## Reflection in the next prime

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• Reflection in the next prime Take any n and next prime p , that is the smallest prime p n, reflect n in p and get a(n)=2p-n; then starting with n=1, we have
Message 1 of 4 , Jul 31, 2003
Reflection in the next prime

Take any n and "next prime p", that is the smallest prime
p>n, "reflect" n in p and get a(n)=2p-n;
then starting with n=1, we have the sequence a(n):
3,4,7,6,9,8,15,14,13,12,15,14,21,20,19,18,21,20,27,26,25,24,35
(A087030?)
Now introduce b(n)=1/0 if a(n)is prime/composite:
1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,
0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,
0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0(A087032?)
Cases of prime a(n) or b(n)=1 are sure less numerous than b(n)=0. Two
(trivial?) observations:
there is no two subsequent ones; number c(n) of zeros in each group
is odd:
1,5,5,11,3,1,5,7,9,3,1,5,5,15,1,5,5,9,1,5,11,5,5,3,7,7,9,5,1,3,5,3,13,
3,1,5,11,7,5,5,9,3,1,5,17,3,1,5,15,15,3,15,5,7,17,23,5(A087033?)
Up to n=100,000, the maximal number of zeros in one group is 75
(starting from which n? - i don't know!).
what is the longest series of subsequent n such that "reflection of n
in the next prime p", 2p-n, is composite, p being the smallest prime
>n?
zak
• Hey Zack: There can not be two ones consecutives because if n is even, 2p-n is even... the same fact explains the oddity of the zeroes chains (groups as you
Message 2 of 4 , Jul 31, 2003
Hey Zack:

There can not be two ones consecutives because if n is even, 2p-n is even... the same fact explains the oddity of the zeroes chains (groups as you call them): if n is even then n+2k is even too, n+(2k-1) could be prime, but 2*nextprime(n+2k) - (n+2k) is certainly composite, so the number c(n) must be odd.

Jose Brox

----- Original Message -----
From: Zak Seidov
Sent: Thursday, July 31, 2003 8:29 PM
Subject: [PrimeNumbers] Reflection in the next prime

Reflection in the next prime

Take any n and "next prime p", that is the smallest prime
p>n, "reflect" n in p and get a(n)=2p-n;
then starting with n=1, we have the sequence a(n):
3,4,7,6,9,8,15,14,13,12,15,14,21,20,19,18,21,20,27,26,25,24,35
(A087030?)
Now introduce b(n)=1/0 if a(n)is prime/composite:
1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,
0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,
0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0(A087032?)
Cases of prime a(n) or b(n)=1 are sure less numerous than b(n)=0. Two
(trivial?) observations:
there is no two subsequent ones; number c(n) of zeros in each group
is odd:
1,5,5,11,3,1,5,7,9,3,1,5,5,15,1,5,5,9,1,5,11,5,5,3,7,7,9,5,1,3,5,3,13,
3,1,5,11,7,5,5,9,3,1,5,17,3,1,5,15,15,3,15,5,7,17,23,5(A087033?)
Up to n=100,000, the maximal number of zeros in one group is 75
(starting from which n? - i don't know!).
what is the longest series of subsequent n such that "reflection of n
in the next prime p", 2p-n, is composite, p being the smallest prime
>n?
zak

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• even (xcus the pun) more fun, take Ulam s Spiral, and place 4 infinite mirrors on it, horizontal, vertical and both diagonals, which primes have the highest
Message 3 of 4 , Jul 31, 2003
'even' (xcus the pun) more fun, take Ulam's Spiral, and place 4 infinite
mirrors on it, horizontal, vertical and both diagonals, which primes have
the highest reflective count?

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths/
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
• Should have written 2*nextprime(n+(2k-1)) - n+(2k-1) could be prime instead of n+(2k-1)... Jose ... From: Jose Ramón Brox To: Prime Numbers Sent:
Message 4 of 4 , Jul 31, 2003
Should have written " 2*nextprime(n+(2k-1)) - n+(2k-1) could be prime " instead of " n+(2k-1)..."

Jose

----- Original Message -----
From: Jose Ramón Brox
To: Prime Numbers
Sent: Thursday, July 31, 2003 9:34 PM
Subject: Re: [PrimeNumbers] Reflection in the next prime

Hey Zack:

There can not be two ones consecutives because if n is even, 2p-n is even... the same fact explains the oddity of the zeroes chains (groups as you call them): if n is even then n+2k is even too, n+(2k-1) could be prime, but 2*nextprime(n+2k) - (n+2k) is certainly composite, so the number c(n) must be odd.

Jose Brox

----- Original Message -----
From: Zak Seidov
Sent: Thursday, July 31, 2003 8:29 PM
Subject: [PrimeNumbers] Reflection in the next prime

Reflection in the next prime

Take any n and "next prime p", that is the smallest prime
p>n, "reflect" n in p and get a(n)=2p-n;
then starting with n=1, we have the sequence a(n):
3,4,7,6,9,8,15,14,13,12,15,14,21,20,19,18,21,20,27,26,25,24,35
(A087030?)
Now introduce b(n)=1/0 if a(n)is prime/composite:
1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,
0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,
0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0(A087032?)
Cases of prime a(n) or b(n)=1 are sure less numerous than b(n)=0. Two
(trivial?) observations:
there is no two subsequent ones; number c(n) of zeros in each group
is odd:
1,5,5,11,3,1,5,7,9,3,1,5,5,15,1,5,5,9,1,5,11,5,5,3,7,7,9,5,1,3,5,3,13,
3,1,5,11,7,5,5,9,3,1,5,17,3,1,5,15,15,3,15,5,7,17,23,5(A087033?)
Up to n=100,000, the maximal number of zeros in one group is 75
(starting from which n? - i don't know!).
what is the longest series of subsequent n such that "reflection of n
in the next prime p", 2p-n, is composite, p being the smallest prime
>n?
zak

Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
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[Non-text portions of this message have been removed]

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