%I
%S 1,1,0,1,0,1,1,0,3,0,1,0,6,0,5,1,0,1,10,0,25,0,1,0,15,0,75,0,61,
%T 1,0,21,0,175,0,427,0,1,0,28,0,350,0,1708,0,1385,1,0,36,0,630,0,
%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, coefficients of polynomials related to the Euler numbers ordered by falling powers.
%C These are the coefficients of the SwissKnife polynomials A153641.  _Peter Luschny_, Jul 21 2012
%C Nonzero diagonals of the triangle are of the form A000364(k)C(n+2k,2k)(1)^k.
%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)
%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 [8] 1, 0, 28, 0, 350, 0, 1708, 0, 1385.
%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^(nk)(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
%Y Row reversed: A119879.
%Y Cf. A000364.
%K easy,sign,tabl
%O 0,9
%A _Paul Barry_, Mar 26 2003
%E Typo in data corrected by _Peter Luschny_, Jul 20 2012.
