## RV: Floor[((p(n+1))2+(pn)2)/(2p(n+1))]=pn

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• Kermit Rose escribió: Fecha: Fri, 30 Mar 2007 10:42:09 -0400 De: Kermit Rose Para: sebi_sebi@yahoo.com Asunto:
Message 1 of 1 , Mar 30, 2007
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Kermit Rose <kermit@...> escribió: Fecha: Fri, 30 Mar 2007 10:42:09 -0400
De: Kermit Rose <kermit@...>
Para: sebi_sebi@...
Asunto: Floor[((p(n+1))2+(pn)2)/(2p(n+1))]=pn

1a. New Conjecture
Posted by: "Sebastian Martin" sebi_sebi@... sebi_sebi
Date: Wed Mar 21, 2007 1:46 pm ((PDT))

I have obtained a New Conjecture for consecutive prime numbers:

Floor[((p(n+1))2+(pn)2)/(2p(n+1))]=pn for all n>=1

I have tested for n=1 to 40000000

To show some prospective for what is going on in this formula, I've replaced
p(n+1) by the largest value which will still generate p(n).

Floor[((p(n+1))^2+(pn)^2)/(2p(n+1))]=pn

n pn p(n+1) sum denom quotient
1 2 5 29 10 2
2 3 6 45 12 3
3 5 9 106 18 5
4 7 11 170 22 7
5 11 16 377 32 11
6 13 19 530 38 13
7 17 23 818 46 17
8 19 26 1037 52 19
9 23 30 1429 60 23
10 29 37 2210 74 29
11 31 39 2482 78 31
12 37 46 3485 92 37
13 41 51 4282 102 41
14 43 53 4658 106 43
15 53 64 6905 128 53
16 59 70 8381 140 59
17 61 73 9050 146 61

There is not anything special about the value of the next prime in this formula.

The value of this conjecture is not for predicting the value of the next prime, but to
place an upper bound for the next prime.

Note that for n = 4, pn = 7, that the upper bound is exactly the next prime.
In all other cases, the upper bound is larger than the next prime.

Kermit < kermit@... >

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