# One more "What does this operator do" question

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...

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

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.