A263233 Triangle read by rows: T(n,k) is the number of partitions of n having k perfect square parts (0<=k<=n).
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
Offset: 0
Examples
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). Triangle starts: 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;
Links
- Alois P. Heinz, Rows n = 0..200, flattened
Programs
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Maple
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.
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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 *)
Formula
G.f.: Product_{i>=1}(1-x^h(i))/((1-x^i)*(1-t*x^h(i))), where h(i) = i^2.
Comments