New Class of primes - proving algorithm

AntonVrba · 2008-10-06, 00:27

Using my new proof for Wagstone numbers I hope to present a proof for a new class of primes.

$\blue \text{Let }p \text{ be any odd prime, and }V_{qsp}=\frac{2^q+sp} {(p-2)}\text{ and } s=\pm1$

$\blue V_{qsp}= \text{ is prime } \Longleftrightarrow \ S_{n} \equiv S_{r\times q } \pmod{V_{qsp}} \text{ , where: } S_0=c \text{ , and } \ S_{i}=S_i^2-2 \ \pmod{V_{q p}} \ .$

$\blue \text{where } n=1 \text{ or }n \text{ is such that } 2^n \text{ divides either } (V_{qsp}+1) \text{ or } (V_{qsp}-1)\text{ and }r \text{ an integer not necessary 1}$

$\blue \text{( and }c\text{ is such that all }S_x\text { are unique )}$

allow a day or two for me to adapt the paper presented in my post of thread

This method will only be of value if one can pick a $S_0=c$ to complete the cycle within $q$ and I am not sure yet if that can be done.

to convince you, you could try:
V_(12+5) = (2^12+5)/3=1367
5 -> 23 -> 527 -> 226 -> 495 -> 330 -> 905 -> 190 -> 556-> 192 -> 1320 -> 840 -> 226
or
19 -> 359 -> 381 -> 257 -> 431 -> 1214 -> 168 -> 882 -> 99 -> 230 -> 952-> 1348 -> 359

now the minus case and that q is prime is coincidental
V_(13-5)= (2^13-5)/3 = 2729
2434 -> 2424 -> 237 -> 1587 -> 2429 -> 2670 -> 750-> 324 -> 1272 -> 2414 -> 979 -> 560 -> 2492 -> 1587
or
2220 -> 2553 -> 955 -> 537 -> 1822 -> 1218 -> 1675 -> 211 -> 855 -> 2380 -> 1723 -> 2304 -> 509-> 2553

or you can try for yourself:

V_(13-23) = (2^13-23)/21
S_0 = 199, then S_2 = 15 = S_39
or
S_0 = 6, then S_1 = 34 = S_13

retina · 2008-10-06, 00:43

Quote:

Originally Posted by AntonVrba

5 -> 23 -> 527 -> 226 -> 495 -> 330 -> 905 -> 190 -> 556-> 192 -> 1320 -> 840 -> 226

Gives a cycle length of 9.

But earlier you say:

Quote:

Originally Posted by AntonVrba

$\ S_{q} \equiv S_{r\times q } \pmod{P}$

Which suggests a cycle length of 12.

Hey this tex thing is interesting, I didn't know we could do that.

AntonVrba · 2008-10-06, 00:53

Quote:

Originally Posted by retina

Gives a cycle length of 9.

But earlier you say:Which suggests a cycle length of 12.

Hey this tex thing is interesting, I didn't know we could do that.

12 includes 5 -> 23 -> 527 ->

Cycle bad terminology and my mistake S_q should have read S_n which has now been corrected in the post.

2008-10-06, 00:27	#1
AntonVrba Jun 2005 2·7² Posts	New Class of primes - proving algorithm Using my new proof for Wagstone numbers I hope to present a proof for a new class of primes. $\blue \text{Let }p \text{ be any odd prime, and }V_{qsp}=\frac{2^q+sp} {(p-2)}\text{ and } s=\pm1$ $\blue V_{qsp}= \text{ is prime } \Longleftrightarrow \ S_{n} \equiv S_{r\times q } \pmod{V_{qsp}} \text{ , where: } S_0=c \text{ , and } \ S_{i}=S_i^2-2 \ \pmod{V_{q p}} \ .$ $\blue \text{where } n=1 \text{ or }n \text{ is such that } 2^n \text{ divides either } (V_{qsp}+1) \text{ or } (V_{qsp}-1)\text{ and }r \text{ an integer not necessary 1}$ $\blue \text{( and }c\text{ is such that all }S_x\text { are unique )}$ allow a day or two for me to adapt the paper presented in my post of thread This method will only be of value if one can pick a $S_0=c$ to complete the cycle within $q$ and I am not sure yet if that can be done. to convince you, you could try: V_(12+5) = (2^12+5)/3=1367 5 -> 23 -> 527 -> 226 -> 495 -> 330 -> 905 -> 190 -> 556-> 192 -> 1320 -> 840 -> 226 or 19 -> 359 -> 381 -> 257 -> 431 -> 1214 -> 168 -> 882 -> 99 -> 230 -> 952-> 1348 -> 359 now the minus case and that q is prime is coincidental V_(13-5)= (2^13-5)/3 = 2729 2434 -> 2424 -> 237 -> 1587 -> 2429 -> 2670 -> 750-> 324 -> 1272 -> 2414 -> 979 -> 560 -> 2492 -> 1587 or 2220 -> 2553 -> 955 -> 537 -> 1822 -> 1218 -> 1675 -> 211 -> 855 -> 2380 -> 1723 -> 2304 -> 509-> 2553 or you can try for yourself: V_(13-23) = (2^13-23)/21 S_0 = 199, then S_2 = 15 = S_39 or S_0 = 6, then S_1 = 34 = S_13 Last fiddled with by AntonVrba on 2008-10-06 at 00:58

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