[seqfan] Re: A151659

Hagen von EItzen math at von-eitzen.de
Sun Jun 21 23:31:46 CEST 2009

```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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```