MC: Ping-Pong Cubes

[Warbucks]

For me the interest is in the series underpinning the n^3 alt parity.

As I would normally enter this in Maple or equivalent, we are moving and I have no access to hardware, so forgive my symbolism.

This seq. is a stair step graph that is 5 wide and 9 groups per decimal radix digit. So if you wanted to know the number of terms < 10^n, you would use:

9*(5^n-1)/4 - 9 n>=2

So if you wanted to now the number of terms<100 (10^2) n=2, there would be 45 terms.

If you wanted to know the number of terms 10^n < #terms < 10^(n+1) we have:

9*5^n

There is indeed a solution. The general case is 2^a(5^b)(j). There are three cases. If anyone is interested, I could do a proof, but not on this phone!

It was said that the next term in the sequence is 21 digits in length, you would start at a number of this form, n is number of digits (21):

101010101010101... or for each digit n >= 1

a sub n = 1/2(1+(-1)^(n+1))

Also if there is interest, for completeness, I can generate the 21 digit cube answer.