[seqfan] Re: An arithmetic conjecture

[David Wilson]

To my knowledge, no one knows how to prove your statement.

Without getting heavily into it, this problem belongs to a class of problems 
with problems like:

Does every sufficiently large power of 2 include the digit 0 in base 10?

which statistically are true with probability 1, but have not to my 
knowledge been proved.

I could go more deeply into this problem. I have a general conjecture 
implying that all sufficiently large m satisfy [2^m / 3^k] mod 6 = 3 for 
some k > 0, but I cannot prove this conjecture nor can I vouch for m = 26 as 
the largest exception.

Explaining my conjecture is too involved for a late night email. I might go 
into it if there is sufficient interest from seqfan (or Tanya).

----- Original Message ----- 
From: "Peter Luschny" <peter.luschny at googlemail.com>
To: <seqfan at list.seqfan.eu>
Sent: Thursday, March 05, 2009 1:31 PM
Subject: [seqfan] An arithmetic conjecture


> Dear all,
>
> an arithmetic conjecture from an old discussion
> in the newsgroup de.sci.mathematik:
>
> ============== Conjecture ===================
>
> For all m > 26 there exist a k > 0 such that
>         [2^m / 3^k] mod 6 = 3.
>
> =============================================
>
> Can someone give a proof?
>
> The attempt of numerical falsification gives
> rise to two sequences, defined by:
>
> Start M[1] = 1, K[1] = 0. For given m > 1
> let s(m) denote the smallest k such that
> the conjecture holds - assuming existence -
> or 0 if not. Further let t(m) be the maximum
> of the s(i) for all i <= m.
>
> Now list those pairs m,t(m) where t(m) increases.
>
> M : 1, 5, 8, 19, 21, 27, 49, 110, 118, 165, 2769, 2837, 3661, 14354,
> 59913, 2712849,
> K : 0, 2, 3,  5, 12, 15, 21,  29,  34,  58,   61,   65,   70,    74,
> 103,     121,
>
> To paraphrase the evidence of these sequences:
> "You can quickly find such a k, even for large m."
>
> Can someone extend the sequences?
>
> Cheers Peter
>
> Maple, in the region of conjectured validity:
> maxK := 1; pow2 := 2^26;
> for m from 27 to 1000 do
>  k := 1; pow3 := 3; pow2 := pow2 + pow2;
>  while modp(iquo(pow2,pow3),6) <> 3 do
>        pow3 := 3*pow3: k := k+1; od;
>  if k > maxK then maxK := k; print(m,maxK); fi;
> od:
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/


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