<html>
<body>
<br>
Something to note about A007733:<br>
For 2 <= n <= 84, A000010(n) / A007733(n) results in the following
sequence: <br>
1 1 2 1 1 2 4 1 1 1 2 1 2 2 8 2 1 1 2 2 1 2 4 1 1 1 4 1 2 6 16 2 2 2 2 1
1 2 4 2 2 3 2 2 2 2 8 2 1 4 2 1 1 2 8 2 1 1 4 1 6 6 32 4 2 1 4 2 2 2 4 8
1 2 2 2 2 2 8 1 2 1 4 <br>
and it seems reasonable to conjecture that A007733 always divides A000010
(<pre>Euler totient function phi(n)).
</pre><a href="http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A000010" eudora="autourl">http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A000010<br>
</a>-- Gerald<br><br>
At 09:32 AM 4/27/2005, Gerald McGarvey wrote:<br><br>
<blockquote type=cite class=cite cite="">A related entry the Mersenne
numbers is A007733 (Period of binary representation of 1/n.)<br>
<a href="http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A007733" eudora="autourl">http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A007733</a><br>
"<br>
<pre>Also sequence of period lengths for n's when you do
primality testing and calculate
"2^k mod n" from k=0 to k=n - Gottfried Helms
(helms(AT)uni-kassel.de), Oct 05 2000"
</pre><font face="Courier New, Courier"></font>Sincerely,<br>
Gerald<br><br>
At 08:45 AM 4/26/2005, Creighton Dement wrote:<br>
<blockquote type=cite class=cite cite="">> Dear Seqfans,<br>
><br>
> I would like to mention two specific examples concerning the
property,<br>
> for m > n : if s | a(n) and s | a(m) then s | a(2*m - n)<br>
><br>
> A proof is given, below, in an attempt to show that the first
example<br>
> has the property, the second doesn't (although, in a way, it
"almost"<br>
> seems to...).<br>
><br>
> Ex. 1.<br>
> a(n) = a(n-1) + 2a(n-2) + 2, a(0) = 0, a(1) = 1.<br>
> Ex. 2.<br>
> a(n) = a(n-1) + 2a(n-2) + 3, a(0) = 0, a(1) = 1.<br>
><br>
> Ex. 1., Mersenne numbers<br>
>
<a href="http://www.research.att.com/projects/OEIS?Anum=A000225" eudora="autourl">http://www.research.att.com/projects/OEIS?Anum=A000225</a><br>
> Sequence:<br>
> [0,1,3,7,15,31,63,127,255,511,1023,2047,4095,8191,16383,<br>
> 32767,65535,131071,262143,524287,1048575,2097151,4194303,<br>
> 8388607,16777215,33554431,67108863,134217727,268435455,<br>
> 536870911,1073741823,2147483647,4294967295]<br>
><br>
> Factorized:<br>
> [0, 1, (3), (7), (3)*(5), (31), (3)^2*(7), (127),
(3)*(5)*(17),<br>
> (7)*(73), (3)*(11)*(31), (23)*(89), (3)^2*(5)*(7)*(13),<br>
> (8191),(3)*(43)*(127), (7)*(31)*(151), (3)*(5)*(17)*(257),
(131071),<br>
> (3)^3*(7)*(19)*(73), (524287), (3)*(5)^2*(11)*(31)*(41),<br>
> (7)^2*(127)*(337), (3)*(23)*(89)*(683), (47)*(178481),<br>
> (3)^2*(5)*(7)*(13)*(17)*(241), (31)*(601)*(1801),
(3)*(8191)*(2731),<br>
> (7)*(73)*(262657), (3)*(5)*(29)*(43)*(113)*(127),
(233)*(1103)*(2089)]<br>
><br>
> Notice, for example, that the number 17 seems to reappear every
8<br>
> terms.<br>
> The number 3 every 2 terms, etc.. (This apparently defines a
function<br>
> f such that f(17) = 3, f(3) = 2, etc. which is beside the point
for<br>
> now.).<br>
> Also note that the fact that the number 17 actually appears for
the<br>
> first time after 8 terms, etc., (i.e. a(8) = 17) is not covered by
the<br>
> property- that seems to be an additional "symmetry" for
this<br>
> particular sequence (which would be in need of an additional
proof!).<br><br>
I assume that the following is the last sentence reformulated as a<br>
conjecture:<br>
for m > n: if s | 2^n - 1 and s | 2^m - 1 then <br>
(m - n) | A002326( (s-1)/2 )<br><br>
<a href="http://www.research.att.com/projects/OEIS?Anum=A002326" eudora="autourl">http://www.research.att.com/projects/OEIS?Anum=A002326</a><br>
( least m such that 2n+1 divides 2^m-1 )<br><br>
Not sure if that conjecture is somehow enlightening, obvious, or
would<br>
only serve to confuse - thus I will hold off on posting it as a
comment<br>
for now.<br><br>
Note: Correction, the above should read f(17) = 8. (I also
accidently<br>
gave the proof for 2^n+1 instead of 2^n - 1; essentially, both
proofs<br>
are the same)<br><br>
Sincerely,<br>
Creighton</blockquote></blockquote></body>
</html>