A278405 a(n) = Sum_{k=0..n} binomial(n,2k)^2*binomial(n-k,k).

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.

Diagonal of the rational function 1 / ((1 + x)*(1 - x)*(1 - y)*(1 - z) - x*y*z). - Ilya Gutkovskiy, Apr 23 2025

Links

Zhi-Wei Sun, Table of n, a(n) for n = 0..200

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

Formula

Recurrence: 16*(n-1)^2*n^2*(158099011*n^10 - 4901671274*n^9 + 67586615859*n^8 - 545625283344*n^7 + 2855189758269*n^6 - 10116688998252*n^5 + 24574792513889*n^4 - 40401528139756*n^3 + 43013843734080*n^2 - 26776492679334*n + 7400750600772)*a(n) = 8*(n-1)^2*(1580990110*n^12 - 50597702850*n^11 + 725456801360*n^10 - 6150020769133*n^9 + 34256126693079*n^8 - 131725795030358*n^7 + 357415342251742*n^6 - 686833714644138*n^5 + 923487632863508*n^4 - 842831927107392*n^3 + 493056019304448*n^2 - 165631844772864*n + 24417641299968)*a(n-1) - 4*(4426772308*n^14 - 154953884904*n^13 + 2468796531103*n^12 - 23715639785381*n^11 + 153368439299140*n^10 - 705994114379114*n^9 + 2384620223615643*n^8 - 6002058365943257*n^7 + 11310425749092354*n^6 - 15878259200550864*n^5 + 16346068908782604*n^4 - 11964179709090432*n^3 + 5882387074623648*n^2 - 1736423411933088*n + 231376410005760)*a(n-2) + 4*(56915643960*n^14 - 2049179878440*n^13 + 33636783111585*n^12 - 333381447319018*n^11 + 2226868226910469*n^10 - 10593990163567062*n^9 + 36974797011964443*n^8 - 96052998294171144*n^7 + 186358897064030787*n^6 - 268274437227521600*n^5 + 281539106062072196*n^4 - 208434428611159680*n^3 + 102694106635953600*n^2 - 30090568321054656*n + 3955971208527360)*a(n-3) + 4*(61026218246*n^14 - 2258202421240*n^13 + 38129367327199*n^12 - 388997452832899*n^11 + 2675948309399848*n^10 - 13114335112594739*n^9 + 47154228395511414*n^8 - 126174901113827190*n^7 + 252042947908215229*n^6 - 373324987945789854*n^5 + 402787447219474882*n^4 - 306292265313242456*n^3 + 154856037732694032*n^2 - 46517929012294272*n + 6263188813728000)*a(n-4) + 2*(n-4)^2*(40157148794*n^12 - 1204867354802*n^11 + 16166453839008*n^10 - 128111700151965*n^9 + 666899291519460*n^8 - 2398912178690784*n^7 + 6104209046516104*n^6 - 11051314432860193*n^5 + 14101651164691002*n^4 - 12344617197884520*n^3 + 7025028138311004*n^2 - 2330934507803988*n + 341312019554880)*a(n-5) + 25*(n-5)^2*(n-4)^2*(158099011*n^10 - 3320681164*n^9 + 30586029888*n^8 - 162420641016*n^7 + 549698424207*n^6 - 1236600540510*n^5 + 1868962301540*n^4 - 1870255835972*n^3 + 1183919373120*n^2 - 427744124424*n + 67084549920)*a(n-6). - Vaclav Kotesovec, Oct 16 2025

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

Cf. A208425, A244973, A275027, A277640, A278415.

Sequence in context: A089659 A396762 A240124 * A285791 A376481 A101253

Adjacent sequences: A278402 A278403 A278404 * A278406 A278407 A278408

Keyword

nonn

Author

Zhi-Wei Sun, Nov 20 2016

Status

approved