Referencing the OEIS

Jonathan Sondow jsondow at alumni.princeton.edu
Thu Jul 24 18:34:32 CEST 2008

```----- Original Message -----
From: "Christian G. Bower" <bowerc at usa.net>
To: "seqfan" <seqfan at ext.jussieu.fr>
Sent: Friday, July 25, 2008 6:45 AM
Subject: Re: Number of m such that n=0 Mod UnitarySigma(m)

>> Hi, Seqfans.
>
>> b(n) = Number of m such that n=0 Mod UnitarySigma(m)
>> b(n) : 1,1,2,2,2,3,1,3,3,3,1,5,1,2,4,3,2,5,1,5
>
>> Ex : 6=0 Mod UnitarySigma(m) , m=1,2,5, so, b(6)=3
>
> Perhaps I don't understand this correctly, but I get b(12)=6 and you
> have b(12)=5
>
> UnitarySigma(1)=1
> UnitarySigma(2)=3
> UnitarySigma(3)=4
> UnitarySigma(5)=6
> UnitarySigma(6)=12
> UnitarySigma(11)=12
>
> giving me 6 values of m : 12==0 Mod UnitarySigma(m)
>
> Also, if I understand correctly, b(n) is inverse moebius transform of
> A063974
> http://www.research.att.com/~njas/sequences/A063974
> and begins 1, 1, 2, 2, 2, 3, 1, 3, 3, 3, 1, 6, 1, 2, 3, 3, 2, 6, 1, 6, 2...
>
> (not in EIS)
>
> ...
>
>> 5 < b(300) = 14
>
> I get b(300)=27
>
>> Yasutoshi
>
> Christian
>
>
>
b(2^2*3*5^5)=27
m=1,2,3,4,5,6,9,11,12,18,19,20,29,36,38,44,45,49,59,76,98,116,180,236,245,269,299

Thank you for correcting my sequence.
You are right.

Yasutoshi

Sequence A140362 (Integers n which are the product of two distinct primes
and which divide the sum of the squares of the divisors of n) is currently
has only three terms: 10, 65, 20737.

For a while I was sure that the sequence was complete. Then I noticed that
the terms could be written as Fib(3)*Fib(5), Fib(5)*Fib(7), and
Fib(11)*Fib(13); that is, the product of prime Fibonacci numbers whose
indices are twin primes!

Using A001605, two additional terms should be Fib(431)*Fib(433) and
Fib(569)*Fib(571):

735108038169226697610336266421235332619480119704052339198145857119174445190576122619635288017445230931072695163057441061367078715257112965183856285090884294459307720873196474208257

and

3523220957390444959595279062040480245884253791540018496569589759612684974224639027640287843213615446328687904372189751725183659047971600027111855728553282782938238390010064604217978755993551604318057918269182928456761611403668577116737601

So the question is: are there other solutions?

Tony

```