A132789 Triangle read by rows: T(n,k) = A007318(n-1, k-1) + A001263(n, k) - 1.
1, 1, 1, 1, 4, 1, 1, 8, 8, 1, 1, 13, 25, 13, 1, 1, 19, 59, 59, 19, 1, 1, 26, 119, 194, 119, 26, 1, 1, 34, 216, 524, 524, 216, 34, 1, 1, 43, 363, 1231, 1833, 1231, 363, 43, 1, 1, 53, 575, 2603, 5417, 5417, 2603, 575, 53, 1, 1, 64, 869, 5069, 14069, 19655, 14069, 5069, 869
Offset: 1
Examples
First few rows of the triangle are: 1; 1, 1; 1, 4, 1; 1, 8, 8, 1; 1, 13, 25, 13, 1; 1, 19, 59, 59, 19, 1; 1, 26, 119, 194, 119, 26, 1; 1, 34, 216, 524, 524, 216, 34, 1; 1, 43, 363, 1231, 1833, 1231, 363, 43, 1; 1, 53, 575, 2603, 5417, 5417, 2603, 575, 53, 1; 1, 64, 869, 5069, 14069, 19655, 14069, 5069, 869, 64, 1; ...
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 50 rows)
Programs
-
Mathematica
<< DiscreteMath`Combinatorica` t[n_, m_, 0] := Binomial[n, m]; t[n_, m_, 1] := Binomial[n, m]*Binomial[n + 1, m]/(m + 1); t[n_, m_, 2] := Eulerian[1 + n, m]; t[n_, m_, q_] := t[n, m, q] = t[n, m, q - 2] + t[n, m, q - 3] - 1; Table[Flatten[Table[Table[t[n, m, q], {m, 0, n}], {n, 0, 10}]], {q, 0, 10}]
-
PARI
T(n,k)={if(k<=n, binomial(n-1, k-1)*(1 + binomial(n, k-1)/k) - 1, 0)} for(n=1, 10, for(k=1, n, print1(T(n, k), ", ")); print); \\ Andrew Howroyd, Sep 08 2018
Formula
A symmetrical triangle recursion: let q=4; t(n,m,0)=Binomial[n,m]; t(n,m,1)=Narayana(n,m); t(n,m,2)=Eulerian(n+1,m); t(n,m,q)=t(n,m,g-2)+t(n,m,q-3).
T(n,k) = binomial(n-1, k-1)*(1 + binomial(n, k-1)/k) - 1. - Andrew Howroyd, Sep 08 2018
Extensions
More terms, Mma program and additional comments from Roger L. Bagula, Apr 20 2010
Edited by N. J. A. Sloane, Apr 21 2010 at the suggestion of R. J. Mathar
Name clarified by Andrew Howroyd, Sep 08 2018