Divisor d is the total number of divisors

[David Wilson]

These are the refactorable numbers, A033950.

----- Original Message ----- 
From: "Eric Angelini" <Eric.Angelini at kntv.be>
To: <seqfan at ext.jussieu.fr>
Sent: Monday, July 23, 2007 12:25 PM
Subject: Divisor d is the total number of divisors


>
>
> Hello SeqFans,
>
> Is this seq of interest?
> If yes could someone check and compute a few more terms?
>
> 1,8,9,12,18,24,36,...
>
> Integers I having one divisor which is also the total number of divisors 
> of I.
>
> 1 has 1 divisor which is 1
> 8 has 4 divs and 4 is one of them
> 9 has 3 divs and 3 is one of them
> 12 has 6 divs and 6 is one of them
> 18 has 6 divs and 6 is one of them
> 24 has 8 divs and 8 is one of them
> 36 has 9 divs and 9 is one of them
> ...
>
> 30 is not a member because 30 has 8 divs but not 8 itself : 
> [1,2,3,5,6,10,15,30]
>
> Best,
> É.
>
>
>
>
>
>
> -- 
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> 7:02 PM
> 




At 10:40 AM -0700 7/23/07, Max Alekseyev wrote:
>On 7/22/07, Paul D. Hanna <pauldhanna at juno.com> wrote:
>>
>>
>> Seqfans,
>>      Consider the nice sequence A006336:
>> a(n) = a(n-1) + a(n-1 - number of even terms so far).
>> http://www.research.att.com/~njas/sequences/A006336
>> begins:
>> [1,2,3,5,8,11,16,21,29,40,51,67,88,109,138,167,207,258,309,376,...].
>>
>> My COMMENT (NOT submitted to OEIS):
>> -----------------------------------------------------------
>> It seems that A006336 can be generated by a rule using the golden ratio:
>>
>> a(n) = a(n-1) + a([n/Phi]) for n>1 with a(1)=1  where Phi = (sqrt(5)+1)/2,
>>
>>
>> i.e., the number of even terms up to position n-1 equals:
>> n-1 - [n/Phi] for n>1 where Phi = (sqrt(5)+1)/2.
>
>To simplify notation, let p = Phi = (sqrt(5)+1)/2.

Nice, Max.

What does this sequence count? It is similar to A00123 and A005704, which
both have a recursion a(n)=a(n-1)+a([n/k]), where k is 2 and 3,
respectively.  Those sequences count "number of partitions of k*n into
powers of k". For sequence A006336, k=Phi.  Does A006336(n) count the
number of partitions of n*Phi into powers of Phi?

Tony