[seqfan] Re: Sigma-related sequence not in OEIS, need 4th term

Jack Brennen jfb at brennen.net
Tue Aug 16 17:40:05 CEST 2011

```Sorry, replace that first line with:

a(k) = smallest integer n>1 such that the first k+1 terms of the iterated

On 8/16/2011 2:41 AM, Benoît Jubin wrote:
>> a(n) = smallest integer n>  1 such that the first n+1 terms of the iterate
>> sigma sequence:
>>
>>    n, ...
>
> There are too many 'n's in here...
>
> Benoit
>
>
> On Mon, Aug 15, 2011 at 12:45 PM, Jack Brennen<jfb at brennen.net>  wrote:
>> a(n) = smallest integer n>  1 such that the first n+1 terms of the iterated
>> sigma sequence:
>>
>>    n, sigma(n), sigma(sigma(n)), ...
>>
>> all have the same number of divisors.
>>
>> a(1) = 2
>> a(2) = 52
>> a(3) = 4112640
>>
>> To illustrate:
>>   n=52; the sequence goes 52, 98, 171, 260, ...
>>
>>   52, 98, and 171 each have 6 divisors, but 260 has 12 divisors.
>>
>> What is a(4)?  It exceeds 2.0793*10^10, determined by exhaustive search to
>> that limit, but I'm thinking that exhaustive search can probably be improved
>> upon through some sort of multi-level sieve...
>>
>> I find it surprising how rare these numbers become -- numdiv(sigma(X)) is a
>> function with a fairly "clumpy" distribution.  And by using consecutive
>> starting numbers rather than iterating, we can easily find even longer
>> chains, such as the seven numbers from 17331728 to 17331734, each of which
>> has numdiv(sigma(X)) equal to 144.  But when iterating, duplication seems to
>> be much harder to come by.
>>
>>
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>>
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>>
>
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>

```