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Expansion of exp(x)/(1 - x^3/3!), where a(n) = 1 + binomial(n,3)*a(n-3).
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%I #8 Feb 02 2019 18:21:20

%S 1,1,1,2,5,11,41,176,617,3445,21121,101806,757901,6040607,37057385,

%T 344844956,3382739921,25199021801,281393484097,3277874983450,

%U 28726884853141,374253333849011,5047927474513001,50875313074912712

%N Expansion of exp(x)/(1 - x^3/3!), where a(n) = 1 + binomial(n,3)*a(n-3).

%e E.g.f.: exp(x)/(1 - x^3/3!) = 1 + x + 1*x^2/2! + 2*x^3/3! + 5*x^4/4! + 11*x^5/5! + 41*x^6/6! + ... + a(n)*x^n/n! + ...

%e where a(n) = 1 + n*(n-1)*(n-2)*a(n-3)/3!.

%p restart: G(x):=2*exp(-x)/(x^3/3!+1): f[0]:=G(x): for n from 1 to 26 do f[n]:=diff(-f[n-1],x) od: x:=0: seq(f[n]/2,n=0..23); # _Zerinvary Lajos_, Apr 03 2009

%o (PARI) a(n)=n!*polcoeff(exp(x+x*O(x^n))/(1-x^3/3! +x*O(x^n)),n)

%Y Cf. A087214, A128230, A128232.

%K nonn

%O 0,4

%A _Paul D. Hanna_, Feb 20 2007