OFFSET
1,1
COMMENTS
a(n) also equals (1/2)*Sum{k=1..n} (k+n)!*((k-1)!*4^k/(2*k)! + 1/(k! * k)).
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..365
FORMULA
Recurrence: n*(2*n + 1)*(216*n^3 - 1062*n^2 + 1191*n - 131)*a(n) = (4320*n^6 - 25560*n^5 + 44808*n^4 - 19756*n^3 - 17153*n^2 + 18363*n - 4380)*a(n-1) - 3*(n-1)*(4752*n^6 - 34380*n^5 + 84072*n^4 - 68855*n^3 - 28120*n^2 + 65537*n - 21356)*a(n-2) + 2*(n-2)*(n-1)*(2*n - 3)*(4320*n^5 - 28800*n^4 + 57102*n^3 - 14749*n^2 - 49592*n + 22788)*a(n-3) - 8*(n-3)^2*(n-2)*(n-1)*(2*n - 5)*(2*n - 3)*(216*n^3 - 414*n^2 - 285*n + 214)*a(n-4). - Vaclav Kotesovec, Mar 02 2014
a(n) ~ sqrt(Pi) * 2^(2*n+1/2) * n^(n-1/2) / exp(n) * (1 + 1/sqrt(n*Pi)). - Vaclav Kotesovec, Mar 02 2014
MAPLE
a:=n->n!*sum(binomial(2*n+1, k)/k, k=1..n): seq(a(n), n=1..20); # Emeric Deutsch, May 27 2007
A129840 := proc(n) n!*add(binomial(2*n+1, k)/k, k=1..n) ; end: for n from 1 to 20 do printf("%d, ", A129840(n)) ; od ; # R. J. Mathar, Jun 08 2007
MATHEMATICA
Table[n!*Sum[Binomial[2n + 1, k]/k, {k, 1, n}], {n, 1, 25}] (* Stefan Steinerberger, May 24 2007 *)
PROG
(PARI) for(n=1, 25, print1(n!*sum(k=1, n, binomial(2*n+1, k)/k), ", ")) \\ G. C. Greubel, Mar 20 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Leroy Quet, May 22 2007
EXTENSIONS
STATUS
approved