A331817 -id:A331817 - cp's OEIS frontend

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

Emeric Deutsch, Oct 14 2006

Row sums yield the double factorial numbers (A001147).

			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;
  ...

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.

Cf. A001147, A000165, A177145, A331817.

/* 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

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

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 *)

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

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)

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

Original entry on oeis.org

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

Peter Bala, Dec 18 2023

Unsigned row reverse of A123516.

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

			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).

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.

# 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;

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).

cp's OEIS Frontend

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

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

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cp's OEIS Frontend

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).

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Magma

Maple

Mathematica

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Formula

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).

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Formula

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

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