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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.
0
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
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
KEYWORD
easy,nonn,tabl
AUTHOR
Markus Neuhauser, Jan 01 2019
STATUS
approved