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

0,9

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

Links

Table of n, a(n) for n=0..83.

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)

Example

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)

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 *)

Prog

(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
(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

Crossrefs

Row reversed: A119879.

Cf. A000364, A046739, A155585, A363393.

Sequence in context: A359364 A073278 A359760 * A187253 A022904 A238341

Adjacent sequences: A081655 A081656 A081657 * A081659 A081660 A081661

Keyword

easy,sign,tabl

Author

Paul Barry, Mar 26 2003

Extensions

Typo in data corrected by Peter Luschny, Jul 20 2012

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

Status

approved