# [seqfan] Re: A combinatorial problem

Alois Heinz heinz at hs-heilbronn.de
Wed Aug 4 15:24:12 CEST 2010

```Vladimir,

http://list.seqfan.eu/pipermail/seqfan/2010-August/005485.html
which has:

%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).

I strongly believe, as I told you in
http://list.seqfan.eu/pipermail/seqfan/2010-August/005517.html
that the correct values of a(p_1*p_2*...*p_n) for n = 1, 2, 3, 4, 5
are  1, 2, 18, 5712, 5859364320

Alois

> 1) I mind partitions with taking account of order of summands (it is better say "compositions").
> 2) I err in the calculation of  Sum{j=1,...,n-1}L_j for n=4. We have L_1=1; since 2=2=1+1, then L_2=2*1+1=3; since 3=3=2+1=1+2=1+1+1, then L_3=3*2+2*1+2*1+1=11 and
> Sum{j=1,...,n-1}L_j=1+3+11=15. Therefore, by the formula
>  a(p_1*p_2*p_3*p_4)=4*18*15=1080 (not 1368).
>
> If anyone can confirm this result by a direct calculation?
>
> Regards,
>
> ----- Original Message -----
> From: Vladimir Shevelev <shevelev at bgu.ac.il>
> Date: Wednesday, August 4, 2010 9:55
> Subject: [seqfan] Re: A combinatorial problem
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
>
>
>> I get a recursion soluton when A is a finite set or,
>> equivalently, I obtained a recursion for
>>  a(p_1*p_2*...*p_n), where  a(n)=A179926(n). It is:
>> a(p_1)=1 and for n>=2 we have
>>
>> a(p_1*p_2*...*p_n)=n*a(p_1*p_2*...*p_(n-1))*Sum{j=1,...,n-1}L_j,
>>
>> where
>>
>> L_j=SumProd{t=1,...,j}(i_t!*(i_t-1)!),
>>
>> where sum is over all partitions of j with parts i_t>=1.
>>
>> For n=1,2,3,4, I get 1,3,18,1368.
>>
>> More terms?
>>
>> Regards,