OFFSET
1,1
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..100
Eric Weisstein's World of Mathematics, Complete Tripartite Graph
Eric Weisstein's World of Mathematics, Matching
Eric Weisstein's World of Mathematics, Maximal Independent Edge Set
FORMULA
From Vaclav Kotesovec, Feb 06 2026: (Start)
Recurrence: (8*n^5 - 164*n^4 + 1270*n^3 - 4543*n^2 + 7500*n - 4572)*a(n) = - (8*n^7 - 188*n^6 + 1850*n^5 - 9669*n^4 + 28259*n^3 - 45091*n^2 + 35160*n - 9612)*a(n-1) + (144*n^8 - 3440*n^7 + 33608*n^6 - 173708*n^5 + 520473*n^4 - 929162*n^3 + 966517*n^2 - 532860*n + 116964)*a(n-2) + 8*(n-2)^2*(16*n^8 - 400*n^7 + 4252*n^6 - 24816*n^5 + 86024*n^4 - 179519*n^3 + 217466*n^2 - 136875*n + 32751)*a(n-3) - 8*(n-3)^3*(n-2)^2*(80*n^6 - 1616*n^5 + 12536*n^4 - 46276*n^3 + 84401*n^2 - 70686*n + 20097)*a(n-4) - 64*(n-4)^3*(n-3)^3*(n-2)^2*(8*n^5 - 124*n^4 + 694*n^3 - 1637*n^2 + 1608*n - 501)*a(n-5).
a(n) ~ 3 * 2^((3*n+1)/2) * exp(sqrt(n/2) - 3*n/2 - 3/8) * n^(3*n/2). (End)
MATHEMATICA
Table[3 n! HypergeometricPFQ[{(1 - n)/2, -n, -n/2}, {1}, -4] - If[Mod[n, 2] == 0, 2 (n!/(n/2)!)^3, 0], {n, 20}]
PROG
(PARI) a(n)={if(n%2==0, binomial(n, n/2)*(n/2)!, 0)^3 + sum(k=0, (n-1)\2, 3*binomial(n, k)^2*binomial(n, 2*k)*binomial(2*k, k)*k!^2*(n-k)!)} \\ Andrew Howroyd, Dec 30 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Eric W. Weisstein, Dec 30 2017
EXTENSIONS
Terms a(6) and beyond from Andrew Howroyd, Dec 30 2017
STATUS
approved
