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Let $\sigma(N)$ denote the sum of divisors of the positive integer $N$.

If $(2N-\sigma(N)) \mid N$, then $N$ is said to be deficient-perfect.

Note that, if $N$ is deficient-perfect, then $N/(2N-\sigma(N))$ is an integer.

Here is my question:

How often is $N/(2N-\sigma(N))$ a palindrome (in base-$10$) if $N$ is deficient-perfect?

MY ATTEMPT

Since powers of $2$ are almost perfect, (in particular, $2(2^r) - \sigma(2^r) = 1$ for integers $r \geq 0$), then they are also deficient-perfect, so that we are led to the related question:

Is there any palindromic power of $2$?

(The accepted answer to the hyperlinked question above references a conjecture due to Simmons, which predicts that the answer is NO for $r \geq 4$.)

We then survey the even deficient-perfect numbers with exactly two distinct prime factors (listed as a subsequence of OEIS sequence A271816), for $N \leq 709784$:

$$\begin{array}{c|c|c|c} \text{$N$} & \text{$N/(2N-\sigma(N))$} & \text{Base-$10$ Palindrome?} & \text{Prime?} \\ \hline 10 & 5 & YES & YES \\ \hline 44 & 11 & YES & YES \\ \hline 136 & 68 & NO & NO \\ \hline 152 & 38 & NO & NO \\ \hline 184 & 23 & NO & YES \\ \hline 752 & 47 & NO & YES \\ \hline 884 & 221 & NO & NO \\ \hline 2144 & 536 & NO & NO \\ \hline 2272 & 284 & NO & NO \\ \hline 2528 & 158 & NO & NO \\ \hline 8384 & 2096 & NO & NO \\ \hline 12224 & 191 & YES & YES \\ \hline 17176 & 113 & NO & YES \\ \hline 18632 & 4658 & NO & NO \\ \hline 18904 & 2363 & NO & NO \\ \hline 32896 & 16448 & NO & NO \\ \hline 33664 & 4208 & NO & NO \\ \hline 34688 & 2168 & NO & NO \\ \hline 49024 & 383 & YES & YES \\ \hline 63248 & 3953 & NO & NO \\ \hline 85936 & 10742 & NO & NO \\ \hline 106928 & 326 & NO & NO \\ \hline 116624 & 29156 & NO & NO \\ \hline 117808 & 7363 & NO & NO \\ \hline 526688 & 32918 & NO & NO \\ \hline 527872 & 65984 & NO & NO \\ \hline 531968 & 33248 & NO & NO \\ \hline 556544 & 8696 & NO & NO \\ \hline 589312 & 4604 & NO & NO \\ \hline 599072 & 1544 & NO & NO \\ \hline 654848 & 2558 & NO & NO \\ \hline 709784 & 578 & NO & NO \\ \hline \end{array}$$

I am therefore led to conjecture the following prediction:

CONJECTURE If $N$ is an even deficient-perfect number and $Q = N/(2N-\sigma(N))$ is a base-$10$ palindrome, then $Q$ is prime.

Notice from the table above that the converse of the CONJECTURE is false.

Note that there is a counterexample to the CONJECTURE if we restrict $N$ to be odd, since if we consider the lone odd deficient-perfect number that we know of (as of June 2019), then we have $$N = {3^2}\cdot{7^2}\cdot{{11}^2}\cdot{{13}^2} = 9018009$$ $$\sigma(N) = 18035199 = {3^2}\cdot{7}\cdot{13}\cdot{{19}^2}\cdot{61}$$ $$2N - \sigma(N) = 2\bigg({3^2}\cdot{7^2}\cdot{{11}^2}\cdot{{13}^2}\bigg) - {3^2}\cdot{7}\cdot{13}\cdot{{19}^2}\cdot{61}$$ $$= {3^2}\cdot{7}\cdot{13}\cdot\bigg(2\cdot{7}\cdot{{11}^2}\cdot{13} - {{19}^2}\cdot{61}\bigg) = {3^2}\cdot{7}\cdot{13} = 819$$ $$\frac{N}{2N-\sigma(N)} = \frac{9018009}{819} = \frac{{3^2}\cdot{7^2}\cdot{{11}^2}\cdot{{13}^2}}{{3^2}\cdot{7}\cdot{13}} = {7}\cdot{{11}^2}\cdot{13} = 11011,$$ which is a palindrome, but is composite.

Finally, it is currently unknown whether there are infinitely many palindromic primes (i.e. numbers that are both palindromes and primes), although it is known that "almost all palindromes are composite".

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