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%I #29 May 21 2021 08:10:02
%S 1,1,5,1,10,37,1,21,120,405,1,42,363,1644,5251,1,85,1117,6814,27405,
%T 84893,1,170,3360,27404,138085,514248,1556535,1,341,10164,111045,
%U 701960,3145848,11133493,33175957,1,682,30520,445132,3521405,18956548,78337448
%N T(n,k) = count of degree k monomials in the elementary symmetric polynomials e(mu,k) summed over all partitions mu of n.
%C T(n,2) = A000975(n+1) because only partitions into parts of 1 and 2 contribute, so that T(n,2) = Sum_{k=0..floor(n/2)} 2^(n-2k). - _Peter J. Taylor_, Mar 01 2017
%H Peter J. Taylor, <a href="/A209669/b209669.txt">Table of n, a(n) for n = 1..5050</a>
%H Peter J. Taylor, <a href="/A209669/a209669.py.txt">Python program to compute terms</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Symmetric_polynomials">Symmetric polynomials</a>
%F T(n,k) = Sum_{lambda} Product_{i} binomial(k, lambda_i) where the sum is over partitions of n. - _Peter J. Taylor_, Mar 01 2017
%e Table starts as
%e 1;
%e 1, 5;
%e 1, 10, 37;
%e 1, 21, 120, 405;
%e 1, 42, 363, 1644, 5251;
%e ...
%e For n = 2, k = 2 the partitions of n are 2 and 1+1, which correspond respectively to (xy) contributing 1 and (x+y)*(x+y) contributing 4 for a total of 5. - _Peter J. Taylor_, Mar 01 2017
%t e[n_, v_] := Tr[Times @@@ Select[Subsets[Table[Subscript[x, j], {j, v}]], Length[#] == n &]]; e[par_?PartitionQ, v_] := Times @@ (e[#, v] & /@ par); Table[Tr[(e[#, k] & /@ Partitions[l]) /. Subscript[x, _] -> 1], {l, 10}, {k, l}]
%o (Python) # See Taylor link
%Y Row sums are A209670, main diagonal is A209671.
%Y Second column is A000975 offset by 1. - _Peter J. Taylor_, Mar 01 2017
%K nonn,tabl
%O 1,3
%A _Wouter Meeussen_, Mar 11 2012
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