A178319 -id:A178319 - cp's OEIS frontend

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

Alois P. Heinz, Jul 03 2021

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

			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, ...
  ...

Alois P. Heinz, Antidiagonals n = 0..43, flattened
Richard Stanley, Proof of the general conjecture, MathOverflow, March 2021.

Columns k=0-3 give: A000007, A000012, A178315, A178319.
Rows n=0-2 give: A000012, A057427, A002061 (for k>0).
Main diagonal gives A342578.
Cf. A000217, A023813.

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);

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).

A342578 a(n) = n! * [x^n] (Sum_{j>=0} n^(j(j+1)/2) x^j/j!)^(1/n) for n > 0, a(0) = 1.

Original entry on oeis.org

1, 1, 3, 199, 249337, 6062674201, 3653786369479951, 65709007885111803731947, 40564683796482484146182142025377, 969773549559254966290998252899999751714721, 999999990999996719397362087568018696141879478712251051, 49037072510879011742983689973641327840345400616866967292640434759551

Offset: 0

Alois P. Heinz, Mar 15 2021

nonn

All terms are odd.

Alois P. Heinz, Table of n, a(n) for n = 0..35
Richard Stanley, Proof of the general conjecture, MathOverflow, March 2021.

Cf. A000217, A023813, A178315, A178319.

Main diagonal of A346061.

a:= n-> `if`(n>0, coeff(series(add(n^binomial(j+1, 2)*
         x^j/j!, j=0..n)^(1/n), x, n+1), x, n)*n!, 1):
seq(a(n), n=0..12);

a(n) == 1 (mod n*(n-1)) for n >= 2 (see "general conjecture" in A178319 and link to proof by Richard Stanley above).

a(n) ~ n^((n^2 + n - 2)/2). - Vaclav Kotesovec, Jul 15 2021

cp's OEIS Frontend

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.

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A342578 a(n) = n! * [x^n] (Sum_{j>=0} n^(j(j+1)/2) x^j/j!)^(1/n) for n > 0, a(0) = 1.

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cp's OEIS Frontend

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Formula

A342578 a(n) = n! * [x^n] (Sum_{j>=0} n^(j*(j+1)/2) * x^j/j!)^(1/n) for n > 0, a(0) = 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Formula

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.

A342578 a(n) = n! * [x^n] (Sum_{j>=0} n^(j(j+1)/2) x^j/j!)^(1/n) for n > 0, a(0) = 1.