OFFSET
0,3
COMMENTS
LINKS
Andrew Howroyd, Table of n, a(n) for n = 0..2713 (rows 0..20)
T. Kyle Petersen and Bridget Eileen Tenner, How to write a permutation as a product of involutions (and why you might care), arXiv:1202.5319 [math.CO], 2012.
EXAMPLE
Triangle begins:
0 | 1;
1 | 1;
2 | 2, 2;
3 | 3, 2, 4;
4 | 4, 3, 6, 4, 10;
5 | 5, 4, 6, 6, 6, 8, 26;
6 | 6, 5, 8, 12, 8, 6, 20, 12, 12, 20, 76;
7 | 7, 6, 10, 12, 10, 8, 12, 18, 16, 12, 20, 30, 24, 52, 232;
...
T(7,7)= 12 since the partition 3;3;1 represents a cycle structure of a permutation that can be decomposed into involutions in 12 ways: 3*3=9 ways by splitting each 3-cycle into a 1-cycle and a 2-cycle, and 3 more ways by combining both 3-cycles to produce three 2-cycles.
MATHEMATICA
Needs["DiscreteMath`Combinatorica`"]; countinvolutions[cyclestructure_List]:= Times@@ ( (Plus@@ Table[(2k)!/k!/2^k Binomial[ #2, 2k] #1^(#2-2k) #1^k, {k, 0, #2/2}]&) @@@ ({First@#, Length@#}& /@ Split[cyclestructure]) ); Table[countinvolutions /@ Reverse/@ Sort[Sort/@ Partitions[n]], {n, 10}]
PROG
(PARI)
B(n, e)=sum(k=0, n\2, binomial(n, 2*k)*e^(n-k)*(2*k)!/(k!*2^k))
C(sig)={my(S=Set(sig)); prod(k=1, #S, my(c=#select(t->t==S[k], sig)); B(c, S[k]))}
Row(n)={apply(C, [Vecrev(p) | p<-partitions(n)])}
{ for(n=0, 7, print(Row(n))) } \\ Andrew Howroyd, Oct 05 2025
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Wouter Meeussen, Aug 13 2009
EXTENSIONS
Typo fixed by Franklin T. Adams-Watters, Aug 29 2009
a(0)=1 prepended by Andrew Howroyd, Oct 05 2025
STATUS
approved
