A119881 Expansion of e.g.f. exp(3*x)*sech(x).
1, 3, 8, 18, 32, 48, 128, 528, 512, -6912, 2048, 357888, 8192, -22351872, 32768, 1903822848, 131072, -209865080832, 524288, 29088886161408, 2097152, -4951498048929792, 8388608, 1015423886523629568, 33554432, -246921480190140874752, 134217728, 70251601603944228323328
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
Programs
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Magma
EulerPoly:= func< n,x | (&+[ (&+[ (-1)^j*Binomial(k,j)*(x+j)^n : j in [0..k]])/2^k: k in [0..n]]) >; A119881:= func< n| (-2)^n*EulerPoly(n,-1) >; [A119881(n): n in [0..40]]; // G. C. Greubel, Jun 07 2023
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Maple
a := proc(n) add(binomial(n,k)*bernoulli(k,1)*2^(n+k)/(n-k+1),k=0..n) end: # Peter Luschny, Dec 14 2008 a := n -> 2^n*abs(euler(n,-1)): # Peter Luschny, Jan 25 2009 P := proc(n,x) option remember; if n = 0 then 1 else (n*x+2*(1-x))*P(n-1,x)+x*(1-x)*diff(P(n-1,x),x); expand(%) fi end: A119881 := n -> subs(x=-1,P(n,x)): seq(A119881(n), n=0..27); # Peter Luschny, Mar 07 2014
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Mathematica
Table[2^(n+1) (Zeta[-n] (2^(n+1)-1)+1), {n,0,27}] (* Peter Luschny, Jul 16 2013 *) Range[0, 30]! CoefficientList[Series[Exp[3 x] Sech[x], {x, 0, 30}], x] (* Vincenzo Librandi, Mar 08 2014 *)
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PARI
my(x='x+O('x^66)); Vec(serlaplace(exp(3*x)/cosh(x))) \\ Joerg Arndt, Apr 20 2013
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Sage
def skp(n, x): A = lambda k: 0 if (k+1)%4 == 0 else (-1)^((k+1)//4)*2^(-(k//2)) return add(A(k)*add((-1)^v*binomial(k,v)*(v+x+1)^n for v in (0..k)) for k in (0..n)) A119881 = lambda n: skp(n,3) [A119881(n) for n in (0..27)] # Peter Luschny, Nov 23 2012
Formula
a(n) = Sum_{k=0..n} A119879(n,k)*3^k.
a(n) = Sum_{k=0..n} binomial(n,k)*B(k,1)*2^(n+k)/(n-k+1). Here B(k,1) are the Bernoulli number A027641(k)/A027642(k) with the exception B(1,1)=1/2. - Peter Luschny, Dec 14 2008
a(n) = 2^n |E(n,-1)| where E(n,x) are the Euler polynomials. - Peter Luschny, Jan 25 2009
The odd part of a(n) = numerator(Euler(n,2)/2) = 1, 3, 1, 9, 1, 3, 1, 33, 1, -27, 1, 699, ... (compare A143074). - Peter Luschny, Nov 23 2012
G.f.: 1/Q(0), where Q(k) = 1 - 2*x - x*(k+1)/(1+x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 19 2013
G.f.: 1/Q(0), where Q(k) = 1 - 4*x + x*(k+1)/(1-x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 20 2013
a(n) = 2^(n+1)*(zeta[-n]*(2^(n+1)-1)+1). - Peter Luschny, Jul 16 2013
E.g.f.: 2/Q(0), where Q(k) = 1 + 2^k/( 1 - 2*x/( 2*x - 2^k*(k+1)/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Dec 16 2013
a(n) = 2^(n+1)*(1+(-1)^n*(2^(n+1)-1)*Bernoulli(n+1)/(n+1)). - Vladimir Reshetnikov, Oct 21 2015
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