Number of m such that n=0 Mod UnitarySigma(m)
koh
zbi74583.boat at orange.zero.jp
Wed Jul 23 05:18:29 CEST 2008
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
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