[seqfan] An arithmetic conjecture
Peter Luschny
peter.luschny at googlemail.com
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] = 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:
More information about the SeqFan
mailing list