A346061 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.
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, 901, 17713, 13729, 1, 0, 1, 1, 31, 2861, 249337, 4572529, 943227, 1, 0, 1, 1, 43, 7291, 1900521, 264273961, 3426693463, 126433847, 1, 0
Offset: 0
Examples
Square array A(n,k) begins: 1, 1, 1, 1, 1, 1, ... 0, 1, 1, 1, 1, 1, ... 0, 1, 3, 7, 13, 21, ... 0, 1, 23, 199, 901, 2861, ... 0, 1, 393, 17713, 249337, 1900521, ... 0, 1, 13729, 4572529, 264273961, 6062674201, ... ...
Links
- Alois P. Heinz, Antidiagonals n = 0..43, flattened
- Richard Stanley, Proof of the general conjecture, MathOverflow, March 2021.
Crossrefs
Programs
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Maple
A:= (n, k)-> `if`(k>0, n!*coeff(series(add(k^(j*(j+1)/2)* x^j/j!, j=0..n)^(1/k), x, n+1), x, n), k^n): seq(seq(A(n, d-n), n=0..d), d=0..10);
Formula
E.g.f. of column k>0: (Sum_{j>=0} k^(j*(j+1)/2) * x^j/j!)^(1/k).
E.g.f. of column k=0: 1.
A(n,k) == 1 (mod k*(k-1)) for k >= 2 (see "general conjecture" in A178319 and link to proof by Richard Stanley above).
Comments