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A155585 a(n) = 2^n*E(n, 1) where E(n, x) are the Euler polynomials. 28

%I

%S 1,1,0,-2,0,16,0,-272,0,7936,0,-353792,0,22368256,0,-1903757312,0,

%T 209865342976,0,-29088885112832,0,4951498053124096,0,

%U -1015423886506852352,0,246921480190207983616,0,-70251601603943959887872,0,23119184187809597841473536,0

%N a(n) = 2^n*E(n, 1) where E(n, x) are the Euler polynomials.

%C Previous name was: a(n) = Sum_{k=0..n-1} (-1)^(k)*C(n-1,k)*a(n-1-k)*a(k) for n>0 with a(0)=1.

%C Factorials have a similar recurrence: f(n) = Sum_{k=0..n-1} C(n-1,k)*f(n-1-k)*f(k), n>0.

%C Related to A102573: letting T(q,r) be the coefficient of n^(r+1) in the polynomial 2^(q-n)/n times Sum_{k=0..n} binomial(n,k)*k^q, then A155585(x) = Sum_{k=0..x-1} T(x,k)*(-1)^k. See Mathematica code below. - _John M. Campbell_, Nov 16 2011

%C For the difference table and the relation to the Seidel triangle see A239005. - _Paul Curtz_, Mar 06 2014

%C From _Tom Copeland_, Sep 29 2015: (Start)

%C Let z(t) = 2/(e^(2t)+1) = 1 + tanh(-t) = e.g.f.(-t) for this sequence = 1 - t + 2 t^3/3! - 16 t^5/5! + ... .

%C dlog(z(t))/dt = -z(-t), so the raising operators that generate Appell polynomials associated with this sequence, A081733, and its reciprocal, A119468, contain z(-d/dx) = e.g.f.(d/dx) as the differential operator component.

%C dz(t)/dt = z*(z-2), so the assorted relations to a Ricatti equation, the Eulerian numbers A008292, and the Bernoulli numbers in the Rzadkowski link hold.

%C From Michael Somos's formula below (drawing on the Edwards link), y(t,1)=1 and x(t,1) = (1-e^(2t))/(1+e^(2t)), giving z(t) = 1 + x(t,1). Compare this to the formulas in my list in A008292 (Sep 14 2014) with a=1 and b=-1,

%C A) A(t,1,-1) = A(t) = -x(t,1) = (e^(2t)-1)/(1+e^(2t)) = tanh(t) = t + -2 t^3/3! + 16 t^5/5! + -272 t^7/7! + ... = e.g.f.(t) - 1 (see A000182 and A000111)

%C B) Ainv(t) = log((1+t)/(1-t))/2 = tanh^(-1)(t) = t + t^3/3 + t^5/5 + ..., the compositional inverse of A(t)

%C C) dA/dt = (1-A^2), relating A(t) to a Weierstrass elliptic function

%C D) ((1-t^2)d/dt)^n t evaluated at t=0, a generator for the sequence A(t)

%C F) FGL(x,y)= (x+y)/(1+xy) = A(Ainv(x) + Ainv(y)), a related formal group law corresponding to the Lorentz FGL (Lorentz transformation--addition of parallel velocities in special relativity) and the Atiyah-Singer signature and the elliptic curve (1-t^2)s = t^3 in Tate coordinates according to the Lenart and Zainoulline link and the Buchstaber and Bunkova link (pp. 35-37) in A008292.

%C A133437 maps the reciprocal odd natural numbers through the refined faces of associahedra to a(n).

%C A145271 links the differential relations to the geometry of flow maps, vector fields, and thereby formal group laws. See Mathworld for links of tanh to other geometries and statistics.

%C Since the a(n) are related to normalized values of the Bernoulli numbers and the Riemann zeta and Dirichlet eta functions, there are links to Witten's work on volumes of manifolds in two-dimensional quantum guage theories and the Kervaire-Milnor formula for homotopy groups of hyperspheres (see my link below).

