100$ prize for the solution of a conjecture

[Dr Sardonicus]

Originally Posted by enzocreti

I am astonished that nobody knows something about this sequence.

Au contraire, there is a lot known about both the whole sequence, and the primes embedded therein. For example: none of the terms are divisible either by 2 or 3. The terms for which n is congruent to 1 (mod 4) are divisible by 5.

For any prime p > 5, the proportion of terms divisible by p can be found (though which terms are divisible by p is not easily predicted).

The rate of growth of the sequence is fairly clear.

There is a formula to calculate the terms of the sequence.

True, the sequence has no obvious divisibility properties like a^n - 1.

And, alas, the sequence has no obvious theoretical importance, so there is little impetus to throw a lot of resources at determining which terms are prp's.

I have no sense of how many prp's one might "expect" to find for n <= X for large X.

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

I suggest you reread this post to this thread. Long story short-- the given sequence isn't random. The sequence of primes isn't random. So their intersection isn't random.

BTW, since the constant log(2)/log(10) = log₁₀2 plays a role in defining the terms of this sequence, I will mention that this number is not only irrational (easily proved by invoking unique factorization), but transcendental (easily proved by invoking the Gelfond-Schneider Theorem).

[CRGreathouse]

Originally Posted by enzocreti

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

What do you think that it means to be random?

[enzocreti]

Originally Posted by Dr Sardonicus

Au contraire, there is a lot known about both the whole sequence, and the primes embedded therein. For example: none of the terms are divisible either by 2 or 3. The terms for which n is congruent to 1 (mod 4) are divisible by 5.

For any prime p > 5, the proportion of terms divisible by p can be found (though which terms are divisible by p is not easily predicted).

The rate of growth of the sequence is fairly clear.

There is a formula to calculate the terms of the sequence.

True, the sequence has no obvious divisibility properties like a^n - 1.

And, alas, the sequence has no obvious theoretical importance, so there is little impetus to throw a lot of resources at determining which terms are prp's.

I have no sense of how many prp's one might "expect" to find for n <= X for large X.


I suggest you reread this post to this thread. Long story short-- the given sequence isn't random. The sequence of primes isn't random. So their intersection isn't random.

BTW, since the constant log(2)/log(10) = log₁₀2 plays a role in defining the terms of this sequence, I will mention that this number is not only irrational (easily proved by invoking unique factorization), but transcendental (easily proved by invoking the Gelfond-Schneider Theorem).

What is the formula for calculating the terms of the sequence?

[enzocreti]

log(2)/log(10) = log102 how does it play a role in the terms of the sequence?

[enzocreti]

Why this sequence there wasn't in Oeis if you say that it is yet known?

[enzocreti]

Originally Posted by CRGreathouse

What do you think that it means to be random?

These numbers seem to be not random at all...why residue 6 didn't yet occur? Why five in a row probable primes with residue 5? Are that all coincidences???

[enzocreti]

Why we found 3 probable primes with residue 1 when the residue 1 occurs with half the frequency of the residue 6?

[Dr Sardonicus]

Originally Posted by enzocreti

log(2)/log(10) = log₁₀2 how does it play a role in the terms of the sequence?

The sequence in which you're looking for primes, as given here, is

f(k) = 10^m * (2^k - 1) + 2^(k-1) - 1,

where m is the number of decimal digits in 2^(k-1).

The value of m is

m = 1 + floor((k-1)*log₁₀(2)), i.e.

m = 1 + floor((k-1)*log(2)/log(10)).

[CRGreathouse]

Originally Posted by enzocreti

These numbers seem to be not random at all...why residue 6 didn't yet occur? Why five in a row probable primes with residue 5? Are that all coincidences???

Probably.
http://mathworld.wolfram.com/StrongL...llNumbers.html
https://www.ime.usp.br/~rbrito/docs/2322249.pdf

[Dr Sardonicus]

Originally Posted by science_man_88

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

The sequence

f(k) = 10^m*(2^k - 1) + 2^(k-1) - 1, where

m = 1 + floor((k-1)*log(2)/log(10)), the number of decimal digits in 2^(k-1),

is 1 (mod 2) for n > 1, and 1 (mod 3) for all n.

Since 10 is divisible by 5, the sequence is also periodic (mod 5) with period 5-1 = 4, the sequence of remainders (mod 5) being 0, 1, 3, 2, 0, 1, 3, 2,...

For p > 5, the sequence is not periodic (mod p), but there is one specific remainder, namely (p - 1)/2, that does recur when n is divisible by the multiplicative order of 2 (mod p), which multiplies the factor 10^m (mod p) by 0 (mod p), taking it out of consideration. (I am too lazy to work out whether this remainder can also occur at other values of n.)

The reason the sequence is not periodic (mod p) for any p > 5 is, first of all, the terms 2^k - 1 and 2^(k-1) - 1 are periodic (mod p), with period equal to the multiplicative order h of 2 (mod p). The non-periodicity of f(k) comes from the non-periodicity of m (mod h). (This problem disappears when p divides 2^k - 1, as noted above.)

To show that m is not periodic modulo any integer N > 1, we show that, for any given positive positive irrational constant c, and any given positive integer L,

floor((n + L)*c) - floor(n*c) is not constant. We use frac(x) to denote x - floor(x).

Clearly, (n + L)*c = floor(n*c) + frac(n*c) + floor(L*c) + frac(L*c), so that

floor((n+L)*c) = floor(n*c) + floor(L*c) if frac(n*c) + frac(L*c) < 1, and floor(n*c) + floor(L*c) + 1, if frac(n*c) + frac(L*c) > 1.

Now it is a well known result that frac(n*c) is uniformly distributed in (0, 1), so both possibilities occur (and, given the value of frac(L*c), we can say how often). Thus,

floor((n + L)*c) - floor(n*c) is not constant, so m is not periodic modulo any integer N greater than 1. Hence f(k) is not periodic (mod p) for any prime p > 5.

[enzocreti]

And so residue 6 mod 7 cannot occur?