OFFSET
1,3
COMMENTS
The Z-transform of the triangle gives the Narayan triangle (A001263).
Every other polynomial, of p(n, x), has a factor of (1+2*x), just like the higher Sierpinski-Pascal Worpitzky form polynomials.
LINKS
G. C. Greubel, Rows n = 1..50 of the triangle, flattened
FORMULA
T(n, k) = [x^k]( p(n,x) ), where p(n, x) = Sum_{k=1..n} A001263(n,k)*binomial(x+k -1, n-1).
From G. C. Greubel, Mar 28 2022: (Start)
Sum_{k=0..n-1} T(n, k) = A001710(n+1).
T(n, 0) = n!. (End)
EXAMPLE
Triangle begins as:
1;
1, 2;
2, 5, 5;
6, 19, 21, 14;
24, 84, 126, 84, 42;
120, 468, 750, 720, 330, 132;
720, 2988, 5496, 5445, 3795, 1287, 429;
5040, 22356, 43120, 50435, 35035, 19019, 5005, 1430;
40320, 186912, 391688, 472472, 398398, 208208, 92092, 19448, 4862;
MATHEMATICA
p[n_, x_]:= p[n, x]= ((-1)^(n+1)/(n+1))*Sum[Binomial[n-1, k-1]*Binomial[n+1, k]*Pochhammer[1-k-x, n-1], {k, n}];
A168391[n_]:= CoefficientList[p[x, n], x];
Table[A168391[n], {n, 12}]//Flatten (* G. C. Greubel, Mar 28 2022 *)
PROG
(Sage)
@CachedFunction
def p(n, x): return ((-1)^(n+1)/(n+1))*sum( binomial(n+1, k)*binomial(n-1, k-1)*rising_factorial(1-k-x, n-1) for k in (1..n) )
def A168391(n, k): return ( p(n, x) ).series(x, n+1).list()[k]
flatten([[A168391(n, k) for k in (0..n-1)] for n in (1..12)]) # G. C. Greubel, Mar 28 2022
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Roger L. Bagula, Nov 24 2009
EXTENSIONS
Edited by G. C. Greubel, Mar 28 2022
STATUS
approved