## ip +/- n = primenumbers

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• - If p = prime numbers 3, 5, 7 or 13; (Se ip = numero primo 3, 5, 7 oppure 13); - if i = n^2 -1 (se i = n^2 -1)... for each n 1
Message 1 of 2 , Apr 2, 2002
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- If p = prime numbers 3, 5, 7 or 13;
(Se ip = numero primo 3, 5, 7 oppure 13);
- if i = n^2 -1
(se i = n^2 -1)...
for each n> 1 <p, ip - n = prime number,
ip + n = prime number,
(per ciascun n>1<p, ip -n= numero primo,
ip+n=numero primo.
-----------------------------------------------------------
p = 3:
n; i; ip; ip-n; ip+n
2; 3; 9; 13; 17;
-------------------------------------
p = 5:
n; i; ip; ip-n; ip+n
2; 3; 15; 13; 17;
3; 8; 40; 37; 43;
4; 15; 75; 71; 79;
--------------------------------------
p = 7:
n; i; ip; ip-n; ip+n
2; 3; 21; 19; 23;
3; 8; 56; 53; 59;
4; 15; 105; 101; 109;
5; 24; 168; 163; 173;
6; 35; 245; 239; 251.
---------------------------------------
p = 13:
n; i; ip; ip-n; ip+n
2; 3; 39; 37; 41;
3; 8; 104; 101; 107;
4; 15; 195; 191; 199;
5; 24; 312; 307; 317;
6; 35; 455; 449; 461;
7; 48; 624; 617; 631;
8; 63; 819; 811; 827;
9; 80; 1040; 1031; 1049;
10; 99; 1287; 1277; 1297;
11; 120; 1560; 1549; 1571;
12; 143; 1859; 1847; 1871.

Regards.
Filippo Giordano.

[Non-text portions of this message have been removed]
• ... [Erk! webmail is misbehaving - unable to quote] Tables of i and n for each prime p, and all 1
Message 2 of 2 , Apr 2, 2002
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--- FILIPPO GIORDANO <fil.giordano@...> wrote:
[Erk! webmail is misbehaving - unable to quote]

Tables of i and n for each prime p, and all 1<n<p, all naturals, such
that i is minimal and i.p+/-n are simultaniously prime.
e.g. p=13, n=12 has i=143, as 1847 and 1871 (1859+/-12) are both
prime.

A little challenge...

Find the i values such that i.p+/-n are _consecutive_ primes.

No particular reason, just a whim, and as the number or results
increases quadratically in the value of the prime chosen, I'm most
interested in the n=p-1 results, as they require the largest
composite run, and are thus heuristically hardest.

Phil

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