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 Summing up the dividers' dividers (Posted on 2014-01-15)
Let c(n) be a function that counts how many positive divisors the positive integer n has, and let

S(n) := summation over d/n { c(d) }

be the sum of the counts c(d) taken over all divisors d of n.

Examples:

Ex1: 15 is divisible by 1, 3, 5, 15, so c(15) = 4. Since c(1) = 1, c(3 )= c(5) = 2, and c(15) = 4, we find that S(15) = 1+2+2+4= 9.

Ex2: 18 is divisible by 1, 2, 3, 6, 9, and 18, so c(18) = 6. Since c(1) = 1, c(2) = c(3) = 2, c(6) = 4 and c(9) = 3, we find that S(18) = 1+2+2+4+3+6 = 18.

Find all roots n of the equation S(n) = n.

Source: Putnam 2002.

 No Solution Yet Submitted by Ady TZIDON No Rating

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 Some sum (spoiler, incomplete proof) Comment 1 of 1
The numbers that work are 1, 3, 18, 36.

If p is prime then S(p)=3, so S(3)=3
If p^a is a prime power S(p^a) = (a+1)(a+2)/2 which is clearly not a prime power if a>1.

For numbers of the form p1^a1 * p2^a2 * ...
S(p1^a1 * p2^a2 * ...) = (a1+1)(a2+1)/2 * (a2+1)(a2+2)/2 * ...

When there are two primes so only a1>0, a2>0
a1=1 and a2=2 gives 3*6=18 which is equal to 2^1*3^2 so we have a match
a1=2 and a2=2 gives 6*6=36 = 2^2*3^2 another match
It is clear that a product of larger triangular numbers will not achieve two exponents that are large enough.

It is also clear to me that a number with 3 or more prime factors will also not be possible but I'm not prepared to explain how the exponents cannot be large enough.

 Posted by Jer on 2014-01-15 14:07:06

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