cp's OEIS Frontend

A263233 Triangle read by rows: T(n,k) is the number of partitions of n having k perfect square parts (0<=k<=n).

%I A263233 #21 Nov 16 2024 16:11:00
%S A263233 1,0,1,1,0,1,1,1,0,1,1,2,1,0,1,2,1,2,1,0,1,3,3,1,2,1,0,1,3,4,3,1,2,1,
%T A263233 0,1,5,4,5,3,1,2,1,0,1,5,8,4,5,3,1,2,1,0,1,8,8,9,4,5,3,1,2,1,0,1,9,12,
%U A263233 9,9,4,5,3,1,2,1,0,1,13,15,13,10,9,4,5,3,1,2,1,0,1
%N A263233 Triangle read by rows: T(n,k) is the number of partitions of n having k perfect square parts (0<=k<=n).
%C A263233 Sum of entries in row n = A000041(n) = number of partitions of n.
%C A263233 T(n,0) = A087153(n).
%C A263233 Sum_{k=0..n}k*T(n,k) = A073336(n) = total number of square parts in all partitions of n.
%H A263233 Alois P. Heinz, <a href="/A263233/b263233.txt">Rows n = 0..200, flattened</a>
%F A263233 G.f.: Product_{i>=1}(1-x^h(i))/((1-x^i)*(1-t*x^h(i))), where h(i) = i^2.
%e A263233 T(8,2) = 5 because we have [6,1,1], [4,4], [4,3,1], [3,3,1,1], [2,2,2,1,1] (the partitions of 8 that have 2 perfect square parts).
%e A263233 Triangle starts:
%e A263233   1;
%e A263233   0, 1;
%e A263233   1, 0, 1;
%e A263233   1, 1, 0, 1;
%e A263233   1, 2, 1, 0, 1;
%e A263233   2, 1, 2, 1, 0, 1;
%e A263233   3, 3, 1, 2, 1, 0, 1;
%e A263233   3, 4, 3, 1, 2, 1, 0, 1;
%e A263233   5, 4, 5, 3, 1, 2, 1, 0, 1;
%p A263233 h:= proc(i) options operator, arrow: i^2 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 A263233 Needs["Combinatorica`"]; Table[Count[Replace[#, n_ /; ! IntegerQ@ Sqrt@ n -> Nothing, {1}] & /@ Combinatorica`Partitions@ n, w_ /; Length@ w == k], {n, 0, 12}, {k, 0, n}] // Flatten (* _Michael De Vlieger_, Dec 19 2015 *)
%Y A263233 Cf. A000041, A073336, A087153.
%K A263233 nonn,tabl
%O A263233 0,12
%A A263233 _Emeric Deutsch_, Nov 12 2015