OFFSET
0,2
COMMENTS
Also, number of scenarios in the Gift Exchange Game when a gift can be stolen at most twice. - N. J. A. Sloane, Jan 25 2017
LINKS
David Applegate and N. J. A. Sloane, 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 [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.
FORMULA
a(n) = Sum_{ b,c >= 0, b+c <= n } (n+b+2c)!/ ((n-b-c)! b! c! 2^b 6^c).
The sum is dominated by the b=0, c=n term, so a(n) ~ constant*(3*n)!/(n!*6^n).
EXAMPLE
a(0) = 1;
a(1) = 3: {1} {12} {123}
a(2) = 31: {1,2} {1,23} {2,13} {3,12} {1,234} {2,134} {3,124} {4,123}
{12,34} {13,24} {14,23} {12,345} {13,245} {14,235} {15,234} {23,145} {24,135}
{25,134} {34,125} {35,124} {45,123} {123,456} {124,356} {125,346} {126,345}
{134,256} {135,246} {136,245} {145,236} {146,235} {156,234}.
MATHEMATICA
t[n_, n_] = 1; t[n_ /; n >= 0, k_] /; 0 <= k <= 3*n := t[n, k] = t[n-1, k-1] + (k-1)*t[n-1, k-2] + (1/2)*(k-1)*(k-2)*t[n-1, k-3]; t[_, _] = 0; a[n_] := Sum[t[n, k], {k, 0, 3*n}]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Feb 18 2017 *)
PROG
(PARI) {a(n) = sum(i=n, 3*n, i!*polcoef(sum(j=1, 3, x^j/j!)^n, i))/n!} \\ Seiichi Manyama, May 22 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
David Applegate and N. J. A. Sloane, Dec 07 2008, Dec 17 2008
STATUS
approved