All About Reversed Slightly Excessive Numbers

In Charles Ashbacher's paper, "On Numbers that are Pseudo-Smarandache and Smarandache Perfect" in Smarandache Notions Journal #14, he discussess the operation of summing the divisors of a number after a function has been applied to those divisors. In this note we will consider the process of applying the reverse function to the divisors of a number and then summing them.

Let srd(n) be the sum of the reversed divisors of n, which produces the following sequence.

n = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, ...

srd(n)= 1, 3, 4, 7, 6, 12, 8, 15, 13, 9, 12, 37, 32, 51, ...

For example, srd(14) = 51 because the divisors of 14 are 1, 2, 7, 14; and reversing and summing gives 1 + 2 + 7 + 41 = 51.

Joseph L. Pe has already investigated some aspects of this function in his paper, "The Picture-Perfect Numbers," which can be found at his web site. The purpose of this note is to provide some results on a new srd(n) problem.

-Background-

According to S. Singh's book, Fermat's Enigma: The Epic Quest to Solve the World's Greatest Mathematical Problem, Pythagoras and his followers were the first to notice they could not find any integers that were equal to one more than the sum of their divisors. That is, they could not find solutions to sigma(n) = 2*n+1, where sigma(n) = sum of divisors of n. They referred to these numbers (still hoping at least one might exist) as "slightly excessive numbers." To this day not a single solution has been found and the problem remains "open." However, a couple of things are known about "slightly excessives." According to the Mathworld web site, if a slightly excessive integer exists, it must be larger than 10³⁵, and have more than 7 distinct prime factors.

Now we will alter the original "slightly excessive" problem in two ways and search for solutions. Our modification will consist of: (1) summing all the divisors of n, not just the proper divisors; (2) reversing the divisors before summing them (i.e., we will use the srd(n) function defined above).

-Solutions-

It is easy to see that palprimes will always be solutions: The only divisors of a prime are 1and itself, and because a palindrome's largest nonproper divisor is already the same as its reversal, the sum of a palprime's divisors will always be equal to n+1. Q.E.D.

Interestingly enough, a computer search revealed that there are also non-palprime solutions to srd(n) = n+1. Namely: 965, 8150, 12966911 all satisfy the equation, with no more solutions found up to 2*10⁷. For example: The divisors of 965 are 1, 5, 193, 965; and reversing and summing produces 1 + 5 + 391 + 569 = 966. Also note there are no obvious patterns in the factorizations of these solutions:

965 = 5 * 193

8150 = 2 * 5 * 5 * 163

12966911 = 19 * 251 * 2719

-Open Questions-

Are there any more non-palprime solutions to srd(n) = n+1? If so, do the solutions have any properties in common? Could there be infinitely many solutions?

I hope so. Because I love the concept of infinity.