[seqfan] Re: An arithmetic conjecture
David Wilson
davidwwilson at comcast.net
Fri Mar 6 06:26:09 CET 2009
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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