A322970 - OEIS

%I #20 Jan 18 2021 09:52:25

%S 1,1,4,1,12,18,1,24,120,96,1,40,420,1200,600,1,60,1080,6720,12600,

%T 4320,1,84,2310,25200,105840,141120,35280,1,112,4368,73920,554400,

%U 1693440,1693440,322560,1,144,7560,183456,2162160,11975040,27941760,21772800,3265920

%N 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.

%H B. Heim, F. Luca, and M. Neuhauser, <a href="https://doi.org/10.1142/S1793042119500726">Recurrence relations for polynomials obtained by arithmetic functions</a>, International Journal of Number Theory, Vol. 15, No. 06, pp. 1291-1303 (2019).

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

%F The rows correspond to the polynomials:

%F P(0,x) = 1;

%F P(1,x) = x + 4;

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

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

%F ...

%F 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).

%e 1;

%e 1,  4;

%e 1,  12,   18;

%e 1,  24,  120,     96;

%e 1,  40,  420,   1200,     600;

%e 1,  60, 1080,   6720,   12600,     4320;

%e 1,  84, 2310,  25200,  105840,   141120,    35280;

%e 1, 112, 4368,  73920,  554400,  1693440,  1693440,   322560;

%e 1, 144, 7560, 183456, 2162160, 11975040, 27941760, 21772800, 3265920

%p 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;

%o (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();)

%Y 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).

%K easy,nonn,tabl

%O 0,3

%A _Markus Neuhauser_, Jan 01 2019