A135085 a(n) = A000110(2^n).
1, 2, 15, 4140, 10480142147, 128064670049908713818925644, 172134143357358850934369963665272571125557575184049758045339873395
Offset: 0
Keywords
Examples
Let S={1,2,3,...,n} be a set of n elements and let SU be the set of all subsets of S including the empty set. The number of elements of SU is |SU| = 2^n. Now form all possible set partitions from SU including the empty set. This gives a set W and its number of elements is |W| = sum((stirling2(2^n,k)), k=0..2^n) = Bell(2^n).
For S={1,2} we have SU = { {}, {1}, {2}, {1,2} } and W =
{
{{{}}, {1}, {2}, {1, 2}},
{{2}, {1, 2}, {{}, {1}}},
{{1}, {1, 2}, {{}, {2}}},
{{1}, {2}, {{}, {1, 2}}},
{{{}}, {1, 2}, {{1}, {2}}},
{{{1}, {2}}, {{}, {1, 2}}},
{{1, 2}, {{}, {1}, {2}}},
{{{}}, {2}, {{1}, {1, 2}}},
{{{1}, {1, 2}}, {{}, {2}}},
{{2}, {{}, {1}, {1, 2}}},
{{{}}, {1}, {{2}, {1, 2}}},
{{{2}, {1, 2}}, {{}, {1}}},
{{1}, {{}, {2}, {1, 2}}},
{{{}}, {{1}, {2}, {1, 2}}},
{{{}, {1}, {2}, {1, 2}}}
}
and |W| = 15.
Programs
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Maple
ZahlDerMengenAusMengeDerZerlegungenEinerMenge:=proc() local n,nend,arg,k,w; nend:=5; for n from 0 to nend do arg:=2^n; w[n]:=sum((stirling2(arg,k)), k=0..arg); od; print(w[0],w[1],w[2],w[3],w[4],w[5],w[6],w[7],w[8],w[9],w[10]); end proc;
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Mathematica
Table[BellB[2^n],{n,0,10}] (* Geoffrey Critzer, Jan 03 2014 *) -
Python
from sympy import bell def A135085(n): return bell(2**n) # Chai Wah Wu, Jun 22 2022
Formula
a(n) = |W| = Sum_{k=0..2^n} Stirling2(2^n,k) = Bell(2^n), where Stirling2(n) is the Stirling number of the second kind and Bell(n) is the Bell number.
a(n) = exp(-1) * Sum_{k>=0} k^(2^n)/k!. - Ilya Gutkovskiy, Jun 13 2019
Comments