Re: Factors of np ( n ) = n^n + (n+1)^(n+1) ; was Square factors of n^n+(n+1)^(n+1)]
- Hi , all ,
Correcting a typo in my message of a few minutes ago :
( n^2 + n + 1 ) / 3
Not even draft yet , let alone published :
still needs a lot of work .
But , I have updated the top of the main table :
with complete factorizations now extending through
np ( 90 ) .
When is np ( n ) prime ?
So far , only for 1 , 2 and 3 .
When n mod 6 = 4 , np ( n ) has at least 3 algebraic
factors , 3 and ( n^2 + n + 1 ) / 3 ( twice ) .
But what about the residue ( primitive part ? ) ?
When is this prime ?
n t n^n + (n+1)^(n+1)
4 4 3 * 7 * 7 * 23
16 6 3 * 7 * 7 * 13 * 13 * 34041259347101651
Any others ?