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A155585
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a(n) = 2^n*E(n, 1) where E(n, x) are the Euler polynomials.
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52
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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
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OFFSET
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0,4
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COMMENTS
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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.
Factorials have a similar recurrence: f(n) = Sum_{k=0..n-1} C(n-1,k)*f(n-1-k)*f(k), n > 0.
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
For the difference table and the relation to the Seidel triangle see A239005. - Paul Curtz, Mar 06 2014
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! + ... .
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.
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.
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,
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)
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) dA/dt = (1-A^2), relating A(t) to a Weierstrass elliptic function
D) ((1-t^2)d/dt)^n t evaluated at t=0, a generator for the sequence A(t)
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.
A133437 maps the reciprocal odd natural numbers through the refined faces of associahedra to a(n).
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.
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 gauge theories and the Kervaire-Milnor formula for homotopy groups of hyperspheres (see my link below).
See A101343, A111593 and A059419 for this and the related generator (1 + t^2) d/dt and associated polynomials. (End)
With the exception of the first term (1), entries are the alternating sums of the rows of the Eulerian triangle, A008292. - Gregory Gerard Wojnar, Sep 29 2018
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LINKS
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FORMULA
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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) = 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
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
a(n) is the binomial transform of A122045(n).
a(n) is the row sum of A081658. For fractional Euler numbers see A238800.
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
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EXAMPLE
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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)) + ...
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MAPLE
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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)):
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MATHEMATICA
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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 *)
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PROG
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(PARI) a(n)=if(n==0, 1, sum(k=0, n-1, (-1)^(k)*binomial(n-1, k)*a(n-1-k)*a(k)))
(PARI) a(n)=local(X=x+x*O(x^n)); n!*polcoeff(exp(X)/cosh(X), n)
(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
(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)
(Sage)
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))
(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
(Python)
from sympy import bernoulli
def A155585(n): return (((2<<(m:=n+1))-2)*bernoulli(m)<<m-2)//(m>>1) if n&1 else (0 if n else 1) # Chai Wah Wu, Apr 14 2023
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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