A109573 E.g.f.: 2*x/(1+exp(-2*x)).

0, 1, 2, 0, -8, 0, 96, 0, -2176, 0, 79360, 0, -4245504, 0, 313155584, 0, -30460116992, 0, 3777576173568, 0, -581777702256640, 0, 108932957168730112, 0, -24370173276164456448, 0, 6419958484945407574016, 0, -1967044844910430876860416, 0, 693575525634287935244206080, 0

Offset: 0

Author

Roger L. Bagula, Jun 27 2005

Keywords

• sign
• easy

Comments

(-1)^n*a(n) = 2^n*(2^n-1)*Bernoulli(n), n >= 1, are integers divisible by n. See the Koecher reference, p. 175, Satz, with a proof, and the J. Worpitzky and L. Kronecker JraM references 94 (1883) 203-232 and 268-269, respectively. - Wolfdieter Lang, Mar 09 2017

References

• Max Koecher, Klassische elementare Analysis, Birkhäuser, Basel, Boston, 1987, p. 175

Links

• Indranil Ghosh, Table of n, a(n) for n = 0..332

Crossrefs

Cf. A027641, A027642.

Programs

Maple

G:=2*x/(1+exp(-2*x)): Gser:=series(G,x=0,35): 0,seq(n!*coeff(Gser,x^n),n=1..32); # yields the signed sequence
A109573 := n -> `if`(n = 0, 0, n*2^(n-1)*euler(n-1,1)): # Peter Luschny, Jan 26 2009

Mathematica

g[x_] = x/(-1 + Sum[(-2)^(n - 1)*x^n/n!, {n, 1, Infinity}]) h[x_, n_] = Dt[g[x], {x, n}] a[x_] = Table[h[x, n], {n, 0, 50}]; b = Abs[a[0]]
X[m_] := m Sum[(-2)^(m-1-k) k! StirlingS2[m-1,k], {k,0,m-1}]; Table[X[i],{i,0,20}] (* Peter Luschny, Apr 29 2009 *)

PARI

for(n=0, 31, print1(n*sum(k=0, n-1, (-2)^(n - 1- k) * k! * stirling(n - 1, k, 2)), ", ")) \\ Indranil Ghosh, Mar 10 2017

Formula

a(0) = 0 and for n > 0 a(n) = n 2^(n-1) E_{n-1}(1) where E_{m}(x) are the Euler polynomials. - Peter Luschny, Jan 26 2009

From Sergei N. Gladkovskii, Apr 04, Nov 23 and Dec 27 2013: (Start)

Continued fractions:

E.g.f.: 2*x/(Q(0)+1) where Q(k) = 1 - 2*x/(2*k+1 - x*(2*k+1)/(x - (k+1)/Q(k+1))).

E.g.f.: 1+2*x + T(0), where T(k) = 4*k-1 - x/(1 + x/( 4*k+1 - x/(1 + x/T(k+1)))).

E.g.f.: 1 + T(0), where T(k) = 4*k-1 + x/(1 - x/( 4*k+1 + x/(1 - x/T(k+1)))).

E.g.f.: x^2*E(0)+x, where E(k) = 1 - x^2/(x^2 + (2*k+1)*(2*k+3)/E(k+1)).

(End)

a(n) = (-1)^n*2^n*(2^n-1)*Bernoulli(n), with Bernoulli(n) = A027641(n)/A027642(n). For |a(n)| see the Koecher reference, p. 175, and a comment above. - Wolfdieter Lang, Mar 09 2017