[seqfan] Re: On A035095

[Vladimir Shevelev]

Thank you, Charles!

Of course, it is dufficient to divide, say, (i^59-1)/(i-1) by A035095(17)=709,
2<=i<=prime(17)=59. But the case of n=4 shows that the smallest
divisor could take the minimum only for i=prime(n) (!) After your remark,
it is sufficient to divide (i^59-1)/(i-1) by 709 only for 2<=i<=19. If you
already have verified this, then a(17) is indeed 20.

Best regards,
Vladimir



________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Charles Greathouse [charles.greathouse at case.edu]
Sent: 01 December 2014 19:42
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: On A035095

Isn't the next term just 20, since (20^59 - 1)/19 has 709 as its least
prime factor?

I hope Peter isn't factoring each F(n,i), that would be a monumental task.

Charles Greathouse
Analyst/Programmer
Case Western Reserve University

On Mon, Dec 1, 2014 at 10:42 AM, Vladimir Shevelev <shevelev at bgu.ac.il>
wrote:

> Dear Seq Fans,
>
> A035095 is a classic sequence which is "a version of the
> "least prime in special arithmetic progressions" problem."[N. Sloane]
> A035095(n) is the smallest prime congruent to 1 mod n-th prime.
> I noted that it is also minimum of the smallest prime factors of
> F(n,i) = (i^prime(n)-1)/(i-1), when i runs all integers in
> [2, prime(n)]. Every prime factor of F(n,i) is congruent
> to 1 modulo prime(n). Indeed, for every considered i,
> F(n,i) is either prime or overpseudoprime to base i (see
> https://cs.uwaterloo.ca/journals/JIS/VOL15/Castillo/castillo2.pdf,
> Theorem 25).
> By handy I found for the first five n the suitable minimal
> values of i=i(n) on which the smallest prime factor of F(n,i(n))
> is A035095(n). Peter Moses extended the sequence {i(n)} up to n=16:
> 2,2,3,7,2,10,8,5,2,3,7,5,3,6,4,3,        (*)
> but he wrote me that "so far, the computer has been working
> on n=17 for over 16 hours!"
> I think that sequence (*) is very interesting and useful  for a further
> research.
>  If anyone could propose a simpler algorithm to calculate numbers (*)?
>
> Best regards,
> Vladimir
>
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>
> Seqfan Mailing list - http://list.seqfan.eu/
>

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