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a(n) = Sum_{i=1..n} Sum_{j=1..n} Sum_{k=1..n} (i+j+k)!/(3!*i!*j!*k!).
8

%I #21 May 19 2019 06:18:36

%S 0,1,37,842,18252,405408,9268549,216864652,5165454442,124762262630,

%T 3047235458767,75109521108771,1865470016184352,46631215889276662,

%U 1172088706950306293,29601905040172054928,750748513858793527974,19110455782881086439234,488057675380082251617235

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

%H Seiichi Manyama, <a href="/A144511/b144511.txt">Table of n, a(n) for n = 0..100</a>

%H Moa Apagodu, David Applegate, N. J. A. Sloane, and Doron Zeilberger, <a href="http://arxiv.org/abs/1701.08394">Analysis of the Gift Exchange Problem</a>, arXiv:1701.08394, 2017.

%H David Applegate and N. J. A. Sloane, <a href="http://arxiv.org/abs/0907.0513">The Gift Exchange Problem</a> (arXiv:0907.0513, 2009)

%F a(n) = (5 + 3*n - 3*binomial(2*n+2, n+1) + A144660(n))/6. - _Vaclav Kotesovec_, Apr 04 2019

%p 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)];

%t Table[Sum[Sum[Sum[(i+j+k)!/i!/j!/k!/6,{i,1,n}],{j,1,n}],{k,1,n}],{n,1,30}]

%t 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 *)

%o (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

%o (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

%Y Column 3 of array in A144510.

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

%K nonn

%O 0,3

%A _David Applegate_ and _N. J. A. Sloane_, Dec 15 2008, Jan 30 2009