# [seqfan] Re: A combinatorial problem

Vladimir Shevelev shevelev at bgu.ac.il
Tue Aug 3 11:00:45 CEST 2010

```   I would like indicate an equivalent multiset formulation of the problem. For a given finite multiset
A we should, beginning with A, to get all submultisets of A, if, by every step, we remove or join 1 element. How ways to do this?
It is interesting even a subproblem, if A is a finite set.
Maybe, anyone met such statement of  a problem?

Regards,

----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Monday, August 2, 2010 22:59
Subject: [seqfan] Re: A combinatorial problem
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> 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).
>
> 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,
>
>
> ----- Original Message -----
> From: Alois Heinz <heinz at hs-heilbronn.de>
> Date: Monday, August 2, 2010 21:57
> Subject: [seqfan] Re: A combinatorial problem
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
>
> >
> > By "arrangements" you probably mean "permutations", see also
> > 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
> >
> > Vladimir Shevelev schrieb:
> > > 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,
> > > Vladimir
> > >
> > >  Shevelev Vladimir‎
> > >
> >
> >
> >
> > _______________________________________________
> >
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> >
>