[seqfan] Re: zig-zag pseudoprimes

[Vladimir Shevelev]

I was successful to complete a proof of the following more general statement:
 
1) if n==1(mod 4) is prime, then  A000111(n)==1(mod n); if n==3(mod 4) is prime, then A000111(n)==-1(mod n);
2) if n is a positive power of 2, then  A000111(n)==1(mod n); 
3) if n is double odd prime, then  A000111(n)==1(mod n).
 
The latter result allows to introduce "zig-zag pseudo-double-primes". They are double odd composite
numbers n, for which we have A000111(n)==1(mod n).
The sequence of them begins from 30, 182.
I think that it is also rather interesting sequence for the consideration.
 
Regards,
Vladimir


----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Thursday, September 2, 2010 16:08
Subject: [seqfan] Re: zig-zag pseudoprimes
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

>    I think, nevertheless, that your conclusion about 
> the statistics of sequence {n: A000111(n)==1
> (mod n)}, maybe, is not true for large n. Indeed, the 92 first 
> terms of this sequence which you presented one can split into 4 
> subsequences: 1) 4*k+1-primes ; 2) positive powers of 2; 3) 
> numbers of the form 2*p, where p is an odd prime; 4)  
> remaider subsequence {30,182,...}.
> Till now I proved that  1) and 2) are indeed subsequences 
> of the considered sequence and it is very plausible ( I did not 
> yet prove this) that 3) is also subsequence. Since, 
> nevertheless, union of these sequences has zero density, then 
> your statement is true only in case when sequence {30,182,...} 
> has a large density. It seems improbable.
>  
> Regards,
> Vladimir
> 
> 
> ----- Original Message -----
> From: Richard Mathar <mathar at strw.leidenuniv.nl>
> Date: Wednesday, September 1, 2010 18:25
> Subject: [seqfan] Re: zig-zag pseudoprimes
> To: seqfan at seqfan.eu
> 
> > 
> > http://list.seqfan.eu/pipermail/seqfan/2010-
> > September/005903.html :
> > vs> Nevertheless, I did not understand if you really obtained 
> > that, e.g., A000111(561)==1(mod 561)?
> > 
> > 
> > Yes. I naively trusted that the following Maple implementation 
> > works 
> > on my computer (one of the two branches of your problem in 
> comments):> 
> > A000111 := proc(n)
> >         2^n*abs( 
> > euler(n,1/2)+euler(n,1)) ;
> > end proc:
> > for n from 1 do
> >         if 
> modp(A000111(n),n) 
> > = 1 and modp(n,4) = 1 and not isprime(n) then
> >         # if 
> > modp(A000111(n),n) = n-1 and modp(n,4) = 3 and not isprime(n) then
> >                 printf("%d,\n",n) ;
> >         end if;
> > end do:
> > 
> > 
> > It is not unusual that A000111(n) == 1 (mod n), if we consider the
> > statistics in the following sequence of A000111(n) mod n, n>=1:
> > 
> > 
> 0,1,2,1,1,1,6,1,7,1,10,5,1,1,2,1,1,7,18,5,19,1,22,17,16,1,2,5,1,1,30,1,31,1,> 12,29,1,1,2,25,1,19,42,5,16,1,46,17,22,21
> > 
> > This is the reason why the first sequence I showed in
> > http://list.seqfan.eu/pipermail/seqfan/2010-September/005898.html
> > is rather dense and filled four lines until reaching 400. 561 
> > comes later
> > in the same sequence.
> > 
> > RJM
> > 
> > 
> > _______________________________________________
> > 
> > Seqfan Mailing list - http://list.seqfan.eu/
> > 
> 
>  Shevelev Vladimir‎
> 
> _______________________________________________
> 
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> 

 Shevelev Vladimir‎