[seqfan] Re: linear recurrence multiples

[Dale Gerdemann]

In the formula a(n) = b*a(n-1) + c*a(n-2), b and c are intended to be
integers. For example if (b,c) = (1,1) you have the Fibonacci sequence and
 here you can find a nice formula  for each positive integer m:

m=2:  2*a(n) = a(n+1) + a(n-2)

m=3: 3*a(n) =  a(n+2) + a(n-2)

m=4: 4*a(n) = a(n+2) + a(n) + a(n-2)

m=5:  5*a(n) =  a(n+3) + a(n-1) + a(n-4)

The interesting thing about these formulae is the relation to golden ratio
base.  If t is the golden ratio, (1+sqrt(5))/2, then:

2 = t^1 + t^{-2}

3 = t^2 + t^{-2}

4 = t^2 + t^0 +t^{-2}

5 = t^3 + t^{-1} + t^{-4}

I experimented with finding simular formulae for other recurrences, i.e.
for other values of b and  c.  The easiest way to do this is to pick a
large value for  n, such as n=100. Then find a greedy sum for m*a(100) such
as 2*a(100) = a(101) + a(98). Then generalize this to 2*a(n) = a(n+1) +
a(n-2). Finally, to be sure that the generalized formulae is correct, it
needs to be checked with some values other than  n=100. Now my
experimentation seems to show that this method yields valid formulae for
recurrences defined with  b <= c, but  the generalized formulae are never
valid when c > b. So what is going on here?

El El dom, 21 jul 2019 a las 0:32, Fred Lunnon <fred.lunnon at gmail.com>
escribió:

>   I'm afraid it is unclear to me what is meant by "formula" in this
> context.
> For example, does it mean "linear recurrence with integer coefficients,
> such that at least one (or exactly one) coefficient equals  m " ?
>
>   In that case the question would reduce to consideration of polynomial
> multiples of a polynomial in a single variable.
>
> WFL
>
>
>
> On 7/19/19, Dale Gerdemann <dale.gerdemann at gmail.com> wrote:
> > I have a question concerning recurrences of the form a(0)=0, a(1)=1,a(n)
> =
> > b*a(n-1)+c*a(n-2). When (b,c) is (3,2), for example, we have A007482 and
> > when (b,c)=(2,3), we have A015518. For A007482, it seems to be possible
> to
> > find formulae for m*a(n) for any m. For example, if m=4, we have 4*a(n) =
> > a(n+1)+a(n-1)+2*a(n-2). For A015518,  however, it seems that no such
> > formulae can be found. It appears that such formulae can be found just in
> > case b >= c. Is this correct?
> >
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> >
>
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