%C See A101343, A111593 and A059419 for this and the related generator (1 + t^2) d/dt and associated polynomials. (End)

%H Vincenzo Librandi, <a href="/A155585/b155585.txt">Table of n, a(n) for n = 0..200</a>

%H P. Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Barry1/barry97r2.html">Riordan Arrays, Orthogonal Polynomials as Moments, and Hankel Transforms</a>, J. Int. Seq. 14 (2011) # 11.2.2, example 28.

%H Tom Copeland, <a href="https://tcjpn.wordpress.com/2015/10/12/the-elliptic-lie-triad-kdv-and-ricattt-equations-infinigens-and-elliptic-genera/">The Elliptic Lie Triad: Ricatti and KdV Equations, Infinigens, and Elliptic Genera</a>

%H T. Copeland, <a href="http://tcjpn.wordpress.com/2015/10/04/the-kervaire-milnor-formula/">The Kervaire-Milnor formula</a>

%H H. M. Edwards, <a href="http://dx.doi.org/10.1090/S0273-0979-07-01153-6">A normal form for elliptic curves</a>, Bull. Amer. Math. Soc. (N.S.) 44 (2007), no. 3, 393-422. see section 12, pp. 410-411.

%H G. Rzadkowski, <a href="http://dx.doi.org/10.1142/S1402925110000635">Bernoulli numbers and solitons-revisited</a>, Jrn. Nonlinear Math. Physics, 1711, pp. 121-126.

%F E.g.f.: exp(x)*sech(x) = exp(x)/cosh(x). (See A009006.) - _Paul Barry_, Mar 15 2006

%F Sequence of absolute values is A009006 (e.g.f. 1+tan(x)).

%F O.g.f.: Sum_{n>=0} n! * x^n / Product_{k=1..n} (1 + 2*k*x). - _Paul D. Hanna_, Jul 20 2011

%F a(n) = 2^n*E_{n}(1) where E_{n}(x) are the Euler polynomials. - _Peter Luschny_, Jan 26 2009

%F a(n) = EL_{n}(-1) where EL_{n}(x) are the Eulerian polynomials. - _Peter Luschny_, Aug 03 2010

%F a(n) = (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

%F 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

%F 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

%F 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

%F 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

%F 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

%F E.g.f. satisfies: A(x) = Sum_{n>=0} Integral( A(-x) dx )^n / n!. - _Paul D. Hanna_, Nov 25 2013

%F a(n) = -2^(n+1)*Li_{-n}(-1). - _Peter Luschny_, Jun 28 2012

%F 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

%F G.f.: 1 + x/T(0) where T(k) = 1 + (k+1)*(k+2)*x^2/T(k+1)); (continued fraction, 1-step). - _Sergei N. Gladkovskii_, Oct 25 2012

%F E.g.f.: exp(x)/cosh(x)= 1 + x/S(0) where S(k)= (2*k+1) + x^2/S(k+1); (continued fraction, 1-step). - _Sergei N. Gladkovskii_, Oct 25 2012

%F 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))));(continued fraction, 4-step). - _Sergei N. Gladkovskii_, Nov 08 2012

%F 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) )); (continued fraction). - _Sergei N. Gladkovskii_, Dec 07 2012

%F G.f.: 1 / (1 - b(1)*x / (1 - b(2)*x / (1 - b(3)*x / ... ))) where b = A001057. - _Michael Somos_, Jan 03 2013

%F 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); (continued fraction due to Stieltjes). - _Sergei N. Gladkovskii_, Jan 13 2013

%F E.g.f.: 1 + x/( G(0) + x ) where G(k) = 1 - 2*x/(1 + (k+1)/G(k+1) ); (continued fraction). - _Sergei N. Gladkovskii_, Feb 02 2013

%F G.f.: 2 - 1/Q(0) where Q(k) = 1 + x*(k+1)/( 1 - x*(k+1)/Q(k+1) ); (continued fraction). - _Sergei N. Gladkovskii_, Apr 02 2013

