Number of m such that n=0 Mod UnitarySigma(m)

[koh]

    Hi, Seqfans.

    b(n) = Number of m such that n=0 Mod UnitarySigma(m)
    b(n) : 1,1,2,2,2,3,1,3,3,3,1,5,1,2,4,3,2,5,1,5

    Ex : 6=0 Mod UnitarySigma(m) , m=1,2,5, so, b(6)=3
    For all n 1<b(n).
    For all prime p b(p)=1, except Fermat primes.

    I met the sequence when I was studying the following equation.

    UnitarySigma(x) = UnitarySigma(y) = 1/k*(x^2+m*y^2)/(x+y)
    Where k < x < y

    Almost all cases the following condition is satisfied.

    {Number of small solutions  whose seed is u,v} < b((u-v)*(1-v))
    If x=c*u, y=c*v, GCD(c,u)=1,GCD(c,v)=1 then {u,v} is called "seed".

    [Example]
    {21,31} is a seed

    k=285, m=519
    x=19*3*7
    y=19*31

    k=290, m=519
    x=29*3*7
    y=29*31

    k=295, m=519
    x=59*3*7
    y=59*31

    k=298, m=519
    x=149*3*7
    y=149*31

    k=299, m=519
    x=299*3*7
    y=299*31

    
    {Number of solutions} = 5

    5 < b(300) = 14




    Yasutoshi