********************** Max Alekseyev's proof of the conjecture (October 23, 2008): From the definition of A145855 it follows that a(n) is equal to the sum of coefficients of x^(n*k)*y^n for k=1,2,... in \prod_{k=1}^{2n-1} (1 + y x^k). The series multisection implies that a(n) equals -1 + 1/n^2 \sum_{i=0}^{n-1} sum_{j=0}^{n-1} \prod_{k=1}^{2n-1} ( 1 + exp(2*Pi*I*(i+k*j)/n) ) where I^2 = -1. It is clear that \prod_{k=1}^{2n-1} ( 1 + exp(2*Pi*I*(i+k*j)/n) ) = [ \prod_{t=0}^{n'-1} ( 1 + exp(2*Pi*I*(i+t*d)/n) ) ]^(2d) / ( 1 + exp(2*Pi*I*i/n) ) = (-1)^(n+d) * 2^(2n) * [ \prod_{t=0}^{n'-1} cos( 2*Pi * (i+t*d)/(2n) ) ]^(2d) / ( 1 + exp(2*Pi*I*i/n) ) where d=gcd(j,n), n'=n/d. Since the result must be a real number, we can take Re() of the above expression, resulting in (-1)^(n+d) * 2^(2n-1) * [ \prod_{t=0}^{n'-1} cos( 2*Pi * (i+t*d)/(2n) ) ]^(2d). Now from de Moivre's identity it is easy to get [ \prod_{t=0}^{n'-1} cos( 2*Pi * (i+t*d)/(2n) ) ]^(2d) = (-1)^n * ( (-1)^n' - cos( 2*Pi*i/d ) )^d = \sum_{s=0}^d binomial(d,s) * (-1)^((n'+1)*s) * cos( 2*Pi*i/d )^s. It is also clear that for 0<=s<d \sum_{i=0}^{n-1} cos( 2*Pi*i/d )^s = n/2^s binomial(s,s/2) if s is even; and 0 otherwise. Similarly, \sum_{i=0}^{n-1} cos( 2*Pi*i/d )^d = n/2^(d-1) if d is odd; and n/2^d * (2 + binomial(d,d/2)) if d is even. Plugging this into original formula, we have a(n) = -1 + 1/n \sum_{j=0}^{n-1} (-1)^(n+d) * ( (-1)^(n+d) + 2^(d-1) * \sum_{s=0}^{[d/2]} binomial(d,2s) * binomial(2s,s) / 2^(2s) ) = -1 + 1/n \sum_{j=0}^{n-1} ( 1 + (-1)^(n+d) * binomial(2d,d)/2 ). Switching to summation over d we have: a(n) = -1 + 1/n \sum_{d|n} phi(n/d) * ( 1 + (-1)^(n+d) * binomial(2d,d)/2 ) = 1/(2n) \sum_{d|n} (-1)^(n+d) * phi(n/d) * binomial(2d,d). **********************