### np ( n ) = n^n + (n+1)^(n+1) ------- 2 prominent questions

This version provides human-readable spacing .

```Date: Tue, 2 Oct 2012 05:11:58 -0000 (UTC)
From: Walter Nissen
Subject: [PrimeNumbers] np ( n ) = n^n + (n+1)^(n+1) ; 2 prominent questions

Greetings , all ,

np ( n ) = n^n + (n+1)^(n+1)  has numerous interesting aspects .
Two aspects of this expression are prominent .

( 1 ) :
The number of prime factors of  np ( n ) .
The first line in the table of  OMEGA  below is :
1   1   1   4   2   3
The first  3  counts in the first line are each  1 ,
because for each of  n = 1 , 2 and 3 ,
np ( n )  =  n^n + (n+1)^(n+1)  is prime .
The last count in the first line is  3 , because
6^6 + 7^7  has  3  prime factors :  11 , 239 , and 331 .

t

OMEGA ( n^n + (n+1)^(n+1) )

n mod 6
1   2   3   4   5   0

1   1   1   4   2   3
2   3   2   5   2   6
4   2   3   6   3   3
4   6   2   7   3   5
5   3   3   5   3   3
3   7   4   5   4   3
2   2   2   6   3   6
4   5   3   8   6   6
3   5   5   5   5   3
3   5   2   8   4   4
6   6   6   8   8   4
3   4   7   7   6   6
4   4   7   9   6   7
5   9   5   7   4   3
5   6   7   9   3   2
5   4   8   9   5   8
2   4   6  11   6   5
4   3   8  11   4   5    ( on this 18th line :  103 <= n <= 108 )

6   7   8  7?   6  3?
6   5  6? 11?   7   6
6?  4?   5   7  13  7?
5   7 10?   9   7   9
4  6?  5? 10?   7   3
___ ___ ___ ___ ___ ___

19  22  15  32  17  23   total of first  6 lines =  6 periods , 1 <= n <= 36
65  79  81 130  77  82   total of first 18 lines = 18 periods , 1 <= n <= 108

3.6 4.4 4.5 7.2 4.3 4.6   average of first 18 lines

It's notable that
when  n mod 6 = 4 , the number of prime factors is relatively large and
when  n mod 6 = 1 , the number of prime factors is relatively small and
when  n mod 6  !=  1 or 4 , the number of prime factors is moderate ,
when  n mod 3  !=  1      , the number of prime factors is moderate .

Perhaps , it's easy to understand the surplus of prime factors for
4 mod 6 , because there are a few known algebraic factors :
3  and  ( n^2 + n + 1 ) / 3  ( twice ) .
E.g. , 4^4 + (4+1)^(4+1)  =  3 * 7 * 7 * 23 ;
Richard Guy mentioned this surplus in a private communication .
But , in the leftmost column ,
why are there so few prime factors for  1 mod 6 ?

( 2 ) :
It's notable that  p | np ( n ) , for certain
n mod ( p ( p - 1 ) ) .
E.g. , for  p = 3 , p(p-1) = 6 , and for  n mod 6  =  4 :
np (  4 )  =  4^4 + (4+1)^(4+1)  =  3 * 7 * 7 * 23
np ( 10 )  =  3 * 37 * 37 * 8017 * 8969
np ( 16 )  =  3 * 7 * 7 * 13 * 13 * 34041259347101651
The top of the  p(p-1)  table is :
(2 | np ( n ) iff n mod   2 =    )
3 | np ( n ) iff n mod   6 =   4
5 | np ( n ) iff n mod  20 =   1   8
7 | np ( n ) iff n mod  42 =   4  12  16  33
11 | np ( n ) iff n mod 110 =   6   7  20  34  61  62  63 101
13 | np ( n ) iff n mod 156 =  16  22  24  53  86  94 100 147
E.g. , np ( 21 )  =  5 * 69454092876521107983605569601
np ( 28 )  =  3 * 5 * 271^2 * 2360926164108571968813424783598971267
To find factors of  np ( n ) , is it possible to compute this table
faster than using ECM or NFS ?

Over the last decade or so , a number of people have contributed to the
pursuit of primality in this expression , to whom I'm quite grateful ;
on the math side , notably David Broadhurst , Mike Oakes and some of the
usual suspects ;  and on the computational side , notably
Rich Dickerson , Lionel Debroux ( recently importantly ) , yoyo@home
( recently importantly ) , Andy Steward , GMP-ECM , Primo , msieve and
ggnfs ( these lists are inadequate ) .

Cheers ,

Walter

---

Here are a few details :
The only known  prime np  are for  n = 1 , 2 and 3 .
np ( 415 ) is easy to factor , it has only 2 factors and one of
them is  29 .   ( ! )
np ( 3754 )  has at least  19  prime factors .
Using many years of CPU , all  np  have been completely factored
for  n <= 111 :
http://upforthecount.com/math/nnp2.txt   ( unpublished )