A155585 a(n) = 2^n*E(n, 1) where E(n, x) are the Euler polynomials.
1, 1, 0, -2, 0, 16, 0, -272, 0, 7936, 0, -353792, 0, 22368256, 0, -1903757312, 0, 209865342976, 0, -29088885112832, 0, 4951498053124096, 0, -1015423886506852352, 0, 246921480190207983616, 0, -70251601603943959887872, 0, 23119184187809597841473536, 0
Offset: 0
Examples
E.g.f.: 1 + x - 2*x^3/3! + 16*x^5/5! - 272*x^7/7! + 7936*x^9/9! -+ ... = exp(x)/cosh(x). O.g.f.: 1 + x - 2*x^3 + 16*x^5 - 272*x^7 + 7936*x^9 - 353792*x^11 +- ... O.g.f.: 1 + x/(1+2*x) + 2!*x^2/((1+2*x)*(1+4*x)) + 3!*x^3/((1+2*x)*(1+4*x)*(1+6*x)) + ...
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- P. Barry, Riordan Arrays, Orthogonal Polynomials as Moments, and Hankel Transforms, J. Int. Seq. 14 (2011) # 11.2.2, example 28.
- Tom Copeland, The Elliptic Lie Triad: Ricatti and KdV Equations, Infinigens, and Elliptic Genera
- Tom Copeland, The Kervaire-Milnor formula
- H. M. Edwards, A normal form for elliptic curves, Bull. Amer. Math. Soc. (N.S.) 44 (2007), no. 3, 393-422. see section 12, pp. 410-411.
- Wenjie Fang, Hsien-Kuei Hwang, and Mihyun Kang, Phase transitions from exp(n^(1/2)) to exp(n^(2/3)) in the asymptotics of banded plane partitions, arXiv:2004.08901 [math.CO], 2020.
- G. Rzadkowski, Bernoulli numbers and solitons-revisited, Journal of Nonlinear Math. Physics, 1711, pp. 121-126.
Crossrefs
Programs
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Maple
A155585 := n -> 2^n*euler(n, 1): # Peter Luschny, Jan 26 2009 a := proc(n) option remember; `if`(n::even, 0^n, -(-1)^n - add((-1)^k*binomial(n,k) *a(n-k), k = 1..n-1)) end: # Peter Luschny, Jun 01 2016 # Or via the recurrence of the Fubini polynomials: F := proc(n) option remember; if n = 0 then return 1 fi; expand(add(binomial(n, k)*F(n-k)*x, k = 1..n)) end: a := n -> (-2)^n*subs(x = -1/2, F(n)): seq(a(n), n = 0..30); # Peter Luschny, May 21 2021
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Mathematica
a[m_] := Sum[(-2)^(m - k) k! StirlingS2[m, k], {k, 0, m}] (* Peter Luschny, Apr 29 2009 *) poly[q_] := 2^(q-n)/n*FunctionExpand[Sum[Binomial[n, k]*k^q, {k, 0, n}]]; T[q_, r_] := First[Take[CoefficientList[poly[q], n], {r+1, r+1}]]; Table[Sum[T[x, k]*(-1)^k, {k, 0, x-1}], {x, 1, 16}] (* John M. Campbell, Nov 16 2011 *) f[n_] := (-1)^n 2^(n+1) PolyLog[-n, -1]; f[0] = -f[0]; Array[f, 27, 0] (* Robert G. Wilson v, Jun 28 2012 *) -
PARI
a(n)=if(n==0,1,sum(k=0,n-1,(-1)^(k)*binomial(n-1,k)*a(n-1-k)*a(k)))
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PARI
a(n)=local(X=x+x*O(x^n));n!*polcoeff(exp(X)/cosh(X),n)
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PARI
a(n)=polcoeff(sum(m=0,n,m!*x^m/prod(k=1,m,1+2*k*x+x*O(x^n))),n) \\ Paul D. Hanna, Jul 20 2011
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PARI
{a(n) = local(A); if( n<0, 0, A = x * O(x^n); n! * polcoeff( 1 + sinh(x + A) / cosh(x + A), n))} /* Michael Somos, Jan 16 2012 */ -
PARI
a(n)=local(A=1+x);for(i=1,n,A=sum(k=0,n,intformal(subst(A,x,-x)+x*O(x^n))^k/k!));n!*polcoeff(A,n) for(n=0,30,print1(a(n),", ")) \\ Paul D. Hanna, Nov 25 2013
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Python
from sympy import bernoulli def A155585(n): return (((2<<(m:=n+1))-2)*bernoulli(m)<
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Sage
def A155585(n) : if n == 0 : return 1 return add(add((-1)^(j+1)*binomial(n+1,k-j)*j^n for j in (0..k)) for k in (1..n)) [A155585(n) for n in (0..26)] # Peter Luschny, Jul 23 2012
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Sage
def A155585_list(n): # Akiyama-Tanigawa algorithm A = [0]*(n+1); R = [] for m in range(n+1) : d = divmod(m+3, 4) A[m] = 0 if d[1] == 0 else (-1)^d[0]/2^(m//2) for j in range(m, 0, -1) : A[j - 1] = j * (A[j - 1] - A[j]) R.append(A[0]) return R A155585_list(30) # Peter Luschny, Mar 09 2014
Formula
E.g.f.: exp(x)*sech(x) = exp(x)/cosh(x). (See A009006.) - Paul Barry, Mar 15 2006
Sequence of absolute values is A009006 (e.g.f. 1+tan(x)).