%F G.f.: 2 - 1/Q(0), where Q(k)= 1 + x*k^2 + x/(1 - x*(k+1)^2/Q(k+1)); (continued fraction). - _Sergei N. Gladkovskii_, Apr 17 2013

%F 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 20 2013

%F G.f.: 1/Q(0), where Q(k)= 1 - x*(k+1)/(1 + x*(k+1)/Q(k+1)); (continued fraction). - _Sergei N. Gladkovskii_, May 20 2013

%F E.g.f.: 1 + x*Q(0), where Q(k) = 1 - x^2/( x^2 + (2*k+1)*(2*k+3)/Q(k+1) ); (continued fraction). - _Sergei N. Gladkovskii_, Oct 10 2013

%F 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) ); (continued fraction). - _Sergei N. Gladkovskii_, Nov 11 2013

%F 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) ; (continued fraction). - _Sergei N. Gladkovskii_, Dec 16 2013

%F From _Paul Curtz_, Mar 06 2014: (Start)

%F a(2n) = A000007(n).

%F a(2n+1) = (-1)^n*A000182(n+1).

%F a(n) is the binomial transform of A122045(n).

%F a(n) is the row sum of A081658. For fractional Euler numbers see A238800.

%F a(n) + A122045(n) = 2, 1, -1, -2, 5, 16,... = -A163982(n).

%F a(n) - A122045(n) = -A163747(n).

%F a(n) is the Akiyama-Tanigawa transform applied to 1, 0, -1/2, -1/2, -1/4, 0,... = A046978(n+3)/A016116(n). (End)

%F 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

%F 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

%F 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

%e 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).

%e O.g.f.: 1 + x - 2*x^3 + 16*x^5 - 272*x^7 + 7936*x^9 - 353792*x^11 +- ...

%e 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)) + ...

%p A155585 := n -> 2^n*euler(n, 1): # _Peter Luschny_, Jan 26 2009

%p 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

%t a[m_] := Sum[(-2)^(m - k) k! StirlingS2[m, k], {k, 0, m}] (* _Peter Luschny_, Apr 29 2009 *)

%t 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 *)

%t 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 *)

%o (PARI) a(n)=if(n==0,1,sum(k=0,n-1,(-1)^(k)*binomial(n-1,k)*a(n-1-k)*a(k)))

%o (PARI) a(n)=local(X=x+x*O(x^n));n!*polcoeff(exp(X)/cosh(X),n)

%o (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

%o (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 */

%o (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)

%o for(n=0,30,print1(a(n),", ")) \\ _Paul D. Hanna_, Nov 25 2013

%o (Sage)

%o def A155585(n) :

%o if n == 0 : return 1

%o return add(add((-1)^(j+1)*binomial(n+1,k-j)*j^n for j in (0..k)) for k in (1..n))

%o [A155585(n) for n in (0..26)] # _Peter Luschny_, Jul 23 2012

%o (Sage)

%o def A155585_list(n) : Akiyama-Tanigawa algorithm

%o A = [0]*(n+1); R = []

%o for m in range(n+1) :

%o d = divmod(m+3, 4)

%o A[m] = 0 if d[1] == 0 else (-1)^d[0]/2^(m//2)

%o for j in range(m, 0, -1) :

%o A[j - 1] = j * (A[j - 1] - A[j])

%o R.append(A[0])

%o return R

%o A155585_list(30) # _Peter Luschny_, Mar 09 2014

%Y Cf. A001057, A009006, A102573.

%Y Equals row sums of A119879. - _Johannes W. Meijer_, Apr 20 2011

%Y Cf. A081733, A119468, A008292, A000182, A000111, A101343, A111593, A059419, A133437, A145271.

%Y (-1)^n*a(n) are the alternating row sums of A123125. - _Wolfdieter Lang_, Jul 12 2017

%K sign,easy

%O 0,4

%A _Paul D. Hanna_, Jan 24 2009

%E New name from _Peter Luschny_, Mar 12 2015

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Last modified February 19 03:22 EST 2018. Contains 299330 sequences. (Running on oeis4.)