A107893 Triangle read by rows, related to A055129 (repunits in base k).

1, 2, 1, 3, 4, 2, 4, 11, 14, 6, 5, 26, 64, 66, 24, 6, 57, 244, 456, 384, 120, 7, 120, 846, 2556, 3744, 2640, 720, 8, 247, 2778, 12762, 28944, 34560, 20880, 5040, 9, 502, 8828, 59382, 195768, 352080, 353520, 186480, 40320, 10, 1013, 27488, 264012, 1216368, 3091320, 4587120, 3966480, 1854720, 362880

Offset

1,2

Comments

Second column of A107893 = Eulerian numbers (A000295) starting with 1: 1, 4, 11, 26, 57, ... Rightmost term in row n = (n-1)!.

Using the Jun 18 2009 formula of Johannes W. Meijer in A028246: Instead of a(n,1)=1 set a(n,1)=n. The result is A107893. - Werner Schulte, Dec 12 2016

Links

Michael De Vlieger, Table of n, a(n) for n = 1..5050 (1 <= n <= 100)

Formula

n-th row = inverse binomial transform of n-th column of A055129, where the latter are generated from f(x) = x^(n-1) + x^(n-2) + ...+ x + 1; (x = 1, 2, 3, ...)

A(n,k) = Sum_{i=1..n} A028246(i,k) for 1 <= k <= n. - Werner Schulte, Dec 08 2016

The polynomials p(n,t) = Sum_{k=1..n} A(n,k)*t^k are given by p(1,t) = t and p(n+1,t) = t + t*(t+1)*(d/dt)p(n,t) for n >= 1. - Werner Schulte, Dec 12 2016

Example

Binomial transform of Row 4 in the form: (4, 11, 14, 6, 0, 0, 0, ...) = Row 4 of A055129: 4, 15, 40, 85, ... which is generated from f(x) = x^3 + x^2 + x + 1; (x = 1,2,3, ...).
Triangle starts:
  1;
  2,   1;
  3,   4,   2;
  4,  11,  14,   6;
  5,  26,  64,  66,  24;
  6,  57, 244, 456, 384, 120;
  ...

Mathematica

Table[Sum[Sum[(-1)^(k - j) Binomial[k, j] j^i, {j, 0, k}]/k, {i, n}], {n, 10}, {k, n}] // Flatten (* Michael De Vlieger, Dec 11 2016, after Jean-François Alcover at A028246 *)

Crossrefs

Cf. A000295, A055129, A028246.

Sequence in context: A086614 A108959 A208750 * A131987 A337226 A120874

Adjacent sequences: A107890 A107891 A107892 * A107894 A107895 A107896

Keyword

nonn,tabl

Author

Gary W. Adamson, May 26 2005

Status

approved