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Definition: A number $$n = \sum_{i = 0}^{k} d_ib^i$$ is called a Cooper number in base $b$, if $$\pi(n) = \prod_{i = 0}^k d_i$$ and $$\pi(\textrm{rev}(n)) = \textrm{rev}(\pi(n)),$$ where $\pi(n)$ is the prime counting function and $\textrm{rev}(n)$ results from reversing the digits of $n$, that is, $$\textrm{rev}(n) = \sum_{i = 0}^{k} d_{k - i} b^i.$$

A Cooper number $n$ is called a Cooper prime if $n$ and $\textrm{rev}(n)$ are prime numbers.

The well known example is $73$ in base $10$, with $\pi(73) = 21$ and $\pi(37) = 12$. It is known that in base $10$ the only Cooper prime is $73$.

Other examples of a Cooper number are $8 = 22_3$ in base $3$, $27 = 33_8$ in base $8$ and $3087 = 21 \times 146 + 21$ in base $146$, since $\pi(3087) = 441 = 21^2$.

I was not able to find any more, but beside $73$ in base $10$ there is a pattern with $8$, $27$ and $3087 $are palindromic in the corresponding base.

Any idea how to prove that there are no more Cooper numbers?

Robert Soupe
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    okay so if it has an even digit it needs pi(n) to be even, forcing something on the first digit of the number of primes less than it's reversal. now apply PNT –  Oct 20 '19 at 21:21
  • doh forgot even last digit only always happens in even bases ... for even numbers –  Oct 21 '19 at 14:23
  • Maybe it helps, if you show us how it was proven in base $10$ – Peter Oct 22 '19 at 14:13
  • I thought the relation (rev())=rev(()) was an interesting reaction beside the original formulation.
  • – Stephan Januar Oct 23 '19 at 15:58
  • I thought the relation (rev())=rev(()) was an interesting reaction beside the original formulation. Applying PNT suggests there are many solutions.
  • My attempt for allowing any base is boring since in base b = n-1 every number $ n = [11]_b $.
  • The proof of the Sheldon Conjecture is in [link] https://math.dartmouth.edu/~carlp/sheldon02132019.pdf
  • Is the sequence of number with (rev())=rev(() interesting?
  • Sorry, sometimes I just play with numbers for recreation.

    – Stephan Januar Oct 23 '19 at 16:07