[seqfan] Re: A combinatorial problem

Vladimir Shevelev shevelev at bgu.ac.il
Tue Aug 3 10:18:04 CEST 2010


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), but I absolutely do not agree that a(n) is calculated "so easily". Try, please, find a recursion for it. Try, please, calculate by handy a(p*q*r*s). So I think that a(n) is very hard calculated (at least, till now).

Vladimir


----- Original Message -----
From: Alois Heinz <heinz at hs-heilbronn.de>
Date: Tuesday, August 3, 2010 0:01
Subject: [seqfan] Re: A combinatorial problem
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> Vladimir Shevelev schrieb:
> > Yes, you are right: here I use "arrangements" in this sense.
> > Of course, a(45)=a(12) since a(n) is function of exponents of 
> prime 
> > power factorization of n only (moreover, it is invariant with 
> respect 
> > of permutations of them).
> So you agree that a(45)=a(12)=3, and not 2 as in your submission?
> 
> And the values of A179926 for n =
> 18, 20, 28, 30, 36, 42, 44  are
>  3,  3,  3, 18,  8, 18,  3, 
> respectively, and not
>  2,  2,  2, 12,  2, 12,  2, as in your 
> submission?
> Please be more careful when you submit a sequence
> that can be computed so easily.
> 
> Alois
> >
> > Let us prove that, for n>=2, a(n)>=1. Let v(n) denote the 
> number of 
> > prime divisors of n. If v(n)=1, n=p^a, then the unique 
> required 
> > arrangement of divisors is p^a, p^(a-1),...,p,1.
> > Suppose that, for m with v(m)=k and with prime divisors 
> p_1,...,p_k, a 
> > required arrangement exists. Let it be d_1=m, d_2,...,d_tau(m).
> > Consider now n=m*p_(k+1)^a. For divisors of n we have an 
> arrangement:> d_1*p_(k+1)^a =n, d_2*p_(k+1)^a,...,d_tau(m)*p_(k+1)^a,
> > d_tau(m)*p_(k+1)^(a-1), d_(tau(m)-1)*p_(k+1)^(a-
> 1),...,d_1*p_(k+1)^(a-1),
> > d_1*p_(k+1)^(a-2),...,d_tau(m)*p_(k+1)^(a-2), ...
> > such that, finally, we obtain a required arrangement for 
> divisors of n.
> >
> > Using this idea, for n=p_1^a_1*p_2^a_2*...*p_k^a_k, we have
> >
> > a(n)>=Sum {i=1,...,k}a(n/p_i^a_i)+Sum 
> > {1<=i<j<=k}a(n/(p_i^a_i*p_j^a_j))*a(p_i^a_i*p_j^a_j)+...
> > Therefore,on every subsequence of the form n_1,n_2,... ,when 
> > v(n_1)<v(n_2)<..., sequence a(n_i) grows very fast.
> >
> > Regards,
> > Vladimir
> >
> 
> 
> 
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> 

 Shevelev Vladimir‎



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