A368235 Triangle read by rows: n-th row polynomial equals the numerator of the rational function (-1)^n*f(x) * (d/dx)^n (1/f(x)), where f(x) = sqrt(x + x^2).

1, 1, 2, 3, 8, 8, 15, 54, 72, 48, 105, 480, 864, 768, 384, 945, 5250, 12000, 14400, 9600, 3840, 10395, 68040, 189000, 288000, 259200, 138240, 46080, 135135, 1018710, 3333960, 6174000, 7056000, 5080320, 2257920, 645120, 2027025, 17297280, 65197440, 142248960, 197568000, 180633600, 108380160, 41287680, 10321920

Offset: 0

Author

Peter Bala, Dec 18 2023

Keywords

• sign
• tabl
• easy

Comments

Unsigned row reverse of A123516.

The row polynomials also occur on repeated integration of 1/sqrt(x + x^2). See the example section.

Examples

			Triangle begins
 n\k |     0       1        2        3        4        5       6
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  0  |     1
  1  |     1       2
  2  |     3       8        8
  3  |    15      54       72       48
  4  |   105     480      864      768      384
  5  |   945    5250    12000    14400     9600     3840
  6  | 10395   68040   189000   288000   259200   138240   46080
 ...
Repeated integration of 1/f(x), where f(x) = sqrt(x + x^2):
Let I denote the integral operator h(x) -> Integral_{t = 0..x} h(t) dt.
Let g(x) = I(1/f(x)) = log(2*x + 1 + 2*f(x)). Then
(2^1 * 1!^2) * I^(2)(1/f(x)) = (2*x + 1)*g(x) - 2*f(x).
(2^2 * 2!^2) * I^(3)(1/f(x)) = (8*x^2 + 8*x + 3)*g(x) - 6*(2*x + 1)*f(x).
(2^3 * 3!^2) * I^(4)(1/f(x)) = (48*x^3 + 72*x^2 + 54*x + 15)*g(x) - 2*(44*x^2 + 44*x + 15)*f(x).
(2^4 * 4!^2) * I^(5)(1/f(x)) = (384*x^4 + 768*x^3 + 864*x^2 + 480*x + 105)*g(x) - 10*(2*x + 1)*(40*x^2 + 40*x + 21)*f(x).
		

Crossrefs

Cf. A001147 (column 1), 2*A161120 (column 2), A000165 (main diagonal) A014479 (first subdiagonal), 3*A286725 (second subdiagonal).

Cf. A059366, A098461, A123516, A156919, A177145, A286724, A331817.

Programs

Maple

# sequence in triangular form
T := (n, k) -> n! * 2^(2*k-n) * binomial(n, k)*binomial(2*n-2*k, n-k):
for n from 0 to 8 do seq(T(n, k), k = 0..n) od;

Formula

T(n, k) = n! * 2^(2*k-n) * binomial(n, k) * binomial(2*n-2*k, n-k).

k*T(n, k) = (2*n^2)*T(n-1, k-1) for k >= 1 with T(n, 0) = (2*n - 1)!! = A001147(n).

T(n, 1) = 2*A161120(n).

T(n, n) = 2^n * n! = A000165(n); T(n+1, n) = 2^n * n! * (n+1)^2 = A014479(n);

T(n+2, n) = 3 * 2^(n-1)*(n+2)!*binomial(n+2, 2) = 3 * A286725(n).

More generally, T(n+r, n) = (2*r - 1)!! * A286724(n+r, r).

E.g.f.: Sum_{k >= 0} (1/2^k)*binomial(2*k, k)*t^k/(1 - 2*t*x)^(k+1) = 1 + (1 + 2*x)*t + (3 + 8*x + 8*x^2)*t^2/2! + (15 + 54*x + 72*x^2 + 48*x^3)*t^3/3! + ....

n-th row polynomial R(n, x) = (-2)^n*(x + x^2)^(n+1/2)*(d/dx)^n (1/sqrt(x + x^2)).

Recurrence for row polynomials:

R(n+1, x) = (2*x + 1)*(2*n + 1)*R(n, x) - 4*x*(x + 1)*n^2*R(n-1, x), with R(0, x) = 1.

R'(n, x) = 2*n^2 * R(n-1, x) for n >= 1.

Functional equation: R(n, -1 - x) = (-1)^n * R(n, x).

Conjecture: the zeros of the polynomial R(n, -x) lie on the vertical line Re(x) = 1/2 in the complex plane.

(-1)^n * x^n * R(n, (- 1 - x)/x) equals the n-th row polynomial of A123516.

(1 - x)^n * R(n, x/(1 - x)) equals the n-th row polynomial of A059366.

Let D denote the operator (1/x)*d/dx. Then D^(n+1)( arcsinh(x) ) = (-1)^n*R(n, x^2)/(x*sqrt(1 + x^2))^(2*n+1).

R(n, 1/2) = A331817(n); R(n, -1/2) = A177145(n+1);

(2^n) * R(n, 1/4) = A098461(n).

Alternating row sums R(n, -1) = (-1)^n * A001147(n).