[seqfan] Re: A151659

[Benoît Jubin]

> 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



2009/6/21 Hagen von EItzen <math at von-eitzen.de>
>
> zbi74583.boat at orange.zero.jp schrieb:
> > Hi, Seqfans
> >
> > A151659 is still “uned”.
> >
> > I suppose Neil might not understand the definition.
> > Could anyone tell me a better description?
> >
> The definition might appear somewhat convoluted due to things explained
> that are not needed for the definition:
>
> > Definition of PrimeFactor_p or PF_p: PrimeFactor_p[n] = largest power
> of p diiding n. It is written as PF_p[n]
>
> Actually, the definition needs only PF_2(n), defineable as PF_2(m*2^k) =
> k if m is odd.
> In fact, I suggest to rather use oddpart(n) = A000256(n) = m if n =
> m*2^k and m odd, see below,
> and let UnitarySigma(n) = A000203(n).
>
>
> > PF_p,q,r[n] = PF_p[n]* PF_q[n]* PF_r[n]
>
> Redundant, apparently not needed below
>
>
> > PF_1[n] = 1
>
> Redundant, apparently not needed below
>
>
> > How to compute c(m):
>
> The target sequence should be called a(n), thus some variables should be
> exchanged for clarity.
>
>
> > Case of Base Primes = {2}{1}
>
> Redundant, apparently not needed below
>
>
> > 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)
>
>
>
> It took be a short while, but finally I understand the definition better
> in the following form (if it really does match what is meant?)
>
>
> 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.
>
> (I see that a(n) is always a power of 2, but is it clear that it is
> alway an integer? Could we not end up with 1/2?)
>
> The following PARI code might also be equivalent:
> a(n) = { b=2^n;t=-n;while(b>1,b=sigma(b);while(b%2==0,b/=2;t++));2^t }
>
> I hope all this sheds some light on the sequence definition.
>
> Hagen
>
> > The factorization of term becomes 2^r
> > My friend who is Japanese understood it.
> >
> > Yasutoshi
> >
> >
> >
> >
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> >
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> >
> >
>
>
>
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