cp's OEIS Frontend

A264391 Triangle read by rows: T(n,k) is the number of partitions of n having k perfect cube parts (0<=k<=n).

%I A264391 #19 Nov 16 2024 18:49:28
%S A264391 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 A264391 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 A264391 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 A264391 Triangle read by rows: T(n,k) is the number of partitions of n having k perfect cube parts (0<=k<=n).
%C A264391 Sum of entries in row n = A000041(n) = number of partitions of n.
%C A264391 T(n,0) = A264393(n).
%C A264391 Sum_{k=0..n}k*T(n,k) = A264392(n) = total number of perfect cube parts in all partitions of n.
%H A264391 Alois P. Heinz, <a href="/A264391/b264391.txt">Rows n = 0..200, flattened</a>
%F A264391 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 A264391 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 A264391 Triangle starts:
%e A264391   1;
%e A264391   0, 1;
%e A264391   1, 0, 1;
%e A264391   1, 1, 0, 1;
%e A264391   2, 1, 1, 0, 1;
%e A264391   2, 2, 1, 1, 0, 1;
%p A264391 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 A264391 # second Maple program:
%p A264391 q:= proc(n) option remember; `if`(n=iroot(n, 3)^3, 1, 0) end:
%p A264391 b:= proc(n, i) option remember; expand(`if`(n=0, 1,
%p A264391       `if`(i<1, 0, b(n, i-1)+x^q(i)*b(n-i, min(i, n-i)))))
%p A264391     end:
%p A264391 T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n$2)):
%p A264391 seq(T(n), n=0..20);  # _Alois P. Heinz_, Nov 14 2020
%t A264391 cnt[P_List] := Count[P, p_ /; IntegerQ[p^(1/3)]];
%t A264391 cnts[n_] := cnts[n] = cnt /@ IntegerPartitions[n];
%t A264391 T[n_, k_] := Count[cnts[n], k];
%t A264391 Table[T[n, k], {n, 0, 18}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Nov 14 2020 *)
%Y A264391 Cf. A000041, A264392, A264393.
%K A264391 nonn,tabl
%O A264391 0,11
%A A264391 _Emeric Deutsch_, Nov 13 2015