A216483 a(n) = Sum_{k=0..n} binomial(n,k)^3 * 4^k.
1, 5, 49, 605, 8065, 113525, 1656145, 24774125, 377601025, 5839329125, 91349718769, 1442580779645, 22959923825281, 367847984671445, 5926784048373265, 95960317086368525, 1560335109283897345, 25466972987548413125, 417048643127042376625, 6850021673230814868125
Offset: 0
Keywords
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- V. Kotesovec, Asymptotic of a sums of powers of binomial coefficients * x^k, 2012.
Programs
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Mathematica
Table[Sum[Binomial[n,k]^3*4^k,{k,0,n}],{n,0,20}]
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PARI
a(n) = sum(k=0, n, binomial(n,k)^3 * 4^k); \\ Gheorghe Coserea, Jul 01 2018
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Sage
A216483 = lambda n: hypergeometric([-n,-n,-n], [1,1], -4) [Integer(A216483(n).n(100)) for n in (0..19)] # Peter Luschny, Sep 23 2014
Formula
Recurrence: (n+3)^2*(3*n+4)*a(n+3) = 5*(9*n^3+57*n^2+116*n+74)*a(n+2) + (99*n^3+528*n^2+938*n+555)*a(n+1) + 125*(3*n+7)*(n+1)^2*a(n).
a(n) ~ (1 + 2^(2/3))^(3*n+2) / (2^(5/3)*sqrt(3)*Pi*n). - Vaclav Kotesovec, Sep 19 2012, simplified Apr 24 2025
G.f.: hypergeom([1/3, 2/3],[1],108*x^2/(1-5*x)^3)/(1-5*x). - Mark van Hoeij, May 02 2013
a(n) = hypergeom([-n,-n,-n],[1,1],-4). - Peter Luschny, Sep 23 2014
G.f. y=A(x) satisfies: 0 = x*(5*x + 2)*(125*x^3 + 33*x^2 + 15*x - 1)*y'' + (1875*x^4 + 1330*x^3 + 273*x^2 + 60*x - 2)*y' + (625*x^3 + 495*x^2 + 42*x + 10)*y. - Gheorghe Coserea, Jul 01 2018
Extensions
Minor edits by Vaclav Kotesovec, Mar 31 2014
Comments