sigma(sigma(sigma(k) - k)))

[Michel Marcus]

Numbers k such that k = sigma(sigma(sigma(k) - k))), where sigma = A000203 has only 3 known terms : 39, 399, 6643.
I searched up to 5*10^9 but found no other terms.

MM

[Geoffrey Caveney]

This is very interesting. To provide some context:

k = sigma(k) - k  is the definition of a perfect number, because sigma(k) includes k itself together with the proper divisors of k.

k = sigma(sigma(k) - k)  is satisfied by every power of 2 that is 1 more than a Mersenne prime. Example: sigma(32) - 32 = 31, and sigma(31) = 32.

The Mersenne primes are also the odd prime factors of the even perfect numbers, such as 31*16 = 496.

Regarding Michel Marcus's numbers, I observe that sigma(39) - 39 = 17 and sigma(399) - 399 = 241 are prime, but sigma(6643) - 6643 = 1645 is composite.

While all known perfect numbers are even, and all powers of 2 are of course even, all three examples of Marcus's numbers are odd. They also have interesting prime factorizations:

39 = 3*13

399 = 3*7*19

6643 = 7*13*73

3 and 7 are Mersenne primes; the other prime factors of these numbers are not.

These numbers may be as close as it is possible to get to the existence of an odd "perfect" number.

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[Jack Brennen]

Problem:  find A such that:

B = sigma(A)-A
C = sigma(B)
A = sigma(C)

Noting that there is a strong tendency for sigma() to result in an even
number, let's look at whether this can happen with A, B, C all being even.

Noting that sigma(2N) >= 3N, for all N>0, it's easy to see that (A,B,C)
all being even is impossible.

That would imply that B/A >= 1/2, C/B >= 3/2, and A/C >= 3/2.

Multiplying those out, you get ABC/ABC >= 9/8, a contradiction.

So at least one of the three numbers must be odd.

sigma(N) is odd only if N is a square or twice a square.

If A is odd, and equal to sigma(C), then C must be a square or
twice a square.

If B is odd, and equal to sigma(A)-A, then either A is odd
(see the previous paragraph), or A is even and sigma(A) is odd,
so A must be a square or twice a square.

If C is odd, and equal to sigma(B), then B must be a square or
twice a square.

So a search strategy would be to enumerate all numbers N^2 and 2N^2,
and try each of them as A, B, and C in the set of three equations.

With that in mind, I searched fairly quickly through 10^12 and
found no other solutions than the three already given.

If we assume that sigma() values have any reasonable spread
of distribution, then the "probability" of any particular N
yielding a solution is O(1/N^2).

The sum of those probabilities converges to a small number,
so heuristically, those three solutions given previously are
likely the only solutions.

On Fri, Aug 29, 2025 at 7:06 AM Michel Marcus <michel.m...@gmail.com> wrote:

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[Michel Marcus]

Thanks.

Forgot to say that this was a sequence which did not get through.

But A387218 might like more terms

MM

Sans virus.www.avast.com

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