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E.g.f. cosh(x)/(1+x).
12

%I #39 Dec 30 2023 10:57:48

%S 1,-1,3,-9,37,-185,1111,-7777,62217,-559953,5599531,-61594841,

%T 739138093,-9608795209,134523132927,-2017846993905,32285551902481,

%U -548854382342177,9879378882159187,-187708198761024553,3754163975220491061

%N E.g.f. cosh(x)/(1+x).

%C Unsigned sequence satisfies a(n)=n*a(n-1)+a(n-2)-(n-2)*a(n-3), a(0)=1,a(1)=1,a(2)=3 with e.g.f. cosh(z)/(1-z). - Mario Catalani (mario.catalani(AT)unito.it), Feb 07 2003

%C (-1)^n*(A000166(n) + A000522(n))/2 = this_sequence, (-1)^n*(A000166(n) - A000522(n))/2 = A009628(n).

%C The positive sequence has e.g.f. cosh(x)/(1-x), with a(n)=sum{k=0..floor(n/2), binomial(n,2k)(n-2k)!}. It is the mean of the binomial and inverse binomial transforms of n!. - _Paul Barry_, May 01 2005

%H Seiichi Manyama, <a href="/A009179/b009179.txt">Table of n, a(n) for n = 0..449</a>

%F a(n) = (-1)^n*floor(n!*cosh(1)). - _Vladeta Jovovic_, Aug 10 2002

%F a(n) = (1+(-1)^n)/2-n*a(n-1). - _Vladeta Jovovic_, Apr 19 2003

%F a(n) = (-1)^n * n! * sum{k=0, [n/2], 1/(2k)!}.

%F E.g.f.: U(0)/(1+x) where U(k)= 1 + x^2/((4*k+1)*(4*k+2) - x^2*(4*k+1)*(4*k+2)/(x^2 + (4*k+3)*(4*k+4)/U(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Oct 22 2012

%F a(n) = (-1)^n*(exp(1)*Gamma(1+n,1) + exp(-1)*Gamma(1+n,-1))/2 - _Peter Luschny_, Dec 18 2017

%p restart: G(x):= cosh(x)/(1+x): f[0]:=G(x): for n from 1 to 21 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=0..20); # _Zerinvary Lajos_, Apr 03 2009

%t a[n_] := (-1)^n (Exp[1] Gamma[1 + n, 1] + Exp[-1] Gamma[1 + n, -1])/2;

%t Table[a[n], {n, 0, 20}] (* _Peter Luschny_, Dec 18 2017 *)

%o (PARI) x='x+O('x^99); Vec(serlaplace(cosh(x)/(1+x))) \\ _Altug Alkan_, Dec 18 2017

%Y Cf. A000166, A000522, A001540, A009628.

%K sign,easy

%O 0,3

%A _R. H. Hardin_

%E Extended with signs by _Olivier GĂ©rard_, Mar 15 1997