[seqfan] Re: A151659
franktaw at netscape.net
franktaw at netscape.net
Mon Jun 22 09:08:37 CEST 2009
First of all, the definition says UnitarySigma, not sigma. See
http://mathworld.wolfram.com/UnitaryDivisor.html for more information.
It is not hard to show that OddPart(UnitarySigma(n)) < n for any odd n
> 1.
As for OddPart(sigma(n)), I would be very surprised if anyone managed
to proved that the sequence from any integer always reaches 1. Because
if n is an odd perfect number, OddPart(sigma(n)) = n. And the
existence or otherwise of odd perfect numbers is still a matter of
conjecture.
I did check a while back; if we take f(n) = OddPart(sigma(n)), the only
n up to (if I remember correctly) 20 million with f(n) >= n and f(f(n))
>= n is 81.
Franklin T. Adams-Watters
-----Original Message-----
From: Hagen von EItzen <math at von-eitzen.de>
Benoît Jubin schrieb:
>> Given n, let b(0) = 2^n and recursively b(k+1) =
>> oddpart(UnitarySigma(b(k))); then b(k) = 1 for all sufficiently big k
>> (or does it?).
>> Define a(n) as the product of all UnitarySigma(b(k))/b(k), k>=0.
>
> Doesn't this give a(3)=4? the associated sequence in this case being
> b(0)=8, b(1)=15, b(2)=3, b(3)=1.
>
> Benoit
>
In the original notation ...
> Let a(0)=2^m, b(0)=2^m
> a(n)=a(n-1)/PF_2[UnitarySigma[b(n-1)]]
> b(n)=UnitarySigma[b(n-1)]/ PF_2 [UnitarySigma[b(n-1)]]
> IF b(k)=1 THEN END
> c(m)=a(k)
> Sequence gives 1/c(m)
... we start (for m=3):
a(0) = 2^3 = 8, b(0) = 8
a(1) = 8
/1 = 8, b(1) = (1+2+4+8)/1 = 15
a(2) = 8/8 =1, b(2) = (1+3+5+15)/8 = 3
a(3) = 1/4, b(3) = (1+3)/4 = 1
c(3) = 1/4
=> A151659(3) = 4.
This matches with my formula, for which you verified the same sequence
of b's and the value
A151659(3) = 15/8 * 24/15 * 4/3 * 1/1 = 4 (this is just a telescope
effect).
Unless I still completely misunderstood the definition, I think the
sequence should start
1, 2, 2, 4, 2, 4, 2, 8, 4, 4, 8, 8, 2, 4, 4, 8, 2, 8, 2, 8
instead of
1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 4, 4, 4, 8, 4.
Could someone clarify this?
BTW, I still wonder why the b sequence always reaches 1, after all we
could have oddpart(sigma(b)) > b repeatedly, e.g. oddpart(sigma(81)) =
121, oddpart(sigma(121)) = 133.
Hagen
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