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%I #16 Jan 29 2016 10:50:29
%S 1,0,1,1,0,1,1,0,0,2,2,0,1,0,2,1,1,1,1,0,3,4,0,1,1,1,0,4,2,2,1,2,1,2,
%T 0,5,6,0,2,1,3,2,2,0,6,5,2,1,4,1,4,2,3,0,8,9,1,3,2,5,2,4,3,3,0,10,7,3,
%U 3,6,2,7,2,6,3,5,0,12,16,0,4,4,7,3,8,3,7,5,5,0,15,11,6,4,8,5,9,3,12,3,10,5,7,0,18
%N Triangle read by rows: T(n,k) is the number of partitions of n in which the sum of the parts of multiplicity 1 is equal to k (0<=k<=n).
%C Row n contains n+1 entries (n>=0).
%C Row sums yield the partition numbers (A000041).
%C T(n,0) = A007690(n).
%C T(n,n) = A000009(n).
%C Sum_{k>=0} k*T(n,k) = A103628(n).
%H Alois P. Heinz, <a href="/A264403/b264403.txt">Rows n = 0..200, flattened</a>
%F G.f.: G(t,x) = Product_{j>=1} (1+t^j*x^j + x^{2*j}/(1 - x^j)).
%e T(7,5) = 2 because we have [3,2,1,1] and [5,1,1].
%e Triangle starts:
%e 1;
%e 0,1;
%e 1,0,1;
%e 1,0,0,2;
%e 2,0,1,0,2;
%p g := product(1+t^j*x^j+x^(2*j)/(1-x^j), j = 1 .. 100): gser := simplify(series(g, x = 0, 25)): for n from 0 to 20 do P[n] := sort(coeff(gser, x, n)) end do: for n from 0 to 20 do seq(coeff(P[n], t, k), k = 0 .. n) end do; # yields sequence in triangular form
%p # second Maple program:
%p b:= proc(n, i) option remember; `if`(n=0, 1,
%p `if`(i<1, 0, add(expand(b(n-i*j, i-1)*
%p `if`(j=1, x^i, 1)), j=0..n/i)))
%p end:
%p T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n$2)):
%p seq(T(n), n=0..15); # _Alois P. Heinz_, Nov 27 2015
%t b[n_, i_] := b[n, i] = If[n==0, 1, If[i<1, 0, Sum[Expand[b[n-i*j, i-1]*If[j == 1, x^i, 1]], {j, 0, n/i}]]]; T[n_] := Function[p,Table[Coefficient[p, x, i], {i, 0, n}]][b[n, n]]; Table[T[n], {n, 0, 15}] // Flatten (* _Jean-François Alcover_, Jan 29 2016, after _Alois P. Heinz_ *)
%Y Cf. A000009, A000041, A007690, A103628.
%K nonn,look,tabl
%O 0,10
%A _Emeric Deutsch_, Nov 27 2015
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