100$ prize for the solution of a conjecture

[enzocreti]

Originally Posted by M344587487

The odds of a very specific thing happening may be low but you're not looking only for that specific thing. It's a trap to think like that. The odds of my exact genetic code being produced through hundreds of generations and my subsequent development leading to being interested in primes enough to be on this forum and talking to you are so vanishingly small it's incredible. The odds of someone reading your comment and taking a similar stance as me if I weren't here are high.




You're essentially looking for the intersection between the set of prime numbers and the set of numbers given by an equation with no randomness. Prime numbers are not random and the equation is not random, so the intersection cannot be random. It's just hideously complex, far more complex than we can unravel. We can tease out some properties based on what we know about the two sets, but we can't predict the next term in the sequence with any degree of accuracy any more than we can predict the next Mersenne prime.

For example I tried to cancatenate in reverse. Instead of 255127, 127255. All residues appear early and there are not coincidences of any sort!

[Dr Sardonicus]

Originally Posted by enzocreti

Ok but would you admit that all that coincidences at once are quite surprising?
[snip]

Given the small number of prp's found, my answer is "no." I had no prior expectations of when or how often prp's might show up in the sequence, or what their residues (mod 7) might be.

I don't know what you mean by "coincidences." I don't know how to estimate the probability of a given n producing a function value that is a prp.

What I did do was, look at the proportion of n's producing multiples of given small primes p. The results were curious, and the determination a bit tricky.

All function values are odd. All values are congruent to 1 (mod 3). Values are divisible by 5 when n is congruent to 1 (mod 4). About 1/18 of all values are divisible by 7 (n has to be congruent to 2 (mod 3), and m has to be congruent to 2 (mod 6)). About 1/20 of n's produce values divisible by 11 (n has to be congruent to 4 (mod 10), and m has to be even). Each value of m repeats 3 or 4 times, so one may assume that the values of m are fairly evenly distributed mod p-1 or whatever.

The above results indicate to me that sieving might have somewhat disappointing results, compared to its application to types of numbers where the proportion divisible by p is usually something like 1/p or an integer multiple of 1/(p-1).

[science_man_88]

Originally Posted by Dr Sardonicus

Given the small number of prp's found, my answer is "no." I had no prior expectations of when or how often prp's might show up in the sequence, or what their residues (mod 7) might be.

I don't know what you mean by "coincidences." I don't know how to estimate the probability of a given n producing a function value that is a prp.

What I did do was, look at the proportion of n's producing multiples of given small primes p. The results were curious, and the determination a bit tricky.

All function values are odd. All values are congruent to 1 (mod 3). Values are divisible by 5 when n is congruent to 1 (mod 4). About 1/18 of all values are divisible by 7 (n has to be congruent to 2 (mod 3), and m has to be congruent to 2 (mod 6)). About 1/20 of n's produce values divisible by 11 (n has to be congruent to 4 (mod 10), and m has to be even). Each value of m repeats 3 or 4 times, so one may assume that the values of m are fairly evenly distributed mod p-1 or whatever.

The above results indicate to me that sieving might have somewhat disappointing results, compared to its application to types of numbers where the proportion divisible by p is usually something like 1/p or an integer multiple of 1/(p-1).

Mod 7 it repeats a pattern every 60 entries. [3,3,6,28,7,6,7] is the distribution mod 7

[enzocreti]

What does it imply that?

[science_man_88]

Originally Posted by enzocreti

What does it imply that?

Mine implies only 3 in 60 ( aka on average 1 in 20) divide by 7. 3 in 60 have a remainder of 1 on division by 7, 6 in 60 ( aka 1 in 10) have a remainder of 2 on division by 7, 28 of 60 ( 7 in 15) have a remainder of 3 on division by 7, 7 in 60 have a remainder have a remainder of 4 on division by 7, 6 in 60( aka 1/10) have a remainder of 5 on division by 7, and 7 in 60 have a remainder of 6 on division by 7.

[Dr Sardonicus]

Unfortunately, the sequence of residues (mod p) is non-repeating. The reason is, the value of the exponent m repeats sometimes 3 times, sometimes 4 times, and the sequence is AFAIK not predictable. It depends on the sequence of integer multiples of

log(2)/log(10) = 0.30102999566+ ,

which is an irrational number. It is known that the fractional parts of the integer multiples of any irrational number are uniformly distributed in (0,1), so we can say here that about 1 - 3*c, or about 9.61% of m-values repeat four times rather than 3. But which 9.61%, who can say?

