Problems & Puzzles: Puzzles

 Puzzle 332. Odd abundance. Joseph L. Pe proposes the following puzzle: I would like to propose the following problem whose answer I don't know. (It might be easy.) Let sigmaO(n) = the sum of the odd (positive) divisors of n. The odd abundance of n is defined similarly to abundance as: oa(n) = sigmaO(n) / n. Question: Is there a (positive) integer N with N > 1 whose odd abundance oa(N) is an integer? I have checked all values of n up to 10^6, but for these values, oa(n) is never an integer.Btw, the problem is very easy if we consider the even or the ordinary abundances.

Contributions came from Faride Firoozbakht, Fred Schneider, Luke Pebody & Dan Dima.

Faride wrote: "It seems that there is no such number. But the proof of this conjecture is likely impossible."

Fred wrote:

Say m> 1 is an odd number:

Since 2 does not contribute to the sum of divisors of sigmaO, for any k>= 0 sigmaO(m*2^k)= sigmaO(m)

So, if want to find sigmaO(n)/n to be an integer, it makes sense that we search only for odd numbers (for any solution n where n= m*2^k, m would also have to be a solution).

Clearly, sigmaO(n)=sigma(n) for odd n.

To find an odd n such that, for sigma(n)/n = 2,  we would have to find an odd perfect number (of which none are known to exist).

In fact for any integers q>=3, there are no solutions for sigma(n)/n=q.  In other words, there are no known odd multiply perfect numbers.

Luke wrote:

Puzzle 332 is equivalent to the following question: "Is there an odd multiply perfect number?" It is a longstanding conjecture that the answer is no.

If n is the largest odd factor of N, then oa(n)=oa(N). Thus if oa(N)/N is an integer, then oa(n)/n=(N/n)*oa(N)/N is an integer. This is the sum of the factors of n, divided by n.

Dan wrote:

Suppose there is a positive integer N with N > 1 whose odd abundance sigmaO(N) / N is an integer. Let N = 2^k * M, where M - odd positive integer. Then: sigmaO(N) = sigmaO(M) = sigma(M) Hence: sigma(M) / M = 2^k * (sigmaO(N) / N) is an integer too, which means that M must be a multiperfect number whose sum of divisors is divided by the number itself:

http://mathworld.wolfram.com/MultiperfectNumber.html