[seqfan] Re: A combinatorial problem

Wed Aug 4 08:36:44 CEST 2010

```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,

----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Tuesday, August 3, 2010 12:30
Subject: [seqfan] Re: A combinatorial problem
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

>    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>
> >
> > >
> > > 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,
> > > >
> > > >  Shevelev Vladimir‎
> > > >
> > >
> > >
> > >
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> > >
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