

A263233


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


1



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, 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, 9, 9, 4, 5, 3, 1, 2, 1, 0, 1, 13, 15, 13, 10, 9, 4, 5, 3, 1, 2, 1, 0, 1
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OFFSET

0,12


COMMENTS

Sum of entries in row n = A000041(n) = number of partitions of n.
T(n,0) = A087153(n).
Sum_{k=0..n}k*T(n,k) = A073336(n) = total number of square parts in all partitions of n.


LINKS

Alois P. Heinz, Rows n = 0..200, flattened


FORMULA

G.f.: Product_{i>=1}(1x^h(i))/((1x^i)*(1t*x^h(i))), where h(i) = i^2.


EXAMPLE

T(8,2) = 6 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.
Triangle starts:
1;
0, 1;
1, 0, 1;
1, 1, 0, 1;
1, 2, 1, 0, 1;
2, 1, 2, 1, 0, 1;


MAPLE

h:= proc(i) options operator, arrow: i^2 end proc: g := product((1x^h(i))/((1x^i)*(1t*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.


MATHEMATICA

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 *)


CROSSREFS

Cf. A000041, A073336, A087153.
Sequence in context: A241062 A284620 A038698 * A300623 A087991 A293439
Adjacent sequences: A263230 A263231 A263232 * A263234 A263235 A263236


KEYWORD

nonn,tabl


AUTHOR

Emeric Deutsch, Nov 12 2015


STATUS

approved



