A065547 Triangle of Salie numbers.

1, 0, 1, 0, -1, 1, 0, 3, -3, 1, 0, -17, 17, -6, 1, 0, 155, -155, 55, -10, 1, 0, -2073, 2073, -736, 135, -15, 1, 0, 38227, -38227, 13573, -2492, 280, -21, 1, 0, -929569, 929569, -330058, 60605, -6818, 518, -28, 1, 0, 28820619, -28820619, 10233219, -1879038, 211419, -16086, 882, -36, 1, 0, -1109652905

Offset

0,8

Comments

Coefficients of polynomials H(n,x) related to Euler polynomials through H(n,x(x-1)) = E(2n,x).

Links

Table of n, a(n) for n=0..56.

D. Dumont and J. Zeng, Polynomes d'Euler et les fractions continues de Stieltjes-Rogers, Ramanujan J. 2 (1998) 3, 387-410.

J. M. Hammersley, An undergraduate exercise in manipulation, Math. Scientist, 14 (1989), 1-23.

Ira M. Gessel and X. G. Viennot, Determinants, paths and plane partitions, 1989, p. 27, eqn 12.1.

A. F. Horadam, Generation of Genocchi polynomials of first order by recurrence relation, Fib. Quart. 2 (1992), 239-243.

Formula

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).

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.

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).

Dumont/Zeng give a continued fraction and other formulas.

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.

Sum_{k=0..n} (-4)^(n-k)*T(n,k) = A000364(n) (Euler numbers). - Philippe Deléham, Oct 25 2006

Example

Triangle begins:
 1;
 0,   1;
 0,  -1,    1;
 0,   3,   -3,  1;
 0, -17,   17, -6,   1;
 0, 155, -155, 55, -10, 1;
 ...

Mathematica

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 *)

Prog

(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

Crossrefs

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.

Column sequences (without leading zeros) give, for k=1..10: A065547 (twice), A095652-9.

Cf. A000795, A005647, A000035.

See A085707 for unsigned and transposed version.

See A098435 for negative values of n, k.

Sequence in context: A264436 A122850 A132062 * A143333 A283798 A065551

Adjacent sequences: A065544 A065545 A065546 * A065548 A065549 A065550

Keyword

sign,tabl

Author

Wouter Meeussen, Dec 02 2001

Extensions

Edited by Ralf Stephan, Sep 08 2004

Status

approved