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
To: primenumbers@yahoogroups.com
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 )
The latest updates :
http://factordb.com/index.php?query=n^n%2B(n%2B1)^(n%2B1)&n=1&perpage=150
Mind-bending ; states non-trivial facts about numbers too large to
represent in the known universe ;
but nevertheless , intended for a general audience :
http://upforthecount.com/math/nnnp1np1.html
Many details :
http://upforthecount.com/math/nnp.html ( unpublished )
http://www.primenumbers.net/prptop/searchform.php?form=(x^x%2By^y)?&action=Search
Many messages right here :
http://tech.groups.yahoo.com/group/primenumbers