OFFSET
1,5
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 50 rows)
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
EXAMPLE
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;
...
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}]
PROG
(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
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Gary W. Adamson, Aug 30 2007
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
STATUS
approved