A264391 - OEIS

%I #16 Nov 14 2020 06:58:48

%S 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,

%T 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,

%U 12,9,6,5,4,2,2,1,1,0,1,19,15,12,9,6,5,4,2,2,1,1,0,1

%N Triangle read by rows: T(n,k) is the number of partitions of n having k perfect cube parts (0<=k<=n).

%C Sum of entries in row n = A000041(n) = number of partitions of n.

%C T(n,0) = A264393(n).

%C Sum_{k=0..n}k*T(n,k) = A264392(n) = total number of perfect cube parts in all partitions of n.

%H Alois P. Heinz, <a href="/A264391/b264391.txt">Rows n = 0..200, flattened</a>

%F 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.

%e 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).

%e Triangle starts:

%e   1;

%e   0, 1;

%e   1, 0, 1;

%e   1, 1, 0, 1;

%e   2, 1, 1, 0, 1;

%e   2, 2, 1, 1, 0, 1;

%p 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

%p # second Maple program:

%p q:= proc(n) option remember; `if`(n=iroot(n, 3)^3, 1, 0) end:

%p b:= proc(n, i) option remember; expand(`if`(n=0, 1,

%p       `if`(i<1, 0, b(n, i-1)+x^q(i)*b(n-i, min(i, n-i)))))

%p     end:

%p T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n$2)):

%p seq(T(n), n=0..20);  # _Alois P. Heinz_, Nov 14 2020

%t cnt[P_List] := Count[P, p_ /; IntegerQ[p^(1/3)]];

%t cnts[n_] := cnts[n] = cnt /@ IntegerPartitions[n];

%t T[n_, k_] := Count[cnts[n], k];

%t Table[T[n, k], {n, 0, 18}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Nov 14 2020 *)

%Y Cf. A000041, A264392, A264393.

%K nonn,tabl

%O 0,11

%A _Emeric Deutsch_, Nov 13 2015