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it is fun to observe that questions like this What does this mathematical operator do? raise more new questions than correct answers!

When trying to solve the original question, I came to this new operator:

0#1 = 32
1#1 = 1
1#2 = 3
1#3 = 37
2#1 = 6
3#1 = 9
5#1 = 4
7#1 = 13

By default we consider that:

For each natural a: a#0 = 0

One useful hint

For each naturals a,b: a#(b+1) > a#b

No complex algorithm here! The answer should not be longer than a 10 to 12-words sentence like "a#b is the ..."

Enjoy solving this. I can provide some extra hints if needed...

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  • $\begingroup$ Can you provide more examples where b>1? Maybe some where a or b > 10? $\endgroup$
    – Lezzup
    Jul 3, 2023 at 9:14

1 Answer 1

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The result of a#b is

the position of the $b$-th occurrence of $a$ in the decimals of $\pi$. You can check here: $\pi = $ 3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651...

But shouldn't a#0 then perhaps be NaN instead of 0?

To @Lezzup's question: I derive from the principle above that 1#10 = 110 and accordingly 10#1 = 49.

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  • $\begingroup$ Well done! I agree with your comments and examples. I initially planned to disclose as an extra hint (if puzzle was not solved so fast!) the value of 15#1 and 159#1 to ring a bell... $\endgroup$ Jul 3, 2023 at 11:15

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