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%I #12 May 25 2018 03:07:57
%S 1,0,1,1,0,1,0,2,0,1,2,0,2,0,1,1,3,0,2,0,1,2,2,4,0,2,0,1,2,4,2,4,0,2,
%T 0,1,4,4,5,2,4,0,2,0,1,4,6,5,6,2,4,0,2,0,1,5,9,8,5,6,2,4,0,2,0,1,6,10,
%U 11,9,5,6,2,4,0,2,0,1,9,13,13,12,10,5,6,2,4,0,2,0,1
%N Triangle read by rows: T(n,k) is the number of partitions of n having k triangular number parts (0<=k<=n).
%C The triangular numbers are i(i+1)/2 (i=0,1,2,3,...) (A000217).
%C Sum of entries in row n = A000041(n) = number of partitions of n.
%C T(n,0) = A225044(n).
%C Sum_{k=0..n} k*T(n,k) = A263235(n) = total number of triangular number parts in all partitions of n.
%H Alois P. Heinz, <a href="/A263234/b263234.txt">Rows n = 0..200, flattened</a>
%F G.f.: Product_{i>0} ((1-x^h(i))/((1-x^i)*(1-t*x^h(i))), where h(i) = i*(i+1)/2.
%e T(6,2) = 4 because we have [4,1,1], [3,3], [3,2,1], and [2,2,1,1] (the partitions of 6 that have 2 triangular number parts).
%e Triangle starts:
%e 1;
%e 0,1;
%e 1,0,1;
%e 0,2,0,1;
%e 2,0,2,0,1;
%e 1,3,0,2,0,1;
%p h := proc (i) options operator, arrow: (1/2)*i*(i+1) end proc: g := product((1-x^h(i))/((1-x^i)*(1-t*x^h(i))), i = 1 .. 80): gser := simplify(series(g, x = 0, 30)): for n from 0 to 18 do P[n] := sort(coeff(gser, x, n)) end do: for n from 0 to 18 do seq(coeff(P[n], t, j), j = 0 .. n) end do; # yields sequence in triangular form
%t max = 15; h[i_] = i*(i + 1)/2; P = Product[(1 - x^h[i])/((1 - x^i)*(1 - t*x^h[i])), {i, 1, max}] + O[x]^max;
%t CoefficientList[#, t]& /@ CoefficientList[P, x] // Flatten (* _Jean-François Alcover_, May 25 2018 *)
%Y Cf. A000041, A000217, A225044, A263235.
%K nonn,tabl
%O 0,8
%A _Emeric Deutsch_, Nov 12 2015
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