%I #103 Apr 14 2023 14:11:15
%S 1,3,25,427,12465,555731,35135945,2990414715,329655706465,
%T 45692713833379,7777794952988025,1595024111042171723,
%U 387863354088927172625,110350957750914345093747,36315529600705266098580265,13687860690719716241164167451,5858139922124796551409938058945
%N E.g.f. x/cos(x) (odd powers only).
%C Expanding x/cosh(x) gives alternated signed values at odd positions.
%C 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
%H R. P. Brent, <a href="http://arxiv.org/abs/1407.3533">Generalising Tuenter's binomial sums</a>, arXiv:1407.3533 [math.CO], 2014.
%H R. B. Brent, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Brent/brent5.html">Generalizing Tuenter's Binomial Sums</a>, J. Int. Seq. 18 (2015) # 15.3.2.
%H Bishal Deb and Alan D. Sokal, <a href="https://arxiv.org/abs/2212.07232">Classical continued fractions for some multivariate polynomials generalizing the Genocchi and median Genocchi numbers</a>, arXiv:2212.07232 [math.CO], 2022. See p. 12.
%H Peter Luschny, <a href="http://oeis.org/wiki/User:Peter_Luschny/TheLostBernoulliNumbers">The lost Bernoulli numbers.</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Bernoulli_polynomials">Bernoulli Polynomials</a>
%F 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
%F Recurrence: a(n) = -(-1)^n*Sum[i=0..n-1, (-1)^i*a(i)*C(2n+1, 2i+1) ]. - _Ralf Stephan_, Feb 24 2005
%F 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
%F a(n) = (2*n+1)! * [x^(2*n+1)] x/cos(x).
%F 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:
%F 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)).
%F 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).
%F 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).
%F 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).
%F 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))).
%F 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)).
%F 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)).
%F 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)
%F 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
%F From _Peter Bala_, Mar 02 2015: (Start)
%F 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.
%F Conjecturally, a(n) = the unsigned numerator of B(2*n+1,1/4).
%F 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)
%F a(n) ~ (2*n+1)! * 2^(2*n+2)/Pi^(2*n+1). - _Vaclav Kotesovec_, Jul 04 2016
%F 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
%e x/cos(x) = x + 1/2*x^3 + 5/24*x^5 + 61/720*x^7 + 277/8064*x^9 + ...
%p seq((2*i+1)!*coeff(series(x/cos(x),x,32),x,2*i+1),i=0..13);
%p A009843 := n -> (-1)^n*(2*n+1)*euler(2*n): # _Peter Luschny_
%t 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 *)
%o (PARI) for(n=0,25,print1(sum(i=0,2*n+1,binomial(2*n+1,i)*bernfrac(i)*4^i),","))
%o (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
%o (Python) # The objective of this implementation is efficiency.
%o # n -> [a(0), a(1), ..., a(n)] for n > 0.
%o def A009843_list(n):
%o S = [0 for i in range(n+1)]
%o S[0] = 1
%o for k in range(1, n+1):
%o S[k] = k*S[k-1]
%o for k in range(1, n+1):
%o for j in range(k, n+1):
%o S[j] = (j-k)*S[j-1]+(j-k+1)*S[j]
%o S[k] = (2*k+1)*S[k]
%o return S
%o print(A009843_list(10)) # _Peter Luschny_, Aug 09 2011
%Y Bisection of A009391, A009392, A065619, A083008.
%Y Cf. A099028. Cf. A002111, A069852, A069994.
%K nonn,easy
%O 0,2
%A _R. H. Hardin_
%E Extended and signs tested by _Olivier GĂ©rard_, Mar 15 1997
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