[seqfan] Re: Smallest prime wich is equal to a multiple of his reversal +/- a prime smaller than him in n ways
njasloane at gmail.com
Tue Mar 13 19:40:30 CET 2012
In fact Hans Havermann replied to my post, although he
didn't indicate that in the subject line. He said, basically, go look at
He is right! I did add a companion, A209063.
On Tue, Mar 13, 2012 at 2:37 PM, Richard Mathar
<mathar at strw.leidenuniv.nl>wrote:
> njas> I guess this asks for the smallest prime p for which there are n
> njas> for n=1,2,3,...
> njas> But p=71 has three solutions:
> njas> 71 = 2*17 +37
> njas> 71 = 4*17 + 3
> njas> 71 = 6*17 - 31,
> njas> so surely the sequence begins 31, 41, 71, 61, 6421, 8501, ... ?
> njas> Or am I missing something? The next term, if it exists, is > 48000.
> The next terms by brute force search is 8116001.
> There is some indication that the sequence is finite: We are searching for
> primes p such that we have n solutions to p = k*R(p)+-q with q<p another
> prime, k=1,2,3...
> So we need p +-q = k*R(p), and the left hand side needs to be <= 2p.
> so k <= (2p)/R(p). As R(p) has the same number of digits as p, 2*p/R(p)
> be much larger than roughly 20 in the best case, supposed that the most
> digit of p is 9 and the least significant digit is 1. So as the number of
> cannot be larger than k, the sequence ought be no longer than (roughly) 20.
> Richard Mathar
> Maxima code:
> Ameller(n) := block(
> L : [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0] ,
> for pi : 1 do (
> p : A000040[pi],
> R : digrev(p),
> s : 0 ,
> for k : 1 thru floor(2*p/R) do (
> r : abs(p-k*R),
> if primep(r) and r < p then
> s : s+1
> if s < length(L) and s> 0 and L[s] = 0 then (
> L[s] : p ,
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