OFFSET
0,11
COMMENTS
LINKS
Alois P. Heinz, Rows n = 0..200, flattened
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.
EXAMPLE
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;
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 *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Emeric Deutsch, Nov 13 2015
STATUS
approved