100$ prize for the solution of a conjecture

[enzocreti]

Originally Posted by Dr Sardonicus

f(k) = 10^m*(2^k - 1) + 2^(k-1) -1, where m is the number of decimal digits in 2^(k-1) - 1.

We have m = 1 + floor((k-1)*log(2)/log(10)). Since 2^3 < 10 < 2^4, a given value of m will occur for either three or four consecutive values of k. Fortuitously, Mod(2,7) has multiplicative order 3, so every possible value of f(k) (mod 7) occurs for any given exponent m.

Since 10 == 3 (mod 7) and Mod(3, 7) has multiplicative order 6, there are 6*3 = 18 possible combinations of Mod(10,7)^m, Mod(2,7)^k. Simply calcuating the remainders in each case gives the proportions indicated in an earlier post. The values of m for which there are 4 consecutive values of k, "shift" the pairings (Mod(m,6), Mod(k,3)) so will cause slight departures from these proportions. The pairs (Mod(m, 6) = 3, Mod(k,3) = 1) and (Mod(m,6) = 4, Mod(k,3) = 2) are the only ones giving f(k) == 6 (mod 7).

By looking at k < 200, the values of k for which f(k) == 6 (mod 7) are k = 10, 11, 14, 28, 32, 49, 53, 70, 71, 74, 88, 92, 109, 113, 130, 131, 148, 152, 169, 173, 190, and 191.

Most (but not all) of the corresponding values of f(k) have small factors. The first k for which the smallest factor is greater than 100 is 11 (factor 479). The first k for which the smallest factor is greater than 1000 is k = 92 (smallest factor 730315371175567). The occurrence of such a large smallest factor makes it seem unlikely that the long run of composite f(k) == 6 (mod 7) could be explained by "covering congruences."

The only possible common factor between 10^m*(2^k - 1) and 2^(k-1) - 1 is 5, when k == 1 (mod 4). If k == 1 (mod 4*5^e) for a positive integer e, the factor 5 can occur e+1 times.

If k =/= 1 (mod 4), one could look for small values of r for which 10^m*(2^k - 1)*r + 2^(k-1) -1 or 10^m*(2^k - 1) + r*(2^(k-1) - 1) is prime.

Amazingly the last five probable primes found are all congruent to 5 mod 7...that's quite incredible!

[enzocreti]

Originally Posted by ATH

Up to k=24K the PRPs are:
k=2,3,4,7,8,12,19,22,36,46,51,67,79,215,359,394,451,1323,2131,3336,3371,6231,19179

with
n%7=3,3,3,2,5,3,1,3,3,2,3,2,1,2,4,5,4,3,1,3,3,3,3

from k=24K to 100K I only checked n%6, no primes.

The last five prp found are all congruent to 5 mod 7 that is surprising

[enzocreti]

I wonder if this sequence is random or not at all.

[science_man_88]

Conditions for it to be 6 mod 7:

k=2 mod 3; m= 4 mod 6 or
k=1 mod 3; m= 3 mod 6

Now to work on conditions of primality or compositeness to contrast with.

[enzocreti]

Yes i know...it is surprising that i found five probable primes in a row congruent to 5 mod 7...i cannot explain that residue 5 occurring only 1/9 of the times

[enzocreti]

Originally Posted by science_man_88

Conditions for it to be 6 mod 7:

k=2 mod 3; m= 4 mod 6 or
k=1 mod 3; m= 3 mod 6

Now to work on conditions of primality or compositeness to contrast with.

For k=92020, 174968, 176006, 181015, 285019 there are five probable primes congruent to 5 mod 7. This is even more surprising than res 6 not appearing. I wonder if this sequence is random but now i believe it is not

[science_man_88]

Originally Posted by enzocreti

Yes i know...it is surprising that i found five probable primes in a row congruent to 5 mod 7...i cannot explain that residue 5 occurring only 1/9 of the times

Mod 7

but these lengths that define m aren't all equally likely. The have: 4,3,3,4,3,... for length m increasing by 1 each entry.

[enzocreti]

So can you explain why occurred five prp in a row?

[Dr Sardonicus]

Originally Posted by enzocreti

The last five prp found are all congruent to 5 mod 7 that is surprising

If I have gotten the data right and have done my sums correctly, the values of n known so far yielding prps are

[2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019]

and the corresponding function values (mod 7) are

[3, 3, 3, 2, 5, 3, 1, 3, 3, 2, 3, 2, 1, 2, 4, 5, 4, 3, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 5, 5, 5, 5, 5].

I note that the nine consecutive values n = 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, and 75894 all yield prp's congruent to 3 (mod 7).

I find this at least as surprising as five consecutive function values congruent to 5 (mod 7).

[enzocreti]

I am astonished that nobody can give me an explanation

[enzocreti]

Do you know some mathematician who can give an explanation to all these coincidences?