(k-+2^m)2^m+-1

[Tomasz Ordowski]

Hello Everyone!  

If 3 | k, then 3 | (k-2^m)2^m+1 and 3 | (k+2^m)2^m-1. 

Let's define: odd numbers k > 1 indivisible by 3 such that 

both (k-2^m)2^m+1 and (k+2^m)2^m-1 are composite 

for every m > 0 with 2^m < k.  I'm asking for data. 

 

Are there any prime numbers among them? 

I do not think such primes exist; hmm...  

My conjecture: if p is an odd prime, 

then there is (at least one) prime 

of the form (p-+2^m)2^m+-1, 

where 1 < 2^m < p. 

Am I wrong? 

 

Best, 

 

Tom Ordo 

[D. S. McNeil]

I think there are lots of such primes, unless I've made a typo:

In [29]: print(counters)
[6599, 9371, 12539, 13109, 25321, 34949, 39839, 40129, 46181, 48821, 49919, 59407, 69473, 69739, 70583, 78167, 87509, 88289, 89189]

For conjectures like this, it's not hard to learn enough coding to test them, and you don't even need to install anything locally these days with open interpreters and notebooks widely available.

Doug

[Tomasz Ordowski]

Doug, thanks! 

Yes, I checked your primes, they are correct.
Forgive me for taking up your precious time.

Until next time, 

Tom 

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[Tomasz Ordowski]

PS. Here's something better for those interested. 

Let a(n) be the smallest odd k indivisible by 3 such that 

(k+2^m)2^m-1 are all composite for m <= n with m >= 1.  

Is the sequence (a) bounded? Namely a(n) = K for n >= N. 

If so, then (K+2^m)2^m-1 are all composite for every m >= 1. 

   Let b(n) be the smallest prime p > 3 such that 

(p+2^m)2^m-1 are all composite for m <= n with m >= 1.

Is the sequence (b) unbounded? It may be so!

   Since, by the dual Riesel conjecture, 

(p+2^m) is not a Riesel number. 

   It's better than nothing.

All right?