[seqfan] Re: A combinatorial problem
Vladimir Shevelev
shevelev at bgu.ac.il
Wed Aug 4 20:27:16 CEST 2010
Unfortunately, maybe, I don't know all possibilities of seqfan list, but I see at the first time that you "told me" in http://list.seqfan.eu/pipermail/seqfan/2010-August/005517.html. Maybe, it is a good idea, and, maybe, I did not take into account anything in my combinatorial construction. If you are right, then, in any case, as I anderstand, there not exist any general formula for A003043(n).
You wonder how do my last results fit to my own submission? I have already written the following:
"Alois, when I by handy calculated a(12) etc., this was concordant with a general model of computation of a(n) which turnes out to be only a lower estimate. I agree that a(12) is, indeed, 3 (not 2)", moreover, I sent Doug's and your corrections into A179926. In its turn, I can wonder: " how do your last conclusions fit to your own opinion that A179926(n) "is so easily calculated""? You state that A179926 contains A003043; so, maybe, you can easily calculate of A003043? In such case,
please, add the sixth term after the Max's fifth one.
Summarizing , we see that the problem is very difficult, but I believe that either you or I or anyone from our colleagues will find an unerring full combinatorial solution of it.
Best regards,
Vladimir
----- Original Message -----
From: Alois Heinz <heinz at hs-heilbronn.de>
Date: Wednesday, August 4, 2010 16:33
Subject: [seqfan] Re: A combinatorial problem
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
>
> Vladimir,
>
> how do your last results fit to your own submission, see
> 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
>
> Vladimir Shevelev schrieb:
> > 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,
> > Vladimir
> >
> > ----- 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,
> >> Vladimir
> >>
> >>
> >>
> >>
>
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>
Shevelev Vladimir
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