A144416 a(n) is the total number of partitions of [1, 2, ..., k] into exactly n blocks, each of size 1, 2 or 3, for 0 <= k <= 3n.

1, 3, 31, 842, 45296, 4061871, 546809243, 103123135501, 25942945219747, 8394104851717686, 3395846808758759686, 1679398297627675722593, 996789456118195908366641, 699283226713639676370419067, 572385833490097906671186099971, 540635257271794961275858251107746, 583630397618757664934692641037584628

Offset: 0

Author

David Applegate and N. J. A. Sloane, Dec 07 2008, Dec 17 2008

Keywords

• nonn

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

Examples

			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}.
		

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.

Crossrefs

Row sums of A144385. Slice sums of A144626.

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

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 *)

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

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).