%I #44 Oct 07 2019 02:40:20

%S 1,1,1,1,5,2,61,17,277,62,50521,1382,540553,21844,199360981,929569,

%T 3878302429,6404582,2404879675441,443861162,14814847529501,

%U 18888466084,69348874393137901,113927491862,238685140977801337,58870668456604,4087072509293123892361

%N Numerators of the coefficients in the Taylor expansion of sec(x) + tan(x) around x=0.

%H Seiichi Manyama, <a href="/A099612/b099612.txt">Table of n, a(n) for n = 0..487</a>

%H L. Euler, <a href="http://arXiv.org/abs/math.HO/0506415">On the sums of series of reciprocals</a>, par. 13, arXiv:math/0506415 [math.HO], 2005-2008.

%H L. Euler, <a href="http://eulerarchive.maa.org/pages/E041.html">De summis serierum reciprocarum</a>, E41, par. 13, Euler Archive.

%H Peter Luschny, <a href="http://oeis.org/wiki/User:Peter_Luschny/TheLostBernoulliNumbers#Formulas_for_Euler.27s_Zeta_numbers">The Euler-Bernoulli diamond and the lost Bernoulli numbers</a>.

%F Let R(x) = (-1)^floor(x/2)*(4^(x+1)-2^(x+1))*((HurwitzZeta(-x,3/4) - HurwitzZeta(-x,1/4)) /(2^(-x)-2)-Zeta(-x))/Gamma(x+1) then a(n) = numerator(R(n)) and A099617(n) = denominator(R(n)) for n>=1. - _Peter Luschny_, Aug 25 2015

%F Let F(x,t) = exp(-I*t*x)*(1+(exp(exp(I*t))-1)/(exp(2*exp(I*t))+1)) and r(x) = ((cos(x*Pi/2)+sin(x*Pi/2))/Pi)*Integral_{t=0..2*Pi} F(x,t) then a(n) = numerator(r(n)) and A099617(n) = denominator(r(n)) for n>=1. - _Peter Luschny_, Aug 25 2015

%F a(n)/A099617(n) = A000111(n)/n!. - _Seiichi Manyama_, Jan 27 2017

%F From _Peter Luschny_, Aug 03 2017: (Start)

%F a(n) = numerator(2*i^(n+1)*PolyLog(-n, -i)/n!) for n>0.

%F a(n) = numerator(2^n*|Euler(n,1/2) - Euler(n,1)|/n!) for n>0 where Euler(n,x) are the Euler polynomials. (End)

%F Conjecture: For n >= 0, (-1)^n * a(n+1) is the numerator of the n-th term of the Taylor expansion of 1/(1 + sin(x)) around x = 0. [This is based on the fact that (sec(x) + tan(x))' = 1/(1 + sin(-x)). _Clark Kimberling_ in A279107 states my conjecture as a fact, but no proof or reference is given.] - _Petros Hadjicostas_, Oct 06 2019

%e 1 + x + 1/2*x^2 + 1/3*x^3 + 5/24*x^4 + 2/15*x^5 + 61/720*x^6 + 17/315*x^7 + ...

%e 1, 1, 1/2, 1/3, 5/24, 2/15, 61/720, 17/315, 277/8064, 62/2835, 50521/3628800, 1382/155925, 540553/95800320, ... = A099612/A099617

%p R := n -> (cos(n*Pi/2)+sin(n*Pi/2))*(4^(n+1)-2^(n+1))*((Zeta(0,-n,3/4)-Zeta(0,-n,1/4))/(2^(-n)-2)-Zeta(-n))/GAMMA(n+1):

%p [1, seq(numer(R(n)), n=1..19)]; # _Peter Luschny_, Aug 25 2015

%p # From _Peter Luschny_, Aug 03 2017: (Start) By recurrence:

%p S := proc(n, k) option remember; if k = 0 then `if`(n = 0, 1, 0) else

%p S(n, k - 1) + S(n - 1, n - k) fi end: A099612 := n -> numer(S(n, n)/n!):

%p seq(A099612(n), n=0..26);

%p # or evaluating polynomials at -i:

%p P := proc(n, x) local k, j; add(add((-1)^j*2^(-k)*binomial(k,j)*(k-2*j)^n*

%p x^(n-k), j=0..k), k=0..n) end: R := n -> `if`(n = 0, 1, P(n-1, -I)/ n!):

%p seq(numer(R(n)), n=0..26);

%p # or with the Euler polynomials:

%p ep := n -> `if`(n=0,1,2^n*(euler(n,1/2)-euler(n,1))*(-1)^iquo(n+1,2)):

%p a := n -> numer(ep(n)/n!): seq(a(n), n=0..26); # (End)

%t nn = 26; Numerator[CoefficientList[Series[Sec[x] + Tan[x], {x, 0, nn}], x]] (* _T. D. Noe_, Jul 24 2013 *)

%t Table[If[n==0,1,2 I^(n+1) PolyLog[-n, -I] / n!], {n,0,26}] // Numerator (* _Peter Luschny_, Aug 03 2017 *)

%t Table[(1 + Mod[n,2])LerchPhi[(-1)^(n+1), n+1, 1/2]/Pi^(n+1), {n, 0, 26}] // Numerator (* _Peter Luschny_, Aug 03 2017 *)

%Y Cf. A000111, A099617, A279107.

%K nonn,frac

%O 0,5

%A _N. J. A. Sloane_, Nov 18 2004