100$ prize for the solution of a conjecture

[axn]

Originally Posted by enzocreti

And so residue 6 mod 7 cannot occur?

Nothing he said has anything to do with whether 6 (mod 7) can occur as a prime. In fact, looking at the factorization of small values of the sequence, I don't think there is any particular reason for 6 to be an impossible value.

My prediction is, if you find enough primes (say another 20 or so), at least one of them will turn out to be 6 mod 7. Or more efficiently, a concerted effort to search all the 6 mod 7 numbers will eventually turn up a prime.

[Dr Sardonicus]

Originally Posted by enzocreti

And so residue 6 mod 7 cannot occur?

As mentioned long, long ago in this thread, the remainder 6 (mod 7) occurs when k == 1 (mod 3) and m == 3 (mod 6), or when k == 2 (mod 3) and m == 4 (mod 6).

The first occurrence of a remainder of 6 (mod 7) in the sequence is with k = 10, m = 3:

f(10) = 1023511 == 6 (mod 7).

The next occurrence is k = 11, m = 4:

f(11) = 20471023 == 6 (mod 7)

The vector of remainders (mod 7) up to k = 100 is

[3, 3, 3, 3, 0, 3, 2, 5, 3, 6, 6, 3, 4, 6, 3, 5, 2, 3, 1, 4, 3, 3, 3, 3, 2, 0, 3, 6, 5, 3, 4, 6, 3, 4, 2, 3, 5, 4, 3, 1, 3, 3, 3, 3, 3, 2, 0, 3, 6, 5, 3, 4, 6, 3, 5, 2, 3, 1, 4, 3, 3, 3, 3, 3, 0, 3, 2, 5, 3, 6, 6, 3, 4, 6, 3, 5, 2, 3, 1, 4, 3, 3, 3, 3, 2, 0, 3, 6, 5, 3, 4, 6, 3, 4, 2, 3, 5, 4, 3, 1]

The remainder 6 (mod 7) occurs for about 1/9 of all k. When an f(k) == 6 (mod 7) that's a prp might show up, is presently unknown.

In answer to my own remark in a previous post, I note that the remainder 3 (mod 7) does in fact occur for values of k that are not divisible by the multiplicative order (3) of 2 mod 7.

[axn]

Originally Posted by Dr Sardonicus

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
;-)

An interesting question is, do these ratios remain the same after trial division by small (say p < 1e6) primes? Obviously, r=0 will drop out of the consideration, but will others maintain the 1:2:8:2:2:2 ratio?

[axn]

Originally Posted by axn

An interesting question is, do these ratios remain the same after trial division by small (say p < 1e6) primes? Obviously, r=0 will drop out of the consideration, but will others maintain the 1:2:8:2:2:2 ratio?

Trying to answer my own question, the count for 100 <= k <= 100000 is [569, 1515, 6560, 1400, 1759, 1684]. More or less what you'd expect.

[enzocreti]

The most surprising fact is that i found five in a row prp congruent to 5 mod 7 for k=92020, 174968, 176006, 181015, 285019. Residue 5 occurs only in 1/9 of cases.

[Dr Sardonicus]

Originally Posted by axn

Nothing he said has anything to do with whether 6 (mod 7) can occur as a prime. In fact, looking at the factorization of small values of the sequence, I don't think there is any particular reason for 6 to be an impossible value.

My prediction is, if you find enough primes (say another 20 or so), at least one of them will turn out to be 6 mod 7. Or more efficiently, a concerted effort to search all the 6 mod 7 numbers will eventually turn up a prime.

That's the way I see it too.

Good grief, the much more straightforward sequence 168451*2^n + 1 went up to n = 19375200 before a prime was found.

[enzocreti]

Exponents k leading to a prime congruent to 5 mod 7 are 8, 394, 92020, 174968, 176006 and 285019. These exponents are congruent to plus or minus 7w mod 43

[enzocreti]

8 is congruent to -35 mod 43 for example

[enzocreti]

curiously if you sum the reciprocals of the exponents leading to a prime (1/2+1/3+1/4+...+1/285019) you get a value 1.64...which is very close to pi^2/6

[CRGreathouse]

Originally Posted by enzocreti

curiously if you sum the reciprocals of the exponents leading to a prime (1/2+1/3+1/4+...+1/285019) you get a value 1.64...which is very close to pi^2/6

1/2 + 1/3 + 1/4 + ... + 1/285019 = 12.1375285428... which is pretty far from Pi^2/6.

1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/285019^2 = 1.6449305583... which is about Pi^2/6 -- it's off by exactly 1/285020^2 + 1/285021^2 + 1/285022^2 + ... = 0.0000035085....

[enzocreti]

Originally Posted by CRGreathouse

1/2 + 1/3 + 1/4 + ... + 1/285019 = 12.1375285428... which is pretty far from Pi^2/6.

1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/285019^2 = 1.6449305583... which is about Pi^2/6 -- it's off by exactly 1/285020^2 + 1/285021^2 + 1/285022^2 + ... = 0.0000035085....

No i mean only the reciprocals of expoments leading to a prime 1/2 yes becasue for k=2 the concatenation gives a prime, k=16 no bevause it gives a composite number