%I
%S 1,1,2,2,1,4,1,5,2,8,2,1,10,4,1,15,5,2,19,8,2,1,27,10,4,1,34,15,5,2,
%T 47,19,8,2,1,59,27,10,4,1,79,34,15,5,2,99,47,19,8,2,1,130,59,27,10,4,
%U 1,162,79,34,15,5,2,209,99,47,19,8,2,1,259,130,59,27,10,4,1
%N Triangle read by rows: T(n,k) is the number of partitions of n having exactly k parts equal to 3 (n >= 0, 0 <= k <= floor(n/3)).
%C Row n has 1+floor(n/3) terms. Row sums are the partition numbers (A000041). T(n,0) = A027337(n). Sum_{k=0..floor(n/3)} k*T(n,k) = A024787(n).
%H Alois P. Heinz, <a href="/A263232/b263232.txt">Rows n = 0..300, flattened</a>
%F G.f.: (1x)*(1x^2)*(1tx^3)*Product_{j>=4} (1x^j).
%e T(7,1) = 4 because we have [4,3], [3,2,2], [3,2,1,1], and [3,1,1,1,1].
%e T(9,2) = 2 because we have [3,3,2,1] and [3,3,1,1,1].
%e Triangle starts:
%e 1;
%e 1;
%e 2;
%e 2, 1;
%e 4, 1;
%e 5, 2;
%e 8, 2, 1.
%p g := 1/((1x)*(1x^2)*(1t*x^3)*(product(1x^j, j = 4 .. 80))): gser := simplify(series(g, x = 0, 30)): for n from 0 to 25 do P[n] := sort(coeff(gser, x, n)) end do: for n from 0 to 25 do seq(coeff(P[n], t, j), j = 0 .. floor((1/3)*n)) end do; # yields sequence in triangular form
%p # second Maple program:
%p b:= proc(n, i) option remember; expand(
%p `if`(n=0, 1, `if`(i<1, 0, `if`(i=3, x, 1)*
%p `if`(i>n, 0, b(ni, i)) +b(n, i1))))
%p end:
%p T:= n> (p> seq(coeff(p, x, i), i=0..degree(p)))(b(n$2)):
%p seq(T(n), n=0..25); # _Alois P. Heinz_, Nov 01 2015
%t b[n_, i_] := b[n, i] = Expand[If[n == 0, 1, If[i < 1, 0, If[i == 3, x, 1]* If[i > n, 0, b[n  i, i]] + b[n, i  1]]]]; T[n_] := Function[p, Table[ Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n, n]]; Table[T[n], {n, 0, 25}] // Flatten (* _JeanFrançois Alcover_, Jan 21 2016, after _Alois P. Heinz_ *)
%Y Cf. A000041, A027337, A024787, A116599.
%K nonn,tabf
%O 0,3
%A _Emeric Deutsch_, Nov 01 2015
