A081658 Triangle read by rows: T(n, k) = (-2)^k*binomial(n, k)*Euler(k, 1/2).

1, 1, 0, 1, 0, -1, 1, 0, -3, 0, 1, 0, -6, 0, 5, 1, 0, -10, 0, 25, 0, 1, 0, -15, 0, 75, 0, -61, 1, 0, -21, 0, 175, 0, -427, 0, 1, 0, -28, 0, 350, 0, -1708, 0, 1385, 1, 0, -36, 0, 630, 0, -5124, 0, 12465, 0, 1, 0, -45, 0, 1050, 0, -12810, 0, 62325, 0, -50521, 1, 0, -55, 0, 1650, 0, -28182, 0, 228525, 0, -555731, 0, 1, 0, -66, 0, 2475, 0

Offset: 0

Author

Paul Barry, Mar 26 2003

Keywords

• easy
• sign
• tabl

Comments

These are the coefficients of the Swiss-Knife polynomials A153641. - Peter Luschny, Jul 21 2012

Nonzero diagonals of the triangle are of the form A000364(k)*binomial(n+2k,2k)*(-1)^k.

A363393 is the dual triangle ('dual' in the sense of Euler-tangent versus Euler-secant numbers). - Peter Luschny, Jun 05 2023

Examples

			The triangle begins
[0] 1;
[1] 1, 0;
[2] 1, 0,  -1;
[3] 1, 0,  -3, 0;
[4] 1, 0,  -6, 0,   5;
[5] 1, 0, -10, 0,  25, 0;
[6] 1, 0, -15, 0,  75, 0,  -61;
[7] 1, 0, -21, 0, 175, 0, -427, 0;
...
From _Peter Luschny_, Sep 17 2021: (Start)
The triangle shows the coefficients of the following polynomials:
[1] 1;
[2] 1 -    x^2;
[3] 1 -  3*x^2;
[4] 1 -  6*x^2 +   5*x^4;
[5] 1 - 10*x^2 +  25*x^4;
[6] 1 - 15*x^2 +  75*x^4 -  61*x^6;
[7] 1 - 21*x^2 + 175*x^4 - 427*x^6;
...
These polynomials are the permanents of the n X n matrices with all entries above the main antidiagonal set to 'x' and all entries below the main antidiagonal set to '-x'. The main antidiagonals consist only of ones. Substituting x <- 1 generates the Euler tangent numbers A155585. (Compare with A046739.)
(End)
		

Crossrefs

Row reversed: A119879.

Cf. A000364, A046739, A155585, A363393.

Programs

Maple

ogf := n -> euler(n) / (1 - x)^(n + 1):
ser := n -> series(ogf(n), x, 16):
T := (n, k) -> coeff(ser(k), x, n - k):
for n from 0 to 9 do seq(T(n, k), k = 0..n) od;  # Peter Luschny, Jun 05 2023
T := (n, k) -> (-2)^k*binomial(n, k)*euler(k, 1/2):
seq(seq(T(n, k), k = 0..n), n = 0..9);  # Peter Luschny, Apr 03 2024

Mathematica

sk[n_, x_] := Sum[Binomial[n, k]*EulerE[k]*x^(n - k), {k, 0, n}];
Table[CoefficientList[sk[n, x], x] // Reverse, {n, 0, 12}] // Flatten (* Jean-François Alcover, Jun 04 2019 *)
Flatten@Table[Binomial[n, k] EulerE[k], {n, 0, 12}, {k, 0, n}] (* Oliver Seipel, Jan 14 2025 *)

Python

from functools import cache
@cache
def T(n: int, k: int) -> int:
    if k == 0: return 1
    if k % 2 == 1:  return 0
    if k == n: return -sum(T(n, j) for j in range(0, n - 1, 2))
    return (T(n - 1, k) * n) // (n - k)
for n in range(10):
    print([T(n, k) for k in range(n + 1)])  # Peter Luschny, Jun 05 2023

Sage

R = PolynomialRing(ZZ, 'x')
@CachedFunction
def p(n, x) :
    if n == 0 : return 1
    return add(p(k, 0)*binomial(n, k)*(x^(n-k)-(n+1)%2) for k in range(n)[::2])
def A081658_row(n) : return [R(p(n,x)).reverse()[i] for i in (0..n)]
for n in (0..8) : print(A081658_row(n)) # Peter Luschny, Jul 20 2012

Formula

Coefficients of the polynomials in k in the binomial transform of the expansion of 2/(exp(kx)+exp(-kx)).

From Peter Luschny, Jul 20 2012: (Start)

p{n}(0) = Signed Euler secant numbers A122045.

p{n}(1) = Signed Euler tangent numbers A155585.

p{n}(2) has e.g.f. 2*exp(x)/(exp(-2*x)+1) A119880.

2^n*p{n}(1/2) = Signed Springer numbers A188458.

3^n*p{n}(1/3) has e.g.f. 2*exp(4*x)/(exp(6*x)+1)

4^n*p{n}(1/4) has e.g.f. 2*exp(5*x)/(exp(8*x)+1).

Row sum: A155585 (cf. A009006). Absolute row sum: A003701.

The GCD of the rows without the first column: A155457. (End)

From Peter Luschny, Jun 05 2023: (Start)

T(n, k) = [x^(n - k)] Euler(k) / (1 - x)^(k + 1).

For a recursion see the Python program.

Conjecture: If n is prime then n divides T(n, k) for 1 <= k <= n-1. (End)

Extensions

Typo in data corrected by Peter Luschny, Jul 20 2012

Error in data corrected and new name by Peter Luschny, Apr 03 2024