A144511 a(n) = Sum_{i=1..n} Sum_{j=1..n} Sum_{k=1..n} (i+j+k)!/(3!*i!*j!*k!).

0, 1, 37, 842, 18252, 405408, 9268549, 216864652, 5165454442, 124762262630, 3047235458767, 75109521108771, 1865470016184352, 46631215889276662, 1172088706950306293, 29601905040172054928, 750748513858793527974, 19110455782881086439234, 488057675380082251617235

Offset: 0

Author

David Applegate and N. J. A. Sloane, Dec 15 2008, Jan 30 2009

Keywords

• nonn

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..100
• Moa Apagodu, David Applegate, N. J. A. Sloane, and Doron Zeilberger, Analysis of the Gift Exchange Problem, arXiv:1701.08394, 2017.
• David Applegate and N. J. A. Sloane, The Gift Exchange Problem (arXiv:0907.0513, 2009)

Crossrefs

Column 3 of array in A144510.

Cf. A144658, A144660 (a very similar sum).

Programs

Maple

f:=n->add( add( add( (i+j+k)!/(3!*i!*j!*k!), i=1..n),j=1..n),k=1..n); [seq(f(n),n=0..20)];

Mathematica

Table[Sum[Sum[Sum[(i+j+k)!/i!/j!/k!/6,{i,1,n}],{j,1,n}],{k,1,n}],{n,1,30}]
Table[(5 + 3*n - 3*Binomial[2*n+2, n+1] + Sum[(1 + k + 2*n)! * HypergeometricPFQ[{1, -1 - k - n, -n}, {-1 - k - 2*n, -k - n}, 1] / ((1 + k + n)*k!*n!^2), {k, 0, n}]) / 6, {n, 0, 20}] (* Vaclav Kotesovec, Apr 04 2019 *)

PARI

{a(n) = sum(i=1, n, sum(j=1, n, sum(k=1, n, (i+j+k)!/(6*i!*j!*k!))))} \\ Seiichi Manyama, Apr 03 2019

PARI

{a(n) = sum(i=3, 3*n, i!*polcoef(sum(j=1, n, x^j/j!)^3, i))/6} \\ Seiichi Manyama, May 19 2019

Formula

a(n) = (5 + 3*n - 3*binomial(2*n+2, n+1) + A144660(n))/6. - Vaclav Kotesovec, Apr 04 2019