cp's OEIS Frontend

A355395 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = Sum_{j=0..n} k^(j*(n-j)) * binomial(n,j).

%I A355395 #78 Aug 24 2023 07:48:58
%S A355395 1,1,2,1,2,2,1,2,4,2,1,2,6,8,2,1,2,8,26,16,2,1,2,10,56,162,32,2,1,2,
%T A355395 12,98,704,1442,64,2,1,2,14,152,2050,15392,18306,128,2,1,2,16,218,
%U A355395 4752,84482,593408,330626,256,2,1,2,18,296,9506,318752,7221250,39691136,8488962,512,2
%N A355395 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = Sum_{j=0..n} k^(j*(n-j)) * binomial(n,j).
%C A355395 The Stanley reference below describes a family of binomial posets whose elements are two colored graphs with vertices labeled on [n] and with edges labeled on [k-1].  T(n,k) is the number of elements in an n-interval of such a binomial poset. - _Geoffrey Critzer_, Aug 21 2023
%D A355395 R. P. Stanley, Enumerative Combinatorics, Volume 1, Second Edition, Example 3.18.3(e), page 323.
%F A355395 E.g.f. of column k: Sum_{j>=0} exp(k^j * x) * x^j/j!.
%F A355395 G.f. of column k: Sum_{j>=0} x^j/(1 - k^j * x)^(j+1).
%F A355395 For k>=1, E(x)^2 = Sum_{n>=0} T(n,k)*x^n/B_k(n) where B_k(n) = n!*k^binomial(n,2) and E(x) = Sum_{n>=0} x^n/b_k(n). - _Geoffrey Critzer_, Aug 21 2023
%e A355395 Square array begins:
%e A355395   1,  1,    1,     1,     1,      1, ...
%e A355395   2,  2,    2,     2,     2,      2, ...
%e A355395   2,  4,    6,     8,    10,     12, ...
%e A355395   2,  8,   26,    56,    98,    152, ...
%e A355395   2, 16,  162,   704,  2050,   4752, ...
%e A355395   2, 32, 1442, 15392, 84482, 318752, ...
%o A355395 (PARI) T(n, k) = sum(j=0, n, k^(j*(n-j))*binomial(n, j));
%Y A355395 Columns k=0..4 give A040000, A000079, A047863, A135079, A355440.
%Y A355395 Main diagonal gives A320287.
%Y A355395 Cf. A009999.
%K A355395 nonn,tabl
%O A355395 0,3
%A A355395 _Seiichi Manyama_, Jul 02 2022