A092472 a(n)=sum(i+j+k=n,(2n)!/(i+j)!/(j+k)!/(k+i)!) 0<=i<=n, 0<=j<=n, 0<=k<=n.

1, 6, 54, 510, 4830, 45486, 425502, 3956238, 36594558, 337038702, 3093092574, 28302208974, 258331692606, 2353101799470, 21397006320030, 194281959853710, 1761880227283710, 15961196057303790, 144466419007648350

Offset: 0

Author

Benoit Cloitre, Mar 25 2004

Keywords

• nonn

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200

Programs

Mathematica

Table[Sum[Sum[Sum[If[i+j+k==n,(2n)!/(i+j)!/(j+k)!/(k+i)!,0],{i,0,n}],{j,0,n}],{k,0,n}],{n,0,20}]
(* or *)
Flatten[{1,6,RecurrenceTable[{(n-3)*n*a[n]==(17*n^2-55*n+24)*a[n-1]-36*(n-2)*(2*n-3)*a[n-2],a[2]==54,a[3]==510},a,{n,2,20}]}] (* Vaclav Kotesovec, Oct 14 2012 *)
Flatten[{1,6,Table[3^(2*n-1)*(2+Sum[2^(k+2)*Binomial[2*k+1,k-1]/3^(2*k+2),{k,1,n-2}]),{n,2,20}]}] (* Vaclav Kotesovec, Oct 28 2012 *)

PARI

a(n)=sum(i=0,n,sum(j=0,n,sum(k=0,n,if(i+j+k-n,0,(2*n)!/(i+j)!/(j+k)!/(k+i)!))))

Formula

Recurrence (for n>3): (n-3)*n*a(n) = (17*n^2-55*n+24)*a(n-1) - 36*(n-2)*(2*n-3)*a(n-2). - Vaclav Kotesovec, Oct 14 2012

a(n) ~ 9^n. - Vaclav Kotesovec, Oct 14 2012

a(n) = 3^(2*n-1) * (2 + Sum_{k=1..n-2} 2^(k+2)*C(2*k+1,k-1)/3^(2*k+2) ), for n>2. - Vaclav Kotesovec, Oct 28 2012