# A computational experiment about identities involving the sum of remainders function

Let $\sigma(m)$ the sum of divisors function and $$S(m)=\sum_{k=1}^m\text{m mod k}$$ the sum of remainders function, then it is know that for integers $m>1$ $$\sigma(m)+S(m)=S(m-1)+2m-1.$$ On the other hand since the sum of divisor functioin is multiplicative, on assumption that there exists an odd perfect number $n$, we have that $\sigma(\sigma(n))=\sigma(2n)=6n$ and thus $n$ satisfies $$2n+S(2n)=S(2n-1)-1.$$

Fact. If $l\geq1$ is and odd integer that satisfies $2l+S(2l)=S(2l-1)-1$, then $l$ is perfect and thus an odd perfect number.

The proof is easy since by previous assumption we've $$-\sigma(2l)+4l-1=-2l-1,$$ and the sum of divisors function is multiplicative.

Computational fact. Seem that there are few even integers $m\geq 1$ satisfying $$2m+S(m)=S(2m-1)-1.$$ The only ones $\leq 10^5=M_0$ are $m=60$ and $336$.

On the other hand if $m=n$ is an odd prefect number, if there is one of them, we can show also that satisfies each of the following similar conditions about positive integers $$\sigma(m)=-S(2m)+S(2m-1)-1$$ $$2m=-S(\sigma(m))+S(2m-1)-1,$$ $$2m=-S(2m)+S(\sigma(m)-1)-1$$ or $$2m=-S(\sigma(m))+S(\sigma(m)-1)-1.$$

For the three first I don't find integers $1<m<10^5$ satisfying one of them, and the last only $m=24$

Question. Can you provide us a summary table with all integers $1<m\leq M$ that you find, for a reasonable upper limit $M$, using your computer for each of the following conditions $$f_i(m)=-S(f_i(m))+S(f_i(m)-1)-1,$$ where $f_i$ are the arithmetical functions $f_1(m)=\sigma(m)$ and $f_2(m)=2\cdot m$? Thanks in advance.

• If there is a case for which there are a lot of examples of one of the previous conditions, say it, without type all examples. Thanks. – user243301 Feb 29 '16 at 11:25
• There was a typo, very thanks much @GerryMyerson – user243301 Feb 29 '16 at 12:58

For $1\leq m\leq 10^5$ then formulas (each of them) have no solutions $$\sigma(m)=-S(2m)+S(2m-1)-1,$$ $$2m=-S(2m)+S(\sigma(2m)-1)-1,$$ $$2m=-S(\sigma(m))+S(2m-1)-1,$$ $$\sigma(m)=-S(\sigma(m))+S(2m-1)-1$$ or $$\sigma(m)=-S(2m)+S(\sigma(m)-1)-1$$ The following have solutions (only in the segment of integers $1\leq m\leq 10^5$ ) $$2m=-S(2m)+S(2m-1)-1,\text{has solutions }60\text{ and }336,$$ $$2m=-S(\sigma(m))+S(\sigma(m)-1)-1,\text{has solutions }24\text{ and }87402,$$ $$\sigma(m)=-S(\sigma(m))+S(\sigma(m)-1)-1$$ $$\text{has solutions }54,56,87, 95, 276, 308, 429, 446, 455, 501, 581\text{ and }611.$$