Fibonacci, a generalization

[Gerald McGarvey]

Emeric and seqfans,

It appears that if we substitute a value of x for the 4 in 1 + 
4*cos(j*Pi/n)^2 below
than we get the sequence defined by the recurrence
a(0) = a(1) = 1, a(n) = a(n-1) + (x/4)*a(n-2).
Is this known?

For example
  0: all 1's
  4: Fibonacci numbers
  8: A001045 Jacobsthal sequence: a(n) = a(n-1) + 2a(n-2), starts as 0, 1, 
1, 3, 5, ...
12: A006130 a(n) = a(n-1) + 3a(n-2), a(0) = a(1) = 1
16: A006131, a(n) = a(n-1) + 4*a(n-2)
20: A015440, Generalized Fibonacci numbers, a(n) = a(n-1) + 5 a(n-2)

I first used this PARI code:
f(n,m) = p=1;for(j=1,ceil(n/2)-1,p=p*(1+m*cos(j*Pi/n)^2));return(p)
for(m=0,12,print1(m,":");for(n=1,10,print1(f(n,m),", "));print(" "))

then this code:
fr(n,m)=round(f(n,m))
for(m=0,24,print1(4*m,":");for(n=1,10,print1(fr(n,4*m),", "));print(" "))

Regards,
Gerald

At 10:40 AM 10/15/2006, Emeric Deutsch wrote:
>I did find it hidden in Fibonacci's Problem Book (M. Bicknell
>and V. E. Hoggatt, Jr, eds), pp. 47-48.
>
>Emeric
>
>
>On Sun, 15 Oct 2006, Emeric Deutsch wrote:
>
>>Dear Seqfan,
>>Is the following known?
>>
>>fibonacci(n)=Product(1 + 4[cos(j*Pi/n)]^2,  j=1..ceil(n/2)-1)
>>
>>Thanks,
>>Emeric