[seqfan] Re: linear recurrence multiples

[Dale Gerdemann]

Sorry: In my last post, I accidentally reversed b and c. If the recurrence
is defined by a(0)=0, a(1)=1, a(n) = b*a(n-1) + c*a(n-2), then as long as c
<= b, formulae for m*a(n) can be generated using the method I described.

El El dom, 21 jul 2019 a las 2:25, Dale Gerdemann <dale.gerdemann at gmail.com>
escribió:

> 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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>>
>