A372118 Square array A(n, k) = ((k+2)^(n+2) - 2 * (k+1)^(n+2) + k^(n+2))/2 for k, n >= 0 read by ascending antidiagonals.

1, 3, 1, 7, 6, 1, 15, 25, 9, 1, 31, 90, 55, 12, 1, 63, 301, 285, 97, 15, 1, 127, 966, 1351, 660, 151, 18, 1, 255, 3025, 6069, 4081, 1275, 217, 21, 1, 511, 9330, 26335, 23772, 9751, 2190, 295, 24, 1, 1023, 28501, 111645, 133057, 70035, 19981, 3465, 385, 27, 1

Offset: 0

Author

Werner Schulte, Apr 19 2024

Keywords

• nonn
• easy
• tabl

Comments

Depending on some fixed integer m >= 0 we define a family of square arrays A(m; n, k) = (Sum_{i=0..m} (-1)^i * binomial(m, i) * (k + m - i)^(n+m)) / m! for k, n >= 0. Special cases are: A004248 (m=0), A343237 (m=1) and this array (m=2). The A(m; n, k) satisfy: A(m; n, k) = (k+m) * A(m; n-1, k) + A(m-1; n, k) with initial values A(0; n, k) = k^n and A(m; 0, k) = 1.

Further properties are conjectures:

(1) O.g.f. of column k is Prod_{i=k..k+m} 1 / (1 - i * t);

(2) E.g.f. of row n is exp(x) * (Sum_{k=0..n} binomial(k+m, m) * A048993(n+m, k+m) * x^k);

(3) The LU decompositions of these arrays are given by the upper triangular matrix U which is the transpose of A007318 and the lower triangular matrix L, where L is defined L(m; n, k) = A048993(n+m, k+m) * (k+m)! / m!, i.e., A(m; n, k) = Sum_{i=0..k} L(m; n, i) * binomial(k, i).

The three conjectures are true, see links. - Sela Fried, Jul 07 2024

Examples

			Square array A(n, k) starts:
n\k :    0     1       2       3        4         5         6         7
=======================================================================
  0 :    1     1       1       1        1         1         1         1
  1 :    3     6       9      12       15        18        21        24
  2 :    7    25      55      97      151       217       295       385
  3 :   15    90     285     660     1275      2190      3465      5160
  4 :   31   301    1351    4081     9751     19981     36751     62401
  5 :   63   966    6069   23772    70035    170898    365001    706104
  6 :  127  3025   26335  133057   481951   1398097   3463615   7628545
  7 :  255  9330  111645  724260  3216795  11075670  31794105  79669320
  etc.
		

Links

• Paolo Xausa, Table of n, a(n) for n = 0..11324 (first 150 antidiagonals, flattened).
• Sela Fried, On an integer sequence related to Euler's formula for the Stirling numbers of the second kind, 2024.

Crossrefs

Rows: A000012 (n=0), A008585 (n=1), A227776 (n=2).

Columns: A000225 (k=0), A000392 (k=1), A016269 (k=2), A016753 (k=3), A016103 (k=4), A019757 (k=5), A020570 (k=6), A020782 (k=7).

Main diagonal: A281596(n+2).

Cf. A007318, A048993, A004248, A343237.

Programs

Mathematica

A372118[n_, k_] := ((k+2)^(n+2) - 2*(k+1)^(n+2) + k^(n+2))/2;
Table[A372118[n-k, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Jul 10 2024 *)

PARI

A(n, k) = ((k+2)^(n+2) - 2 * (k+1)^(n+2) + k^(n+2))/2

Formula

A(n, k) = (k+2) * A(n-1, k) + (k+1)^(n+1) - k^(n+1) for n > 0.

Conjectures:

(1) O.g.f. of column k is Prod_{i=k..k+2} 1 / (1 - i * t);

(2) E.g.f. of row n is exp(x) * (Sum_{k=0..n} binomial(k+2, 2) * A048993(n+2, k+2) * x^k);

(3) The LU decomposition of this array is given by the upper triangular matrix U which is the transpose of A007318 and the lower triangular matrix L, where L is defined L(n, k) = A048993(n+2, k+2) * (k+2)! / 2!, i.e., A(n, k) = Sum_{i=0..k} L(n, i) * binomial(k, i).

The three conjectures are true. See comments. - Sela Fried, Jul 09 2024