A322970 Coefficient triangle of polynomials recursively defined by P(n,x) = (n+1)*(n+1)! + x*Sum_{k=1..n} k^2*n!/(n+1-k)!*P(n-k,x) with P(0,x) = 1.

1, 1, 4, 1, 12, 18, 1, 24, 120, 96, 1, 40, 420, 1200, 600, 1, 60, 1080, 6720, 12600, 4320, 1, 84, 2310, 25200, 105840, 141120, 35280, 1, 112, 4368, 73920, 554400, 1693440, 1693440, 322560, 1, 144, 7560, 183456, 2162160, 11975040, 27941760, 21772800, 3265920

Offset

0,3

Links

Table of n, a(n) for n=0..44.

B. Heim, F. Luca, and M. Neuhauser, Recurrence relations for polynomials obtained by arithmetic functions, International Journal of Number Theory, Vol. 15, No. 06, pp. 1291-1303 (2019).

Formula

A(n,k) = (n+1)!*(n+1+k)!/((k+1)!*(2k+1)!*(n-k)!) (proved);

The rows correspond to the polynomials:

P(0,x) = 1;

P(1,x) = x + 4;

P(2,x) = x^2 + 12*x + 18;

P(3,x) = x^3 + 24*x^2 + 120*x + 96;

...

They satisfy the recurrence relation P(n+1,x) = (x+3*n+3)*P(n,x) + (n+1)*(x-3*n)*P(n-1,x) + (n+1)*n*(n-1)*P(n-2,x) with P(0,x) = 1, P(1,x) = (x+3)*P(0,x) + 1, P(2,x) = (x+6)*P(1,x) + 2*(x-3)*P(0,x) (proved).

Example

1;
1,  4;
1,  12,   18;
1,  24,  120,     96;
1,  40,  420,   1200,     600;
1,  60, 1080,   6720,   12600,     4320;
1,  84, 2310,  25200,  105840,   141120,    35280;
1, 112, 4368,  73920,  554400,  1693440,  1693440,   322560;
1, 144, 7560, 183456, 2162160, 11975040, 27941760, 21772800, 3265920

Maple

for n from 0 to nn do for k from 0 to n do printf("%g, ", (n+1)!*binomial(2*n+1-k, 2*(n-k)+1)/(n-k+1)!); end do; printf("\n"); end do;

Prog

(PARI) tabl = (nn)->for(n=0, nn, for(k=0, n, print1((n+1)!*binomial(2*n+1-k, 2*(n-k)+1)/(n-k+1)!, ", "); ); print(); )

Crossrefs

Cf. A089231 (polynomials satisfy a similar recurrence relation with k instead of k^2 and (n+1)! instead of (n+1)*(n+1)! (proved)), A001563 (right diagonal).

Sequence in context: A370129 A187541 A117413 * A157384 A173621 A274087

Adjacent sequences: A322967 A322968 A322969 * A322971 A322972 A322973

Keyword

easy,nonn,tabl

Author

Markus Neuhauser, Jan 01 2019

Status

approved