (10k+9)*a(k) == p(4(-2/p)+5) (mod p^2) for any prime p > 3.
LINKS
Zhi-Wei Sun, Open Conjectures on Congruences, preprint, arXiv:0911.5665 [math.NT], 2009-2011.
Zhi-Wei Sun, On Apery numbers and generalized central trinomial coefficients, preprint, arXiv:1006.2776 [math.NT], 2010-2011.
EXAMPLE
For n=2 we have a(2) = 1 + 2^2*81 = 325.
MATHEMATICA
a[n_]:=Sum[Binomial[n, k]^2Binomial[n-k, k]^2*81^k, {k, 0, n}] Table[a[n], {n, 0, 25}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jul 18 2010
STATUS
approved