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A(n,k) = n! * [x^n] (Sum_{j=0..n} k^(j*(j+1)/2) * x^j/j!)^(1/k) if k>0, A(n,0) = 0^n; square array A(n,k), n>=0, k>=0, read by antidiagonals.
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%I #25 Jul 05 2021 12:02:50

%S 1,1,0,1,1,0,1,1,1,0,1,1,3,1,0,1,1,7,23,1,0,1,1,13,199,393,1,0,1,1,21,

%T 901,17713,13729,1,0,1,1,31,2861,249337,4572529,943227,1,0,1,1,43,

%U 7291,1900521,264273961,3426693463,126433847,1,0

%N A(n,k) = n! * [x^n] (Sum_{j=0..n} k^(j*(j+1)/2) * x^j/j!)^(1/k) if k>0, A(n,0) = 0^n; square array A(n,k), n>=0, k>=0, read by antidiagonals.

%C A(n,k) is odd if k >= 1 or n = 0.

%H Alois P. Heinz, <a href="/A346061/b346061.txt">Antidiagonals n = 0..43, flattened</a>

%H Richard Stanley, <a href="https://mathoverflow.net/q/385402">Proof of the general conjecture</a>, MathOverflow, March 2021.

%F E.g.f. of column k>0: (Sum_{j>=0} k^(j*(j+1)/2) * x^j/j!)^(1/k).

%F E.g.f. of column k=0: 1.

%F A(n,k) == 1 (mod k*(k-1)) for k >= 2 (see "general conjecture" in A178319 and link to proof by _Richard Stanley_ above).

%e Square array A(n,k) begins:

%e 1, 1, 1, 1, 1, 1, ...

%e 0, 1, 1, 1, 1, 1, ...

%e 0, 1, 3, 7, 13, 21, ...

%e 0, 1, 23, 199, 901, 2861, ...

%e 0, 1, 393, 17713, 249337, 1900521, ...

%e 0, 1, 13729, 4572529, 264273961, 6062674201, ...

%e ...

%p A:= (n, k)-> `if`(k>0, n!*coeff(series(add(k^(j*(j+1)/2)*

%p x^j/j!, j=0..n)^(1/k), x, n+1), x, n), k^n):

%p seq(seq(A(n, d-n), n=0..d), d=0..10);

%Y Columns k=0-3 give: A000007, A000012, A178315, A178319.

%Y Rows n=0-2 give: A000012, A057427, A002061 (for k>0).

%Y Main diagonal gives A342578.

%Y Cf. A000217, A023813.

%K nonn,tabl

%O 0,13

%A _Alois P. Heinz_, Jul 03 2021