A144660 a(n) = Sum_{i=0..n} Sum_{j=0..n} Sum_{k=0..n} (i+j+k)!/(i!*j!*k!).

1, 16, 271, 5248, 110251, 2435200, 55621567, 1301226496, 30992872483, 748574130016, 18283414868863, 450657134765056, 11192820128307871, 279787295456009728, 7032532242167190271, 177611430242835570688, 4504491083159761986451, 114662734697313744041248

Offset: 0

Author

N. J. A. Sloane, Jan 31 2009, Feb 01 2009

Keywords

• nonn

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..100
• Vidunas, Raimundas Counting derangements and Nash equilibria Ann. Comb. 21, No. 1, 131-152 (2017).

Crossrefs

Cf. A030662, A144661, A307318. This sum is very close to that in A144511.

Programs

Maple

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

Mathematica

Table[Sum[(i + j + k)!/(i!*j!*k!), {i, 0, n}, {j, 0, n}, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Apr 02 2019 *)
Table[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}], {n, 0, 20}] (* Vaclav Kotesovec, Apr 04 2019 *)

PARI

{a(n) = sum(i=0, n, sum(j=0, n, sum(k=0, n, (i+j+k)!/(i!*j!*k!))))} \\ Seiichi Manyama, Apr 02 2019

Formula

From Vaclav Kotesovec, Apr 02 2019: (Start)

Recurrence: n^2*(2*n + 1)*(91*n^4 - 478*n^3 + 917*n^2 - 755*n + 222)*a(n) = 3*(2*n - 3)*(3*n - 5)*(3*n - 4)*(91*n^4 - 114*n^3 + 29*n^2 + 9*n - 3)*a(n-1) + n^2*(2*n + 1)*(91*n^4 - 478*n^3 + 917*n^2 - 755*n + 222)*a(n-2) - 3*(2*n - 3)*(3*n - 5)*(3*n - 4)*(91*n^4 - 114*n^3 + 29*n^2 + 9*n - 3)*a(n-3).

a(n) ~ 3^(3*n + 7/2) / (16*Pi*n). (End)