OFFSET
0,2
COMMENTS
Row sums yield the double factorial numbers (A001147).
LINKS
G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
B. T. Gill, Problem 1729Math. Magazine, vol. 79, No. 4, 2006, p. 313, problem 1729.
FORMULA
T(n,0) = 2^n * n! = A000165(n).
T(n,n) = (-1)^n*A001147(n).
From Peter Bala, Aug 09 2024: (Start)
The polynomial P(n, x) = Sum_{k = 0..n} T(n, k)*x^(n-k) satisfies the functional equation P(n, 1 - x) = (-1)^n*P(n, x).
P(n, x) = (2*n - 1)*(2*x - 1)*P(n-1, x) + 4*(n - 1)^2*x*(1 - x)*P(n-2, x) with P(0, x) = 1 and P(1, x) = 2*x - 1.
Conjecture 1: for n >= 1, the zeros of P(n, x) lie on the vertical line Re(x) = 1/2 in the complex plane; that is, the family of polynomials {P(n, x) : n >= 1} satisfies a Riemann hypothesis.
Set u = x^2 and define p(n, u) = P(n, 1/2 + x) if n is even, else p(n, x) = (1/x)* P(n, 1/2 + x). The first few polynomials are p(0, u) = 1, p(1, u) = 2, p(2, u) = 8*u + 1, p(3, u) = 48*u + 18 and p(4, u) = 384*u^2 + 288*u + 9.
Conjecture 2: for n >= 2, the zeros of p(n+1, u) are negative and interlace the zeros of p(n, u). (End)
EXAMPLE
Triangle begins:
1;
2, -1;
8, -8, 3;
48, -72, 54, -15;
384, -768, 864, -480, 105;
3840, -9600, 14400, -12000, 5250, -945;
46080, -138240, 259200, -288000, 189000, -68040, 10395;
...
MAPLE
T:=(n, k)->(-1)^k*n!*2^(n-2*k)*binomial(n, k)*binomial(2*k, k): for n from 0 to 8 do seq(T(n, k), k=0..n) od; # yields sequence in triangular form
MATHEMATICA
Table[(-1)^k*n! 2^(n - 2 k)*Binomial[n, k]*Binomial[2*k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Oct 14 2017 *)
PROG
(PARI) for(n=0, 10, for(k=0, n, print1((-1)^k*n!*2^(n-2*k)*binomial(n, k)* binomial(2*k, k), ", "))) \\ G. C. Greubel, Oct 14 2017
(Magma) /* As triangle */ [[(-1)^k*Factorial(n)*2^(n-2*k)* Binomial(n, k)*Binomial(2*k, k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 15 2017
CROSSREFS
KEYWORD
sign,tabl
AUTHOR
Emeric Deutsch, Oct 14 2006
STATUS
approved