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
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.
Sylvie Corteel, Alexander Lazar, and Anna Vanden Wyngaerd, Decorated square paths at q = -1, arXiv:2408.10640 [math.CO], 2024. See p. 3.
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
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
Extended and signs tested by Olivier Gérard, Mar 15 1997
STATUS
approved