100$ prize for the solution of a conjecture
 

I offer 100$ for the solution of this conjecture (I mean if you find a counter-example or give a proof):
[URL="https://math.stackexchange.com/questions/2635516/a-conjecture-about-numbers-of-the-form-10m2k%e2%88%9212k-1%e2%88%921-where-m-is"]https://math.stackexchange.com/questions/2635516/a-conjecture-about-numbers-of-the-form-10m2k%e2%88%9212k-1%e2%88%921-where-m-is[/URL]

Not a counterexample or proof but since nearly all terms are 1 mod 6, 6 mod 7 is equivalent to 13 mod 42.

number of decimal digits is such an arbitrary thing, the divisibility/primality questions with it are gonna be very arbitrary too, I'm afraid.

About the remainders mod 7...
 

The numbers in question, when viewed as strings of decimal digits, are obtained by concatenating the decimal digits of 2^k - 1 and 2^(k-1) - 1. This gives the first few terms

10, 31, 73, 157, 3115, 6331, 12763,...

Since the question concerned those of the numbers which happened to be congruent to 6 (mod 7), I computed the remainders (mod 7) out to k = one million, and found the following approximate proportions of k giving these remainders:

r = 0: 1/18
r = 1: 1/18
r = 2: 1/9
r = 3: 4/9
r = 4: 1/9
r = 5: 1/9
r = 6: 1/9

Perhaps someone might like to have a go at investigating this
;-)

Counter example: k=19179 k%7=6:

[CODE]10^5774*(2^19179-1)+2^(19179-1)-1 is 2-PRP! (1.6956s+0.0005s)
10^5774*(2^19179-1)+2^(19179-1)-1 is 3-PRP! (1.6811s+0.0005s)
10^5774*(2^19179-1)+2^(19179-1)-1 is 5-PRP! (1.6777s+0.0005s)
10^5774*(2^19179-1)+2^(19179-1)-1 is 7-PRP! (1.6963s+0.0005s)
10^5774*(2^19179-1)+2^(19179-1)-1 is 11-PRP! (1.6753s+0.0005s)
10^5774*(2^19179-1)+2^(19179-1)-1 is 13-PRP! (1.6836s+0.0005s)
10^5774*(2^19179-1)+2^(19179-1)-1 is 17-PRP! (1.6772s+0.0005s)
[/CODE]

EDIT: Nevermind it is n%7 that has to be 6, not k%7.

[QUOTE=Dr Sardonicus;488349]The numbers in question, when viewed as strings of decimal digits, are obtained by concatenating the decimal digits of 2^k - 1 and 2^(k-1) - 1. This gives the first few terms

10, 31, 73, 157, 3115, 6331, 12763,...

Since the question concerned those of the numbers which happened to be congruent to 6 (mod 7), I computed the remainders (mod 7) out to k = one million, and found the following approximate proportions of k giving these remainders:

r = 0: 1/18
r = 1: 1/18
r = 2: 1/9
r = 3: 4/9
r = 4: 1/9
r = 5: 1/9
r = 6: 1/9

Perhaps someone might like to have a go at investigating this
;-)[/QUOTE]

Yes if you read.the post on mathexchange or mathoverflow you will find in comments exactly that distribution of residues. And the strange thing is that i found 3 primes with residue 1 which occurs with half the frequency of residue 6 and none with residue 6

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.

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.

Find a prime
 

[QUOTE=ATH;488408]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
I checked up to 400.000 no prime found but a doublecheck would be apreciated

vector of exponents leading to a prime
 

Here the vector of exponents <=200.000 leading to a prime:
[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]
Can you find other primes in the range 200.000-400.000 (i mean exponents)?

new prime found
 

New prime found at 285019!