OFFSET
0,5
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..487
L. Euler, On the sums of series of reciprocals, par. 13, arXiv:math/0506415 [math.HO], 2005-2008.
L. Euler, De summis serierum reciprocarum, E41, par. 13, Euler Archive.
Peter Luschny, The Euler-Bernoulli diamond and the lost Bernoulli numbers.
FORMULA
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
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
From Peter Luschny, Aug 03 2017: (Start)
a(n) = numerator(2*i^(n+1)*PolyLog(-n, -i)/n!) for n>0.
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)
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
EXAMPLE
MAPLE
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):
[1, seq(numer(R(n)), n=1..19)]; # Peter Luschny, Aug 25 2015
# From Peter Luschny, Aug 03 2017: (Start) By recurrence:
S := proc(n, k) option remember; if k = 0 then `if`(n = 0, 1, 0) else
S(n, k - 1) + S(n - 1, n - k) fi end: A099612 := n -> numer(S(n, n)/n!):
seq(A099612(n), n=0..26);
# or evaluating polynomials at -i:
P := proc(n, x) local k, j; add(add((-1)^j*2^(-k)*binomial(k, j)*(k-2*j)^n*
x^(n-k), j=0..k), k=0..n) end: R := n -> `if`(n = 0, 1, P(n-1, -I)/ n!):
seq(numer(R(n)), n=0..26);
# or with the Euler polynomials:
ep := n -> `if`(n=0, 1, 2^n*(euler(n, 1/2)-euler(n, 1))*(-1)^iquo(n+1, 2)):
a := n -> numer(ep(n)/n!): seq(a(n), n=0..26); # (End)
MATHEMATICA
nn = 26; Numerator[CoefficientList[Series[Sec[x] + Tan[x], {x, 0, nn}], x]] (* T. D. Noe, Jul 24 2013 *)
Table[If[n==0, 1, 2 I^(n+1) PolyLog[-n, -I] / n!], {n, 0, 26}] // Numerator (* Peter Luschny, Aug 03 2017 *)
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 *)
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
N. J. A. Sloane, Nov 18 2004
STATUS
approved