A123516 Triangle read by rows: T(n,k) = (-1)^k * n! * 2^(n-2*k) * binomial(n,k) * binomial(2*k,k) (0<=k<=n).

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, 645120, -2257920, 5080320, -7056000, 6174000, -3333960, 1018710, -135135, 10321920, -41287680, 108380160, -180633600, 197568000

Offset: 0

Author

Emeric Deutsch, Oct 14 2006

Keywords

• sign
• tabl

Comments

Row sums yield the double factorial numbers (A001147).

Examples

			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;
  ...
		

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.

Crossrefs

Cf. A001147, A000165, A177145, A331817.

Programs

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

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 *)

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

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.

P(n, 1/2) = A177145(n+1); P(n, -1/2) = (-1)^n*A331817(n).

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)