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A264391
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Triangle read by rows: T(n,k) is the number of partitions of n having k perfect cube parts (0<=k<=n).
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3
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1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 2, 2, 1, 1, 0, 1, 4, 2, 2, 1, 1, 0, 1, 4, 4, 2, 2, 1, 1, 0, 1, 6, 5, 4, 2, 2, 1, 1, 0, 1, 8, 6, 5, 4, 2, 2, 1, 1, 0, 1, 11, 9, 6, 5, 4, 2, 2, 1, 1, 0, 1, 13, 12, 9, 6, 5, 4, 2, 2, 1, 1, 0, 1, 19, 15, 12, 9, 6, 5, 4, 2, 2, 1, 1, 0, 1
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OFFSET
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0,11
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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) = A264392(n) = total number of perfect cube parts in all partitions of n.
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LINKS
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FORMULA
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G.f.: G(t,x) = Product_i>=1}(1-x^h(i))/((1-x^i)*(1-t*x^h(i))), where h(i) = i^3.
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EXAMPLE
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T(7,1) = 4 because we have [6,1],[4,2,1],[3,3,1], and [2,2,2,1] (the partitions of 7 that have 1 perfect cube part).
Triangle starts:
1;
0, 1;
1, 0, 1;
1, 1, 0, 1;
2, 1, 1, 0, 1;
2, 2, 1, 1, 0, 1;
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MAPLE
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h := proc (i) options operator, arrow: i^3 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
# second Maple program:
q:= proc(n) option remember; `if`(n=iroot(n, 3)^3, 1, 0) end:
b:= proc(n, i) option remember; expand(`if`(n=0, 1,
`if`(i<1, 0, b(n, i-1)+x^q(i)*b(n-i, min(i, n-i)))))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n$2)):
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MATHEMATICA
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cnt[P_List] := Count[P, p_ /; IntegerQ[p^(1/3)]];
cnts[n_] := cnts[n] = cnt /@ IntegerPartitions[n];
T[n_, k_] := Count[cnts[n], k];
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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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