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

Author

Wouter Meeussen, Dec 02 2001

Keywords

• sign
• tabl

Comments

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

Examples

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

Links

• 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.

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.

Programs

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 *)

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(nRalf Stephan

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

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

Extensions

Edited by Ralf Stephan, Sep 08 2004

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