OFFSET
1,3
COMMENTS
It seems the nonzero coefficients are powers of 2.
This function is multiplicative with a(p^n) = Re(I^(p^n+1) - I^(p^(n-1)+1)).
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Wikipedia, Dirichlet beta function
Wikipedia, Riemann zeta function
FORMULA
a(p^n) = (-1)^n*2, for prime p=3 mod 4.
a(p^n) = 0, for prime p=1 mod 4.
a(2) = -1, a(2^n) = 0 for n > 1.
a(n) = -Re(I*Sum_{d|n}(mobius(d)*I^(n/d))).
Dirichlet g.f.: Sum_{n >= 1} a(n)/n^s = beta(s)/zeta(s), where beta(s) and zeta(s) are Dirichlet Beta and Riemann zeta functions accordingly.
Sum_{n >= 1} a(n)/n^s = (1-2^(-s))*Product_{p=3 mod 4}(p^s-1)/(p^s+1), where p runs over prime numbers.
Sum_{n>=1} mobius(n)/(z^n-I) = Sum_{n >= 1} b(n)/z^n. a(n)=Re(b(n)).
Sum_{n>=1} a(n)/(z^n-1) = z/(z^2+1)
Sum_{d|n} a(d) = A101455(n). - Gevorg Hmayakyan, Dec 27 2017
EXAMPLE
a(4) = -Re(I*(mobius(1)*I^4 + mobius(2)*I^2 + mobius(4)*I)) = Re((I^4-I^2)*I) = Re(2*I) = 0.
MAPLE
a(n):=-Re(I*add(numtheory:-mobius(d)*I^(n/d), d = numtheory:-divisors(n))).
MATHEMATICA
Table[-Re[I * Sum[MoebiusMu[d] * (I^(n/d)), {d, Divisors[n]}]], {n, 81}] (* Indranil Ghosh, Mar 19 2017 *)
PROG
(PARI) a(n)=my(f=factor(n)); prod(i=1, #f~, if(f[i, 1]==2, if(f[i, 2]==1, -1, 0), if(f[i, 1]%4==3, 2*(-1)^f[i, 2], 0))) \\ Charles R Greathouse IV, Mar 22 2017
CROSSREFS
KEYWORD
sign,mult
AUTHOR
Gevorg Hmayakyan, Mar 19 2017
STATUS
approved