[science_man_88]

Originally Posted by Dr Sardonicus

Unfortunately, the sequence of residues (mod p) is non-repeating. The reason is, the value of the exponent m repeats sometimes 3 times, sometimes 4 times, and the sequence is AFAIK not predictable. It depends on the sequence of integer multiples of

log(2)/log(10) = 0.30102999566+ ,

which is an irrational number. It is known that the fractional parts of the integer multiples of any irrational number are uniformly distributed in (0,1), so we can say here that about 1 - 3*c, or about 9.61% of m-values repeat four times rather than 3. But which 9,61%, who can say?

I can ... mod 7, 10^m has period 6 containg under it 20 entries per loop, the mersennes used have period 3, theses match up again every 60 entries. Edit: similar arguments with a twist show periodicity mod 11 every 300 entries.

[enzocreti]

If you want and it doesnt take too much time you could search for other primes with exponent >300.000

[science_man_88]

Originally Posted by enzocreti

If you want and it doesnt take too much time you could search for other primes with exponent >300.000

forprime(p=1,1000,print(lcm(lcm(p-1,3)*10,p-1)","p))
30,2
60,3
120,5
60,7
300,11
120,13
480,17
180,19
660,23
840,29
300,31
360,37
1200,41
420,43
1380,47
1560,53
1740,59
600,61
660,67
2100,71
720,73
780,79
2460,83
2640,89
960,97
3000,101
1020,103
3180,107
1080,109
3360,113
1260,127
3900,131
4080,137
1380,139
4440,149
1500,151
1560,157
1620,163
4980,167
5160,173
5340,179
1800,181
5700,191
1920,193
5880,197
1980,199
2100,211
2220,223
6780,227
2280,229
6960,233
7140,239
2400,241
7500,251
7680,257
7860,263
8040,269
2700,271
2760,277
8400,281
2820,283
8760,293
3060,307
9300,311
3120,313
9480,317
3300,331
3360,337
10380,347
3480,349
10560,353
10740,359
3660,367
3720,373
3780,379
11460,383
11640,389
3960,397
12000,401
4080,409
12540,419
4200,421
12900,431
4320,433
4380,439
13260,443
13440,449
4560,457
13800,461
4620,463
13980,467
14340,479
4860,487
14700,491
4980,499
15060,503
15240,509
15600,521
5220,523
5400,541
5460,547
16680,557
16860,563
17040,569
5700,571
5760,577
17580,587
17760,593
17940,599
6000,601
6060,607
6120,613
18480,617
6180,619
6300,631
19200,641
6420,643
19380,647
19560,653
19740,659
6600,661
6720,673
20280,677
20460,683
6900,691
21000,701
7080,709
21540,719
7260,727
7320,733
7380,739
22260,743
7500,751
7560,757
22800,761
7680,769
23160,773
7860,787
23880,797
24240,809
8100,811
24600,821
8220,823
24780,827
8280,829
25140,839
8520,853
25680,857
8580,859
25860,863
8760,877
26400,881
8820,883
26580,887
9060,907
27300,911
9180,919
27840,929
9360,937
28200,941
28380,947
28560,953
9660,967
29100,971
29280,977
29460,983
9900,991
9960,997

Not necessarily correct for all of them as for example p=23 divides into a mersenne number with index less than p-1

[Dr Sardonicus]

Originally Posted by science_man_88

I can ... mod 7, 10^m has period 6 containg under it 20 entries per loop, the mersennes used have period 3, theses match up again every 60 entries.
[snip]

I tabulated the values of n up to 400 giving a new value of m that repeats 4 times. The values are regularly spaced, differing by 10 -- until they aren't. The lengths of the AP's with common difference 10 also vary...

So, like I said, there's no periodicity in m (mod p).

[1, 11, 21, 31, 41, 51, 61, 71, 81, 91, 104, 114, 124, 134, 144, 154, 164, 174, 184, 197, 207, 217, 227, 237, 247, 257, 267, 277, 287, 300, 310, 320, 330, 340, 350, 360, 370, 380, 393]

[enzocreti]

I am astonished that nobody knows something about this sequence. Only a person told me that this sequence is NOT random, but he didn't explain why it is NOT random...I am very angry for that