m = k^2 *j, j<=k, GCD(k,j)=1

[Leroy Quet]

Answer to puzzle below copied original post.


>Let a(m) be the number of ways that (ie. the number of (k,j)'s where)
>
>m = k^2 *j,
>j <= k,
>GCD(k,j)=1,
>j and k= positive integers.
>
>So we have the sequence:
>
>{a(m)}: 1,0,0,1,0,0,0,0,1,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,1,0,0,...
>
>(Not every term is 0 or 1. a(100) = 2.)
>
>And we can also have the sequence, {b(n)}, of m's where a(m) >= 1:
>
>{b(n)}: 1, 4, 9, 16, 18, 25, 36, 48, 49, 50,...
>
>Neither sequence is in the EIS.
>
>I know these sequences might seem a tad arbitrary.
>But here is a puzzle (for fun) which might somewhat justify the 
>a-sequence:
>
>Evaluate the sum:
>
>sum{m=1 to oo} a(m)/m.
>
>
>(I'll give the answer in a few days if no one gets it sooner. But someone 
>surely will get the {somewhat surprising} answer soon, since this puzzle 
>is not extremely difficult.)




I don't think anyone will try to solve this, so I will give the answer 
now.


sum{m=1 to oo} a(m)/m   =   2.


Basic idea behind proof:

sum{m=1 to oo} a(m)/m =

sum{m=1 to oo} sum{1<=k<=m, GCD(k,m)=1} 1/(k m^2).

Now sum{1<=k<=m, GCD(k,m)=1} 1/k =

sum{k|m} mu(k) H(m/k) /k,

where H(n) = sum{j=1 to n} 1/j, the nth harmonic number,
and mu(k) is the Moebius function.

So, putting this together, we get

sum{m=1 to oo} a(m)/m =

sum{m=1 to oo} sum{k|m} mu(k) H(m/k) /(k m^2) =

sum{m=1 to oo} sum{k=1 to oo} mu(k) H(m) /(k^3 m^2)

= (sum{m=1 to oo} H(m)/m^2) / zeta(3)

= 2.

(If I did not make any errors)

thanks,
Leroy Quet