A278415 a(n) = Sum_{k=0..n} binomial(n, 2k)*binomial(n-k, k)*(-1)^k.

1, 1, 0, -5, -16, -24, 15, 197, 576, 724, -1200, -8832, -22801, -21293, 76440, 408795, 922368, 499104, -4446588, -19025060, -37012416, -1673992, 245604832, 880263936, 1441226991, -908700649, -13088509200, -40222012703, -52991533744, 88167061704, 678172355415, 1805175708261, 1747974632448, -6237554623536, -34300087628480

Offset

0,4

Comments

Conjecture: For any prime p > 3 and positive integer n, the number (a(p*n)-a(n))/(p*n)^2 is always a p-adic integer.

We are able to show that for any prime p > 3 and positive integer n the number (a(p*n)-a(n))/(p^2*n) is always a p-adic integer.

See also A275027 and A278405 for similar conjectures.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 0..400

Zhi-Wei Sun, Supercongruences involving Lucas sequences, arXiv:1610.03384 [math.NT], 2016.

Example

a(3) = -5 since a(3) = C(3, 2*0)*C(3-0, 0)(-1)^0 + C(3,2*1)*C(3-1,1)(-1)^1 = 1 - 6 = -5.

Mathematica

a[n_]:=Sum[Binomial[n, 2k]Binomial[n-k, k](-1)^k, {k, 0, n}]
Table[a[n], {n, 0, 34}]

Crossrefs

Cf. A208425, A244973, A277640, A275027, A278405.

Sequence in context: A029450 A225539 A299124 * A063243 A063232 A087747

Adjacent sequences: A278412 A278413 A278414 * A278416 A278417 A278418

Keyword

sign

Author

Zhi-Wei Sun, Nov 21 2016

Status

approved