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• ## Re: Factors of np ( n ) = n^n + (n+1)^(n+1) ; was Square factors of n^n+(n+1)^(n+1)

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• Hi , all , Not even draft yet , let alone published : http://upforthecount.com/math/nnp.html still needs a lot of work . But , I have updated the top of the
Message 1 of 81 , Nov 29, 2011
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Hi , all ,

Not even draft yet , let alone published :
http://upforthecount.com/math/nnp.html
still needs a lot of work .
But , I have updated the top of the main table :
http://upforthecount.com/math/nnp2.txt
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 ( 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 ?

Cheers ,

Walter
• Hi , all , Not even draft yet , let alone published : http://upforthecount.com/math/nnp.html still needs a lot of work . But , I have updated the top of the
Message 81 of 81 , Nov 29, 2011
View Source
Hi , all ,

Not even draft yet , let alone published :
http://upforthecount.com/math/nnp.html
still needs a lot of work .
But , I have updated the top of the main table :
http://upforthecount.com/math/nnp2.txt
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 ( 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 ?

Cheers ,

Walter
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