A189886 a(n) is the number of compositions of the set {1, 2, ..., n} into blocks, each of size 1, 2 or 3 (n >= 0).
1, 1, 3, 13, 74, 530, 4550, 45570, 521640, 6717480, 96117000, 1512819000, 25975395600, 483169486800, 9678799930800, 207733600074000, 4755768505488000, 115681418156304000, 2979408725813520000, 80998627977002736000, 2317937034142810080000, 69649003197501567840000, 2192459412316607834400000, 72152830779716793506400000, 2477756318984329979756160000
Offset: 0
Keywords
Examples
a(3) = 13 because all compositions of set {a,b,c} into blocks of size 1, 2, or 3 are: 1: ({a,b,c}), 2: ({a},{b,c}), 3: ({b,c},{a}), 4: ({b},{a,c}), 5: ({a,c},{b}), 6: ({c},{a,b}), 7: ({a,b},{c}), 8: ({a},{b},{c}), 9: ({a},{c},{b}), 10: ({b},{a},{c}), 11: ({b},{c},{a}), 12: ({c},{a},{b}), 13: ({c},{b},{a}).
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..424
- Moa Apagodu, David Applegate, N. J. A. Sloane, and Doron Zeilberger, Analysis of the Gift Exchange Problem, arXiv:1701.08394, 2017.
- David Applegate and N. J. A. Sloane, The Gift Exchange Problem (arXiv:0907.0513, 2009)
- Adi Dani, Compositions and partitions of sets
Programs
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Maple
A189886 := proc(n) local m, j; add(add(2^(2*m-n-j)*3^(j-m)*n! *binomial(m,j)*binomial(j,2*j-(3*m-n)),j=0..3*m-n),m=0..n) end: seq(A189886(n),n=0..24); # Peter Luschny, May 02 2011 # second Maple program: a:= proc(n) option remember; `if`(n=0, 1, add( a(n-i)*binomial(n, i), i=1..min(n, 3))) end: seq(a(n), n=0..25); # Alois P. Heinz, Sep 22 2016 # third Maple program: a:= n-> n! * (<<0|1|0>, <0|0|1>, <1/6|1/2|1>>^n)[3, 3]: seq(a(n), n=0..25); # Alois P. Heinz, Sep 22 2016
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Mathematica
Table[Sum[n!/(2^(n+j-2m)3^(m-j))*Binomial[m,j]*Binomial[j,n+2j-3m], {m,0,n},{j,0,3m-n}],{n,0,15}]
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PARI
a(n)=sum(m=0,n, sum(j=0,3*m-n, n!/(2^(n+j-2*m) *3^(m-j)) *binomial(m,j) *binomial(j,n+2*j-3*m))); /* Joerg Arndt, May 03 2011 */
Formula
a(n) = sum(m=0..n, sum(j=0..3*m-n, n!/(2^(n+j-2*m) * 3^(m-j)) * C(m,j) * C(j,n+2*j-3*m))) where C(n,k) is the binomial coefficient.
a(n) = n * a(n-1) + n*(n-1)/2 * a(n-2) + n*(n-1)*(n-2)/6 * a(n-3). - Istvan Mezo, Jun 06 2013
E.g.f.: 1/(1 - x - x^2/2 - x^3/6). - Geoffrey Critzer, Dec 04 2012
Comments