# [seqfan] Re: A151659

Hagen von EItzen math at von-eitzen.de
Mon Jun 22 07:51:43 CEST 2009

```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
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

>
>
> 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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```