[seqfan] An arithmetic conjecture

Thu Mar 5 19:31:40 CET 2009

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, K = 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: