A281360 Number of scenarios in the Gift Exchange Game when a gift can be stolen at most 8 times.

1, 9, 92368, 124762262630, 2774049143394729653, 476872353039366288373555323, 414678423576860263798348331987688320, 1383884737648788823775562903922773021277571568, 14584126149704606223764458141727351569547933381159988406, 419715170056359079715862408734598208208707081189266290220651371206

Offset: 0

Author

N. J. A. Sloane, Jan 25 2017

Keywords

• nonn

Comments

More than the usual number of terms are shown in the DATA field because there are the initial values needed for one of the recurrences.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..69
• Moa Apagodu, David Applegate, N. J. A. Sloane, and Doron Zeilberger, Analysis of the Gift Exchange Problem, arXiv:1701.08394 [math.CO], 2017.
• Moa Apagodu, David Applegate, N. J. A. Sloane, and Doron Zeilberger, On-Line Appendix I to "Analysis of the gift exchange problem", giving Type D recurrences for G_1(n) through G_15(n) (see A001515, A144416, A144508, A144509, A149187, A281358-A281361)
• Moa Apagodu, David Applegate, N. J. A. Sloane, and Doron Zeilberger, On-Line Appendix II to "Analysis of the gift exchange problem", giving Type C recurrences for G_1(n) through G_15(n) (see A001515, A144416, A144508, A144509, A149187, A281358-A281361)
• David Applegate and N. J. A. Sloane, The Gift Exchange Problem, arXiv:0907.0513 [math.CO], 2009.

Crossrefs

The gift scenarios sequences when a gift can be stolen at most s times, for s = 1..9, are A001515, A144416, A144508, A144509, A149187, A281358, A281359, A281360, A281361.

Programs

Maple

with(combinat):
b:= proc(n, i, t) option remember; `if`(t*i add(b(k, 9, n), k=0..9*n):
seq(a(n), n=0..12);  # Alois P. Heinz, Feb 01 2017

Mathematica

t[n_, n_] = 1; t[n_ /; n >= 0, k_] /; 0 <= k <= 9*n := t[n, k] = Sum[(1/j!)*Product[k - m, {m, 1, j}]*t[n - 1, k - j - 1], {j, 0, 8}]; t[, ] = 0; a[n_] := Sum[t[n, k], {k, 0, 9*n}]; Table[a[n], {n, 0, 10}] (* Jean-François Alcover, Feb 18 2017 *)

PARI

{a(n) = sum(i=n, 9*n, i!*polcoef(sum(j=1, 9, x^j/j!)^n, i))/n!} \\ Seiichi Manyama, May 22 2019