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A107893
Triangle read by rows, related to A055129 (repunits in base k).
1
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
KEYWORD
nonn,tabl
AUTHOR
Gary W. Adamson, May 26 2005
STATUS
approved