(In reply to
Computer exploration by Charlie)
Of course the numbers in your table are all due to the fact that
1 + 2^3 = 3^2 so for all n,
3^(3n+2) = 3^(3n)*3^2 = 3^(3n)(1+2^3) = 3^(3n)+2^3*3^(3n) = (3^n)^3+(2*3^n)^3
So if the coprime requirement is lifted your answer would become
The (3n+2) powers of 3 can be written as the sum of two k=3rd powers of integers (with a common factor of 3^n)
Edited on August 26, 2016, 10:15 am

Posted by Jer
on 20160825 21:27:27 