%I #45 Sep 04 2023 22:19:36
%S 1,0,1,0,-1,1,0,3,-3,1,0,-17,17,-6,1,0,155,-155,55,-10,1,0,-2073,2073,
%T -736,135,-15,1,0,38227,-38227,13573,-2492,280,-21,1,0,-929569,929569,
%U -330058,60605,-6818,518,-28,1,0,28820619,-28820619,10233219,-1879038,211419,-16086,882,-36,1,0,-1109652905
%N Triangle of Salie numbers.
%C Coefficients of polynomials H(n,x) related to Euler polynomials through H(n,x(x-1)) = E(2n,x).
%H D. Dumont and J. Zeng, <a href="http://math.univ-lyon1.fr/homes-www/zeng/public_html/paper/publication.html">Polynomes d'Euler et les fractions continues de Stieltjes-Rogers</a>, Ramanujan J. 2 (1998) 3, 387-410.
%H J. M. Hammersley, <a href="http://www.appliedprobability.org/data/files/TMS%20articles/14_1_1.pdf">An undergraduate exercise in manipulation</a>, Math. Scientist, 14 (1989), 1-23.
%H Ira M. Gessel and X. G. Viennot, <a href="http://people.brandeis.edu/~gessel/homepage/papers/pp.pdf">Determinants, paths and plane partitions</a>, 1989, p. 27, eqn 12.1.
%H A. F. Horadam, <a href="http://www.fq.math.ca/Scanned/30-3/horadam.pdf">Generation of Genocchi polynomials of first order by recurrence relation</a>, Fib. Quart. 2 (1992), 239-243.
%F E.g.f.: Sum_{n, k=0..oo} T(n, k) t^k x^(2n)/(2n)! = cosh(sqrt(1+4t) x/2) / cosh(x/2).
%F T(k, n) = Sum_{i=0..n-k} A028296(i)/4^(n-k)*C(2n, 2i)*C(n-i, n-k-i), or 0 if n<k.
%F Polynomial recurrences: x^n = Sum_{0<=2i<=n} C(n, 2i)*H(n-i, x); (1/4+x)^n = Sum_{m=0..n} C(2n, 2m)*(1/4)^(n-m)*H(m, x).
%F Dumont/Zeng give a continued fraction and other formulas.
%F Triangle T(n, k) read by rows; given by [0, -1, -2, -4, -6, -9, -12, -16, ...] DELTA A000035, where DELTA is Deléham's operator defined in A084938.
%F Sum_{k=0..n} (-4)^(n-k)*T(n,k) = A000364(n) (Euler numbers). - _Philippe Deléham_, Oct 25 2006
%e Triangle begins:
%e 1;
%e 0, 1;
%e 0, -1, 1;
%e 0, 3, -3, 1;
%e 0, -17, 17, -6, 1;
%e 0, 155, -155, 55, -10, 1;
%e ...
%t h[n_, x_] := Sum[c[k]*x^k, {k, 0, n}]; eq[n_] := SolveAlways[h[n, x*(x - 1)] == EulerE[2*n, x], x]; row[n_] := Table[c[k], {k, 0, n}] /. eq[n] // First; Table[row[n], {n, 0, 10}] // Flatten (* _Jean-François Alcover_, Oct 02 2013 *)
%o (PARI) { S2(n, k) = (1/k!)*sum(i=0,k,(-1)^(k-i)*binomial(k,i)*i^n) }{ Eu(n) = sum(m=0,n,(-1)^m*m!*S2(n+1,m+1)*(-1)^floor(m/4)*2^-floor(m/2)*((m+1)%4!=0)) } T(n,k)=if(n<k,0,sum(l=0,n-k,Eu(2*l)/2^(2*(n-k))*binomial(2*n,2*l)*binomial(n-l,n-k-l))) \\ _Ralf Stephan_
%Y Sum_{k>=0} (-1)^(n+k)*2^(n-k)*T(n, k) = A005647(n). Sum_{k>=0} (-1)^(n+k)*2^(2n-k)*T(n, k) = A000795(n). Sum_{k>=0} (-1)^(n+k)*T(n, k) = A006846(n), where A006846 = Hammersley's polynomial p_n(1). - _Philippe Deléham_, Feb 26 2004.
%Y Column sequences (without leading zeros) give, for k=1..10: A065547 (twice), A095652-9.
%Y Cf. A000795, A005647, A000035.
%Y See A085707 for unsigned and transposed version.
%Y See A098435 for negative values of n, k.
%K sign,tabl
%O 0,8
%A _Wouter Meeussen_, Dec 02 2001
%E Edited by _Ralf Stephan_, Sep 08 2004