A215341 Expansion of series_reversion( x/(1+x^4*sum(k>=0, x^k)) ) / x.
1, 0, 0, 0, 1, 1, 1, 1, 5, 10, 16, 23, 53, 118, 232, 411, 813, 1718, 3568, 7012, 13925, 28603, 59533, 121878, 247915, 509136, 1057278, 2194138, 4536943, 9394145, 19552639, 40803472, 85131237, 177640486, 371426592, 778275264, 1632420197, 3425607187, 7195476245, 15134138683, 31866093569
Offset: 0
Keywords
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
Crossrefs
Programs
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Maple
b:= proc(x, y, t) option remember; `if`(y
0 and t in [0, 4], b(x-1, y, 0), 0) +b(x, y-1, min(t+1, 4)))) end: a:= n-> b(n, n, 0): seq(a(n), n=0..50); # Alois P. Heinz, Apr 16 2013 -
Mathematica
InverseSeries[x/(1+x^4/(1-x)) + O[x]^50] // CoefficientList[#, x]& // Rest (* Jean-François Alcover, Mar 29 2017 *)
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Maxima
a(n):=sum(binomial(n+1,i)*binomial(n-3*i-1,n-4*i),i,0,floor(n/4))/(n+1); /* Vladimir Kruchinin, Apr 01 2019 */
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PARI
N=66; Vec( serreverse(x/(1+x^4*sum(k=0,N,x^k))+O(x^N)) / x )
Formula
G.f. A(x) satisfies 0 = -x^4*A(x)^4 - x*A(x)^2 + (x + 1)*A(x) - 1. [Joerg Arndt, Mar 01 2014]
Recurrence: 2*n*(n+1)*(2*n+3)*(16204*n^4 - 82948*n^3 + 139973*n^2 - 85643*n + 10674)*a(n) = - (n-1)*n*(307876*n^5 - 960260*n^4 + 288863*n^3 + 582749*n^2 + 5406*n + 12696)*a(n-1) + 4*(n-2)*(129632*n^6 - 469136*n^5 + 354226*n^4 + 317255*n^3 - 469674*n^2 + 176517*n - 21420)*a(n-2) - 2*(n-3)*(n-2)*(16204*n^5 - 34336*n^4 + 82943*n^3 - 208775*n^2 + 192120*n - 40656)*a(n-3) + 6*(n-3)*(n-2)*(97224*n^5 - 351852*n^4 + 179198*n^3 + 540009*n^2 - 571727*n + 92968)*a(n-4) + 229*(n-4)*(n-3)*(n-2)*(16204*n^4 - 18132*n^3 - 11647*n^2 + 10275*n - 1740)*a(n-5). - Vaclav Kotesovec, Mar 22 2014
a(n) ~ sqrt((s-1)*s^3/(6-8*s+3*s^2)) / (2*sqrt(Pi)*n^(3/2)*r^n), where r = 0.4577644245749322..., s = 1.232809919151165... are roots of the system of equations 1 + r*s^2 + r^4*s^4 = (1+r)*s, 1+r = 2*r*s + 4*r^4*s^3. - Vaclav Kotesovec, Mar 22 2014
a(n) = (1/(n+1)) * Sum_{i=0..floor(n/4)} C(n+1,i) * C(n-3*i-1,n-4*i). - Vladimir Kruchinin, Apr 01 2019
Extensions
Modified definition to obtain offset 0 for combinatorial interpretation, Joerg Arndt, Apr 16 2013
Comments