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A278405
a(n) = Sum_{k=0..n} binomial(n,2k)^2*binomial(n-k,k).
2
1, 1, 2, 19, 110, 476, 2477, 15093, 86830, 485290, 2826902, 16857116, 100034453, 594833357, 3574477090, 21611465819, 130955824174, 796195223398, 4860425688176, 29760574848750, 182655048136510, 1123720751229858, 6929124085148938, 42811398244528788
OFFSET
0,3
COMMENTS
Conjecture: For any prime p > 5 and positive integer n, the number (a(p*n)-a(n))/(p*n)^3 is always a p-adic integer.
We have proved that for any prime p > 5 and positive integer n the number (a(p*n)-a(n))/(p^3*n^2) is always a p-adic integer.
LINKS
Zhi-Wei Sun, Supercongruences involving Lucas sequences, arXiv:1610.03384 [math.NT], 2016.
EXAMPLE
a(3) = 19 since a(3) = C(3,2*0)^2*C(3-0,0) + C(3,2*1)^2*C(3-1,1) = 1 + 3^2*2 = 19.
G.f. = 1 + x + 2*x^2 + 19*x^3 + 110*x^4 + 476*x^5 + 2477*x^6 + 15093*x^7 + ...
MATHEMATICA
a[n_]:=a[n]=Sum[Binomial[n, 2k]^2*Binomial[n-k, k], {k, 0, n/2}]
Table[a[n], {n, 0, 27}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Nov 20 2016
STATUS
approved