[seqfan] Re: A possible new Keyword

Max Alekseyev maxale at gmail.com
Thu Sep 24 20:03:24 CEST 2015


Notice that there is also a way to enter coefficients of polynomials in
infinite number of variables without referring to partitions.
This is what I call prime factorization order:

For n with the prime factorization n = p1^k1 * p2^k2 * ..., where p1=2,
p2=3, ... are the primes in their natural order (so only a finite number of
ki are nonzero),
we let a(n) be the coefficient of u[1]^k1 * u[2]^k2 * ....

E.g., multinomial coefficients C(k1+k2+...; k1, k2, ...) arranged this way
give http://oeis.org/A008480

Regards,
Max



On Thu, Sep 24, 2015 at 11:47 AM, Juan Arias de Reyna <arias at us.es> wrote:

>
> Frequently we consider  sequences of  polynomials in an infinite number of
> variables.
>
> For example the fourth polynomial in a given sequence I find is
>
> >  - u[1]^4 + 4 u[1]^2 u[2] - 2 u[2]^2 - 4 u[1] u[3] + 4 u[4]
>
> There is a coefficient for each partition of  n = 4. In this case each
> coefficient is
> related to a partition of 4
>
> 4 = 1 + 1 + 1 + 1       coef  -1
>   = 1 + 1 + 2           coef   4
>   = 1 + 3               coef  -4
>   = 2 + 2               coef  -2
>   = 4                   coef   4
>
>
> To include the sequence of coefficients we want to give a canonical order
> to
> all possible partitions.
> I will consider useful that all partition of the number n precede all
> partitions of n+1.
>
> But how to order the partitions of a number n?
> Perhaps writing the summands in increasing order and then by lexicografic
> order  as  I have
> written above the partitions of 4.
>
> Maybe somebody here in OEIS has a better election.
> I will like to fix a particular order,
> then in all cases that a sequence of
> this type is included in OEIS this same order should be used.  Also a new
> keyword as the
> actual tabf and tabl should be added to the Keywords file, to indicate
> this type of sequence.
>
>
> Somebody know sequences of this type already included in OEIS?
> Has this problem appeared before?
>
> Best regards,
> Juan Arias de Reyna
>
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>
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>


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