A324013 Number of self-complementary set partitions of {1, ..., n} with no singletons.

1, 0, 1, 1, 4, 3, 15, 16, 75, 89, 428, 571, 2781, 4060, 20093, 31697, 159340, 268791, 1372163, 2455804, 12725447, 24012697, 126238060, 249880687, 1332071241, 2754348360, 14881206473, 32029000641, 175297058228, 391548016475, 2169832010759

Offset: 0

Author

Gus Wiseman, Feb 12 2019

Keywords

• nonn

Comments

The complement of a set partition pi of {1, ..., n} is defined as n + 1 - pi (elementwise) on page 3 of Callan. For example, the complement of {{1,5},{2},{3,6},{4}} is {{1,4},{2,6},{3},{5}}.

Examples

			The  a(3) = 1 through a(6) = 15 self-complementary set partitions with no singletons:
  {{123}}  {{1234}}    {{12345}}    {{123456}}
           {{12}{34}}  {{135}{24}}  {{123}{456}}
           {{13}{24}}  {{15}{234}}  {{124}{356}}
           {{14}{23}}               {{1256}{34}}
                                    {{1346}{25}}
                                    {{135}{246}}
                                    {{145}{236}}
                                    {{16}{2345}}
                                    {{12}{34}{56}}
                                    {{13}{25}{46}}
                                    {{14}{25}{36}}
                                    {{15}{26}{34}}
                                    {{16}{23}{45}}
                                    {{16}{24}{35}}
                                    {{16}{25}{34}}
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..500
• David Callan, On conjugates for set partitions and integer compositions, arXiv:math/0508052 [math.CO], 2005.

Crossrefs

Cf. A000110, A000296, A080107 (self-complementary), A086365, A124323, A324012 (self-conjugate).

Programs

Mathematica

sps[{}]:={{}};sps[set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@sps[Complement[set,s]]]/@Cases[Subsets[set],{i,_}];
cmp[stn_]:=Union[Sort[Max@@Join@@stn+1-#]&/@stn];
Table[Select[sps[Range[n]],And[cmp[#]==Sort[#],Count[#,{_}]==0]&]//Length,{n,0,10}]

PARI

seq(n)={my(x=x+O(x*x^(n\2)), p=exp((exp(2*x)-3)/2-x+exp(x)), q=(exp(x)-1)*p); vector(n+1, n, my(c=(n-1)\2); c!*polcoef(if(n%2, p, q), c))} \\ Andrew Howroyd, Feb 16 2022

Formula

From Andrew Howroyd, Feb 16 2022: (Start)

a(2*n) = A086365(n-1) for n > 0.

a(2*n) = n!*[x^n] exp((exp(2*x) - 3)/2 - x + exp(x));

a(2*n+1) = n!*[x^n] (exp(x) - 1)*exp((exp(2*x) - 3)/2 - x + exp(x)).

(End)

Extensions

Terms a(13) and beyond from Andrew Howroyd, Feb 16 2022