# [seqfan] Re: A combinatorial problem

Robert G. Wilson v rgwv at rgwv.com
Mon Aug 2 22:10:12 CEST 2010

```Et al,

If you are going to rearrange the divisors, which is contrary to the
definition but is worth studying,
then I get the following sequence:
{1, 1, 1, 1, 1, 4, 1, 1, 1, 4, 1, 8, 1, 4, 4, 1, 1, 8, 1, 8, 4, 4, 1, 14,
1, 4, 1, 8, 1, 72, 1, 1, 4, 4, 4, 20, 1, 4, 4, 14, 1, 72, 1, 8, 8, 4, 1, 22,
1, 8, 4, 8, 1, 14, 4, 14, 4, 4, 1}

Since we are using 12 as an example, I get the following as permutations
which satisfy the criterion:
{1,2,4,12,6,3}
{1,3,6,2,4,12}
{1,3,6,12,4,2}
{2,1,3,6,12,4}
{2,4,12,6,3,1}
{3,1,2,4,12,6}
{3,1,2,6,12,4}
{3,6,12,4,2,1}

Several of the above sets are just rotation of others.

Sincerely yours, Bob.

--------------------------------------------------
From: "Alois Heinz" <heinz at hs-heilbronn.de>
Sent: Monday, August 02, 2010 2:27 PM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: A combinatorial problem

>
> http://mathworld.wolfram.com/Arrangement.html
>
> I computed different values of A179926(n) for the following n:
> 12, 18, 20, 28, 30, 36, 42, 44, 45
>
> First example: A179926(12)=3:
> [12, 6, 3, 1, 2, 4]
> [12, 4, 2, 6, 3, 1]
> [12, 4, 2, 1, 3, 6]
> Last example: A179926(45)=3:
> [45, 15, 5, 1, 3, 9]
> [45, 9, 3, 15, 5, 1]
> [45, 9, 3, 1, 5, 15]
>
> seq (A179926(n), n=1..120);
> 0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 2,
> 1, 1, 3, 1, 3, 2, 2, 1, 4, 1, 2, 1, 3, 1, 18,
> 1, 1, 2, 2, 2, 8, 1, 2, 2, 4, 1, 18, 1, 3, 3,
> 2, 1, 5, 1, 3, 2, 3, 1, 4, 2, 4, 2, 2, 1, 106,
> 1, 2, 3, 1, 2, 18, 1, 3, 2, 18, 1, 17, 1, 2,
> 3, 3, 2, 18, 1, 5, 1, 2, 1, 106, 2, 2, 2, 4,
> 1, 106, 2, 3, 2, 2, 2, 6, 1, 3, 3, 8, 1, 18,
> 1, 4, 18, 2, 1, 17, 1, 18, 2, 5, 1, 18, 2, 3,
> 3, 2, 2, 572
>
> Alois
>
>> Dear SeqFans,
>>
>> I have submitted the following sequence:
>>
>> %I A179926
>> %S A179926
>> 0,1,1,1,1,2,1,1,1,2,1,2,1,2,2,1,1,2,1,2,2,2,1,4,1,2,1,2,1,12,1,1,2,2,2,
>> %T A179926 2,1,2,2,4,1,12,1,2,2,2,1
>> %N A179926 a(n) is the number of arrangements of all divisors of n of the
>> form d_1=n, d_2, d_3,...,d_tau(n) such that d_(i+1)/d_i is prime or
>> 1/prime
>> %C A179926 In view of formulas given below, there are many common first
>> terms with A001221.
>> %F A179926 a(p^k)=1, a(p*q)=a(p^2*q)=a(p^2*q^2)=2, a(p^3*q)=4, a(pqr)=12
>> (here p,q,r are distinct primes, k>=1).
>> %Y A179926 A000005 A001221
>> %K A179926 nonn
>> %O A179926 1,1
>>
>> More terms? More formulas? Corrections?
>>
>> Regards,
>>
>>
>
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>

```