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%I #44 Apr 03 2024 10:07:32
%S 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,
%T 0,-21,0,175,0,-427,0,1,0,-28,0,350,0,-1708,0,1385,1,0,-36,0,630,0,
%U -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
%N Triangle read by rows: T(n, k) = (-2)^k*binomial(n, k)*Euler(k, 1/2).
%C These are the coefficients of the Swiss-Knife polynomials A153641. - _Peter Luschny_, Jul 21 2012
%C Nonzero diagonals of the triangle are of the form A000364(k)*binomial(n+2k,2k)*(-1)^k.
%C A363393 is the dual triangle ('dual' in the sense of Euler-tangent versus Euler-secant numbers). - _Peter Luschny_, Jun 05 2023
%F Coefficients of the polynomials in k in the binomial transform of the expansion of 2/(exp(kx)+exp(-kx)).
%F From _Peter Luschny_, Jul 20 2012: (Start)
%F p{n}(0) = Signed Euler secant numbers A122045.
%F p{n}(1) = Signed Euler tangent numbers A155585.
%F p{n}(2) has e.g.f. 2*exp(x)/(exp(-2*x)+1) A119880.
%F 2^n*p{n}(1/2) = Signed Springer numbers A188458.
%F 3^n*p{n}(1/3) has e.g.f. 2*exp(4*x)/(exp(6*x)+1)
%F 4^n*p{n}(1/4) has e.g.f. 2*exp(5*x)/(exp(8*x)+1).
%F Row sum: A155585 (cf. A009006). Absolute row sum: A003701.
%F The GCD of the rows without the first column: A155457. (End)
%F From _Peter Luschny_, Jun 05 2023: (Start)
%F T(n, k) = [x^(n - k)] Euler(k) / (1 - x)^(k + 1).
%F For a recursion see the Python program.
%F Conjecture: If n is prime then n divides T(n, k) for 1 <= k <= n-1. (End)
%e The triangle begins
%e [0] 1;
%e [1] 1, 0;
%e [2] 1, 0, -1;
%e [3] 1, 0, -3, 0;
%e [4] 1, 0, -6, 0, 5;
%e [5] 1, 0, -10, 0, 25, 0;
%e [6] 1, 0, -15, 0, 75, 0, -61;
%e [7] 1, 0, -21, 0, 175, 0, -427, 0;
%e ...
%e From _Peter Luschny_, Sep 17 2021: (Start)
%e The triangle shows the coefficients of the following polynomials:
%e [1] 1;
%e [2] 1 - x^2;
%e [3] 1 - 3*x^2;
%e [4] 1 - 6*x^2 + 5*x^4;
%e [5] 1 - 10*x^2 + 25*x^4;
%e [6] 1 - 15*x^2 + 75*x^4 - 61*x^6;
%e [7] 1 - 21*x^2 + 175*x^4 - 427*x^6;
%e ...
%e 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.)
%e (End)
%p ogf := n -> euler(n) / (1 - x)^(n + 1):
%p ser := n -> series(ogf(n), x, 16):
%p T := (n, k) -> coeff(ser(k), x, n - k):
%p for n from 0 to 9 do seq(T(n, k), k = 0..n) od; # _Peter Luschny_, Jun 05 2023
%p T := (n, k) -> (-2)^k*binomial(n, k)*euler(k, 1/2):
%p seq(seq(T(n, k), k = 0..n), n = 0..9); # _Peter Luschny_, Apr 03 2024
%t sk[n_, x_] := Sum[Binomial[n, k]*EulerE[k]*x^(n - k), {k, 0, n}];
%t Table[CoefficientList[sk[n, x], x] // Reverse, {n, 0, 12}] // Flatten (* _Jean-François Alcover_, Jun 04 2019 *)
%o (Sage)
%o R = PolynomialRing(ZZ, 'x')
%o @CachedFunction
%o def p(n, x) :
%o if n == 0 : return 1
%o return add(p(k, 0)*binomial(n, k)*(x^(n-k)-(n+1)%2) for k in range(n)[::2])
%o def A081658_row(n) : return [R(p(n,x)).reverse()[i] for i in (0..n)]
%o for n in (0..8) : print(A081658_row(n)) # _Peter Luschny_, Jul 20 2012
%o (Python)
%o from functools import cache
%o @cache
%o def T(n: int, k: int) -> int:
%o if k == 0: return 1
%o if k % 2 == 1: return 0
%o if k == n: return -sum(T(n, j) for j in range(0, n - 1, 2))
%o return (T(n - 1, k) * n) // (n - k)
%o for n in range(10):
%o print([T(n, k) for k in range(n + 1)]) # _Peter Luschny_, Jun 05 2023
%Y Row reversed: A119879.
%Y Cf. A000364, A046739, A155585, A363393.
%K easy,sign,tabl
%O 0,9
%A _Paul Barry_, Mar 26 2003
%E Typo in data corrected by _Peter Luschny_, Jul 20 2012
%E Error in data corrected and new name by _Peter Luschny_, Apr 03 2024
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