O.g.f.: Sum_{n>=0} n! * x^n / Product_{k=1..n} (1 + 2*k*x). - Paul D. Hanna, Jul 20 2011
a(n) = 2^n*E_{n}(1) where E_{n}(x) are the Euler polynomials. - Peter Luschny, Jan 26 2009
a(n) = EL_{n}(-1) where EL_{n}(x) are the Eulerian polynomials. - Peter Luschny, Aug 03 2010
a(n+1) = (4^n-2^n)*B_n(1)/n, where B_{n}(x) are the Bernoulli polynomials (B_n(1) = B_n for n <> 1). - Peter Luschny, Apr 22 2009
G.f.: 1/(1-x+x^2/(1-x+4*x^2/(1-x+9*x^2/(1-x+16*x^2/(1-...))))) (continued fraction). - Paul Barry, Mar 30 2010
G.f.: -log(x/(exp(x)-1))/x = Sum_{n>=0} a(n)*x^n/(2^(n+1)*(2^(n+1)-1)*n!). - Vladimir Kruchinin, Nov 05 2011
E.g.f.: exp(x)/cosh(x) = 2/(1+exp(-2*x)) = 2/(G(0) + 1); G(k) = 1 - 2*x/(2*k + 1 - x*(2*k+1)/(x - (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 10 2011
E.g.f. is x(t,1) + y(t,1) where x(t,a) and y(t,a) satisfy y(t,a)^2 = (a^2 - x(t,a)^2) / (1 - a^2 * x(t,a)^2) and dx(t,a) / dt = y(t,a) * (1 - a * x(t,a)^2) and are the elliptic functions of Edwards. - Michael Somos, Jan 16 2012
E.g.f.: 1/(1 - x/(1+x/(1 - x/(3+x/(1 - x/(5+x/(1 - x/(7+x/(1 - x/(9+x/(1 +...))))))))))), a continued fraction. - Paul D. Hanna, Feb 11 2012
E.g.f. satisfies: A(x) = Sum_{n>=0} Integral( A(-x) dx )^n / n!. - Paul D. Hanna, Nov 25 2013
a(n) = -2^(n+1)*Li_{-n}(-1). - Peter Luschny, Jun 28 2012
a(n) = Sum_{k=1..n} Sum_{j=0..k} (-1)^(j+1)*binomial(n+1,k-j)*j^n for n > 0. - Peter Luschny, Jul 23 2012
From Sergei N. Gladkovskii, Oct 25 2012 to Dec 16 2013: (Start)
Continued fractions:
G.f.: 1 + x/T(0) where T(k) = 1 + (k+1)*(k+2)*x^2/T(k+1).
E.g.f.: exp(x)/cosh(x) = 1 + x/S(0) where S(k) = (2*k+1) + x^2/S(k+1).
E.g.f.: 1 + x/(U(0)+x) where U(k) = 4*k+1 - x/(1 + x/(4*k+3 - x/(1 + x/U(k+1)))).
E.g.f.: 1 + tanh(x) = 4*x/(G(0)+2*x) where G(k) = 1 - (k+1)/(1 - 2*x/(2*x + (k+1)^2/G(k+1)));
G.f.: 1 + x/G(0) where G(k) = 1 + 2*x^2*(2*k+1)^2 - x^4*(2*k+1)*(2*k+2)^2*(2*k+3)/G(k+1) (due to Stieltjes).
E.g.f.: 1 + x/(G(0) + x) where G(k) = 1 - 2*x/(1 + (k+1)/G(k+1)).
G.f.: 2 - 1/Q(0) where Q(k) = 1 + x*(k+1)/( 1 - x*(k+1)/Q(k+1)).
G.f.: 2 - 1/Q(0) where Q(k) = 1 + x*k^2 + x/(1 - x*(k+1)^2/Q(k+1)).
G.f.: 1/Q(0) where Q(k) = 1 - 2*x + x*(k+1)/(1-x*(k+1)/Q(k+1)).
G.f.: 1/Q(0) where Q(k) = 1 - x*(k+1)/(1 + x*(k+1)/Q(k+1)).
E.g.f.: 1 + x*Q(0) where Q(k) = 1 - x^2/( x^2 + (2*k+1)*(2*k+3)/Q(k+1)).
G.f.: 2 - T(0)/(1+x) where T(k) = 1 - x^2*(k+1)^2/(x^2*(k+1)^2 + (1+x)^2/T(k+1)).
E.g.f.: 1/(x - Q(0)) where Q(k) = 4*k^2 - 1 + 2*x + x^2*(2*k-1)*(2*k+3)/Q(k+1). (End)
G.f.: 1 / (1 - b(1)*x / (1 - b(2)*x / (1 - b(3)*x / ... ))) where b = A001057. - Michael Somos, Jan 03 2013
From Paul Curtz, Mar 06 2014: (Start)
a(2n) = A000007(n).
a(2n+1) = (-1)^n*A000182(n+1).
a(n) is the binomial transform of A122045(n).
a(n) is the Akiyama-Tanigawa transform applied to 1, 0, -1/2, -1/2, -1/4, 0, ... = A046978(n+3)/A016116(n). (End)
a(n) = 2^(2*n+1)*(zeta(-n,1/2) - zeta(-n, 1)), where zeta(a, z) is the generalized Riemann zeta function. - Peter Luschny, Mar 11 2015
a(n) = 2^(n + 1)*(2^(n + 1) - 1)*Bernoulli(n + 1, 1)/(n + 1). (From Bill Gosper, Oct 28 2015) - N. J. A. Sloane, Oct 28 2015 [See the above comment from Peter Luschny, Apr 22 2009.]
a(n) = -(n mod 2)*((-1)^n + Sum_{k=1..n-1} (-1)^k*C(n,k)*a(n-k)) for n >= 1. - Peter Luschny, Jun 01 2016
a(n) = (-2)^n*F_{n}(-1/2), where F_{n}(x) is the Fubini polynomial. - Peter Luschny, May 21 2021
Extensions
New name from Peter Luschny, Mar 12 2015
Comments