Re: Conjecture: 2n+1= 2^i+p ; 6k-2 or 6k+2 = 3^i+p

On 3月4日, 下午2时21分, Gerry Myerson <ge...@xxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
In article
<2cb32c1f-d320-4478-946f-b6d0d5260...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Tangerine Luo <tangerine....@xxxxxxxxx> wrote:

I have a conjecture:
Any odd positive number is the sum of 2 to an i-th power and a
(negative) prime.
2n+1 = 2^i+p

It's false.

Mladen V Vassilev-Missana, Note on extraordinary primes, Notes
Number Theory Discrete Math. 1 (1995) 111-113, MR 97g:11004,
proved that there are infinitely many primes q such that
2^k - q is composite for all k. For any such q, you can't
find k and (positive) prime p such that 2^k - q = p, so you can't
write q as 2^k - p.

--
Gerry Myerson (ge...@xxxxxxxxxxxxxxx) (i -> u for email)

Thank you very much!
It's a pity that I can't get the Mladen V Vassilev-Missana' article.
Fortunatly, I found that

In 1950 P.Erdos proved that if x= 2036812(mod5592405)and x= 3(mod62)
then x is not of the form 2^n + p where n is a nonnegative integer and
p is a prime.

Proceedings of the Edinburgh Mathematical Society (2002), Series 2,
45:155-160 Cambridge University Press
Copyright (c) Edinburgh Mathematical Society 2002
doi:10.1017/S0013091500000924

Research Article
A NOTE ON INTEGERS OF THE FORM 2n+cp

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