A009843 E.g.f. x/cos(x) (odd powers only).

1, 3, 25, 427, 12465, 555731, 35135945, 2990414715, 329655706465, 45692713833379, 7777794952988025, 1595024111042171723, 387863354088927172625, 110350957750914345093747, 36315529600705266098580265, 13687860690719716241164167451, 5858139922124796551409938058945

Offset

0,2

Comments

Expanding x/cosh(x) gives alternated signed values at odd positions.

Related to the formulas sum(k>0,sin(kx)/k^(2n+1))=(-1)^(n+1)/2*x^(2n+1)/(2n+1)!*sum(i=0,2n,(2Pi/x)^i*B(i)*C(2n+1,i)) and if x=Pi/2 sum(k>0,(-1)^(k+1)/k^(2n+1))=(-1)^n*E(2n)*Pi^(2n+1)/2^(2n+2)/(2n)!. - Benoit Cloitre, May 01 2002

Links

Table of n, a(n) for n=0..16.

R. P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014.

R. B. Brent, Generalizing Tuenter's Binomial Sums, J. Int. Seq. 18 (2015) # 15.3.2.

Bishal Deb and Alan D. Sokal, Classical continued fractions for some multivariate polynomials generalizing the Genocchi and median Genocchi numbers, arXiv:2212.07232 [math.CO], 2022. See p. 12.

Peter Luschny, The lost Bernoulli numbers.

Wikipedia, Bernoulli Polynomials

Formula

a(n) = (2n+1)*A000364(n) = sum(i=0, 2n, B(i)*C(2n+1, i)*4^i)=(2n+1)*E(2n) where B(i) are the Bernoulli numbers, C(2n, i) the binomial numbers and E(2n) the Euler numbers. - Benoit Cloitre, May 01 2002

Recurrence: a(n) = -(-1)^n*Sum[i=0..n-1, (-1)^i*a(i)*C(2n+1, 2i+1) ]. - Ralf Stephan, Feb 24 2005

a(n) = 4^n |E_{2n}(1/2)+E_{2n}(1)| (2n+1) for n > 0; E_{n}(x) Euler polynomial. - Peter Luschny, Nov 25 2010

a(n) = (2*n+1)! * [x^(2*n+1)] x/cos(x).

From Sergei N. Gladkovskii, Nov 15 2011, Oct 19 2012, Nov 10 2012, Jan 14 2013, Apr 10 2013, Oct 13 2013, Dec 01 2013: (Start) Continued fractions:

E.g.f.: x / cos(x) = x+x^3/Q(0); Q(k) = 8k+2-x^2/(1+(2k+1)*(2k+2)/Q(k+1)).

E.g.f.: x + x^3/U(0) where U(k) = (2*k+1)*(2*k+2) - x^2 + x^2*(2*k+1)*(2*k+2)/U(k+1).

G.f.: x/T(0) where T(k) = 1 + x^2 - x^2*(2*k+2)^2/(1 - x^2*(2*k+2)^2/T(k+1).

G.f.: 1/G(0) where G(k) = 1 - x*(8*k^2+8*k+3)-16*x^2*(k+1)^4/G(k+1).

E.g.f.: 2*x/(Q(0) + 1) where Q(k)= 1 - x/(2*k+1)/(2*k+2)/(1 - 1/(1 + 1/Q(k+1))).

Let A(x) = S_{n>=0}a(n)*x^n/(2*n+1)! then A(x) = 1 + Q(0)*x/(2-x) where Q(k) = 1 - x*(2*k+1)*(2*k+2)/(x*(2*k+1)*(2*k+2) + ((2*k+1)*(2*k+2) - x)*((2*k+3)*(2*k+4) - x)/Q(k+1)).

G.f.: T(0)/(1-3*x) where T(k) = 1 - 16*x^2*(k+1)^4/(16*x^2*(k+1)^4 - (1 - x*(8*k^2 +8*k+3)) *(1 - x*(8*k^2+24*k+19))/T(k+1)).

G.f.: 1/(T(0) where T(k) = 1 + x - x*(2*k+2)^2/(1 - x*(2*k+2)^2/T(k+1). (End)

a(n) = (-1)^n*2^(4*n+1)*(2*n+1)*(zeta(-2*n,1/4)-zeta(-2*n,3/4)). - Peter Luschny, Jul 22 2013

From Peter Bala, Mar 02 2015: (Start)

a(n) = (-1)^(n+1)*4^(2*n + 1)*B(2*n + 1,1/4), where B(n,x) denotes the n-th Bernoulli polynomial. Cf. A002111, A069852 and A069994.

Conjecturally, a(n) = the unsigned numerator of B(2*n+1,1/4).

G.f. for signed version of sequence: Sum_{n >= 0} { 1/(n + 1) * Sum_{k = 0..n} (-1)^k*binomial(n,k)/( (1 - (4*k + 1)*x)*(1 - (4*k + 3)*x) ) } = 1 - 3*x^2 + 25*x^4 - 427*x^6 + .... (End)

a(n) ~ (2*n+1)! * 2^(2*n+2)/Pi^(2*n+1). - Vaclav Kotesovec, Jul 04 2016

G.f.: 1/(1 + x - 4*x/(1 - 4*x/(1 + x - 16*x/(1 - 16*x/(1 + x - 36*x/(1 - 36*x/(1 + x - ...))))))). Cf. A005439. - Peter Bala, May 07 2017

Example

x/cos(x) = x + 1/2*x^3 + 5/24*x^5 + 61/720*x^7 + 277/8064*x^9 + ...

Maple

seq((2*i+1)!*coeff(series(x/cos(x), x, 32), x, 2*i+1), i=0..13);
A009843 := n -> (-1)^n*(2*n+1)*euler(2*n): # Peter Luschny

Mathematica

c = CoefficientList[Series[1/MittagLefflerE[2, z^2], {z, 0, 40}], z]; Table[(-1)^n* Factorial[2*n+1]*c[[2*n+1]], {n, 0, 16}] (* Peter Luschny, Jul 03 2016 *)

Prog

(PARI) for(n=0, 25, print1(sum(i=0, 2*n+1, binomial(2*n+1, i)*bernfrac(i)*4^i), ", "))
(PARI) a(n)=subst(bernpol(2*n+1), 'x, 1/4)*4^(2*n+1)*(-1)^(n+1) \\ Charles R Greathouse IV, Dec 10 2014
(Python) # The objective of this implementation is efficiency.
# n -> [a(0), a(1), ..., a(n)] for n > 0.
def A009843_list(n):
    S = [0 for i in range(n+1)]
    S[0] = 1
    for k in range(1, n+1):
        S[k] = k*S[k-1]
    for k in range(1, n+1):
        for j in range(k, n+1):
            S[j] = (j-k)*S[j-1]+(j-k+1)*S[j]
        S[k] = (2*k+1)*S[k]
    return S
print(A009843_list(10)) # Peter Luschny, Aug 09 2011

Crossrefs

Bisection of A009391, A009392, A065619, A083008.

Cf. A099028. Cf. A002111, A069852, A069994.

Sequence in context: A074708 A323217 A160143 * A366007 A182962 A223076

Adjacent sequences: A009840 A009841 A009842 * A009844 A009845 A009846

Keyword

nonn,easy

Author

R. H. Hardin

Extensions

Extended and signs tested by Olivier Gérard, Mar 15 1997

Status

approved