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A263233
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Triangle read by rows: T(n,k) is the number of partitions of n having k perfect square parts (0<=k<=n).
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1
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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
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0,12
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COMMENTS
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Sum of entries in row n = A000041(n) = number of partitions of n.
Sum_{k=0..n}k*T(n,k) = A073336(n) = total number of square parts in all partitions of n.
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LINKS
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FORMULA
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G.f.: Product_{i>=1}(1-x^h(i))/((1-x^i)*(1-t*x^h(i))), where h(i) = i^2.
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EXAMPLE
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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;
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MAPLE
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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
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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 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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