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%I #23 Jan 21 2023 02:23:26
%S 1,0,0,0,0,1,0,0,0,1,0,0,1,0,1,0,0,0,0,0,1,0,0,1,1,6,0,1,0,0,0,0,0,10,
%T 0,1,0,0,1,0,7,10,15,0,1,0,0,0,1,0,10,20,21,0,1,0,0,1,0,12,1,30,35,28,
%U 0,1,0,0,0,0,0,20,0,56,56,36,0,1,0,0,1,1,13,10,126,21,98,84,45,0,1
%N Triangular array read by rows: T(n,k) is the number of compositions of n into exactly k parts in which no part is unique (each part occurs at least twice).
%C Row sums = A240085.
%D S. Heubach and T. Mansour, Combinatorics of Compositions and Words, Chapman and Hall, 2009 page 87.
%H Alois P. Heinz, <a href="/A240315/b240315.txt">Rows n = 0..140, flattened</a>
%F Product_{i>=1} exp(x^i*y) - x^i*y = Sum_{k>=0} A_k(x)*y^k/k!, where A_k(x) is the o.g.f. for the number of compositions of n into k parts in which no part is unique. In other words, A_k(x) is the o.g.f. for column k.
%e Triangle begins:
%e 1;
%e 0, 0;
%e 0, 0, 1;
%e 0, 0, 0, 1;
%e 0, 0, 1, 0, 1;
%e 0, 0, 0, 0, 0, 1;
%e 0, 0, 1, 1, 6, 0, 1;
%e 0, 0, 0, 0, 0, 10, 0, 1;
%e 0, 0, 1, 0, 7, 10, 15, 0, 1;
%e 0, 0, 0, 1, 0, 10, 20, 21, 0, 1;
%e 0, 0, 1, 0, 12, 1, 30, 35, 28, 0, 1;
%e 0, 0, 0, 0, 0, 20, 0, 56, 56, 36, 0, 1;
%e 0, 0, 1, 1, 13, 10, 126, 21, 98, 84, 45, 0, 1;
%e ...
%e T(8,4) = 7 because we have: 3+3+1+1, 3+1+3+1, 3+1+1+3, 1+3+3+1, 1+3+1+3, 1+1+3+3, 2+2+2+2.
%p b:= proc(n, i, t) option remember; `if`(n=0, t!, `if`(i<1, 0,
%p expand(b(n, i-1, t)+add(x^j*b(n-i*j, i-1, t+j)/j!, j=2..n/i))))
%p end:
%p T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n$2, 0)):
%p seq(T(n), n=0..14); # _Alois P. Heinz_, Apr 03 2014
%t nn=10;Table[Take[Transpose[Range[0,nn]!CoefficientList[Series[ Product[Exp[x^i y]-x^i y,{i,1,nn}],{y,0,nn}],{y,x}]],nn+1][[j,Range[1,j]]],{j,1,nn}]//Grid
%K nonn,tabl
%O 0,26
%A _Geoffrey Critzer_, Apr 03 2014
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