A264391 Triangle read by rows: T(n,k) is the number of partitions of n having k perfect cube parts (0<=k<=n).
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
Offset: 0
Examples
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;
Links
- Alois P. Heinz, Rows n = 0..200, flattened
Programs
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Maple
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)): seq(T(n), n=0..20); # Alois P. Heinz, Nov 14 2020 -
Mathematica
cnt[P_List] := Count[P, p_ /; IntegerQ[p^(1/3)]]; cnts[n_] := cnts[n] = cnt /@ IntegerPartitions[n]; T[n_, k_] := Count[cnts[n], k]; Table[T[n, k], {n, 0, 18}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 14 2020 *)
Formula
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.
Comments