# [seqfan] For reflection and calculations

Tomasz Ordowski tomaszordowski at gmail.com
Mon Sep 21 20:00:44 CEST 2020

```Dear readers!

By the dual Sierpinski conjecture,
there are no odd primes p such that (p-2^m)2^n+1 is composite for every
1<2^m<p and n>0.
Are there odd primes p such that (p-2^m)2^n+1 is composite for every
1<2^m<p and 1<2^n<p ?
Find odd numbers k>1 such that (k-2^m)2^n+1 is composite for every 2^m<k
and 2^n<k.
Note that such a number k (if it exists) must be composite (put m=n=0).

By the dual Riesel conjecture,
there are no odd primes p such that (p+2^m)2^n-1 is composite for every m>0
and n>0.
Are there odd primes p such that (p+2^m)2^n-1 is composite for every
1<2^m<p and 1<2^n<p ?
Find odd numbers k>1 such that (k+2^m)2^n-1 is composite for every 2^m<k
and 2^n<k.
Note that such a number k (if it exists) must be composite (put m=n=0).

Here are my conjectures in three combinations:

Conjecture 1.
(a) There are infinitely many odd numbers k>1 such that (k-2^m)2^n-1 is
composite for every 2^m<k and 2^n<k. (*)
However, (b) there are no odd numbers k>1 such that (k-2^m)2^n+1 is
composite for every 2^m<k and 2^n<k.

Conjecture 2.
(a) There are infinitely many odd numbers k>1 such that (k+2^m)2^n+1 is
composite for every 2^m<k and 2^n<k. (**)
However, (b) there are no odd numbers k>1 such that (k+2^m)2^n-1 is
composite for every 2^m<k and 2^n<k.

Conjecture 3.
(a) There are infinitely many odd naturals k that are not of the form
p+2^m+2^n. (***)
(b) There are infinitely many odd integers k that are not of the form
p-2^m-2^n.
However, (c) every odd integer k is of the form p+2^m-2^n or -p-2^m+2^n.
Where p is a prime.
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Note: for m>=0 and n>=0 in all three of the above conjectures.

These conjectures are an attempt to solve the problems
and answer the questions from my previous post.

Any comments are welcome!

Best regards,

Thomas Ordowski
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(*) There are odd numbers k>1 such that
(k-2^m)2^n-1 is composite for every 2^m<k and 2^n<k,
namely 65535 and 13975275 [found by Amiram Eldar].
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(**) There are odd numbers k>1 such that
(k+2^m)2^n+1 is composite for every 2^m<k and 2^n<k,
namely 14568915 [Amiram Eldar found only this one].
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