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A092472
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a(n)=sum(i+j+k=n,(2n)!/(i+j)!/(j+k)!/(k+i)!) 0<=i<=n, 0<=j<=n, 0<=k<=n.
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4
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1, 6, 54, 510, 4830, 45486, 425502, 3956238, 36594558, 337038702, 3093092574, 28302208974, 258331692606, 2353101799470, 21397006320030, 194281959853710, 1761880227283710, 15961196057303790, 144466419007648350
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OFFSET
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0,2
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LINKS
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FORMULA
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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) = 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
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MATHEMATICA
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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 *)
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PROG
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(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)!))))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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