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Another function.

Rules

For starters, redefine: 10^^0 = 1

Then:

0₳b = (b+1)₳0₳(b-1)

a₳0 = 10^^a

a₳1 = (a+1)(a-1)

a₳b = (((((a-1)₳b)₳b)₳b...)₳b)₳b (b nests)

These are to be solved from left to right.

examples

1₳0 = 10

0₳1 = 0₳0₳0 = 1₳0 = 10

0₳2 = 3₳0₳1 = (10^^3)₳1 = ((10^^3)-1)((10^^3)-1)

Conclusion

I have a feeling this notation is extremely powerful, yet can not be sure wherever it will ever even resolve.


Lol.

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