Linear() -> Lucas in pfgw

[CRGreathouse]

I'm searching for primes in a Fibonacci-like recurrence:
a(1) = 8193, a(2) = 28618, a(n) = a(n-1) + a(n-2).

I entered an ABC2 file like so

ABC2 Linear(8193, 28618, 36811, 65429, $a)
a: from 1 to 100000

and pfgw seems to search it, but it gives the message with each a value:

Linear() -> Lucas P=1,Q=-1

so I think it wants me to enter it differently. What's the right way?

[paulunderwood]

C.3.3 Linear
The Linear() function will attempt to find a two-term recurrence
relation with constant coefficients to fit the input data. The
coefficients it discovers are output to the screen and may be
used in lucasU or lucasV. The two Lucas sequences are 'primitive'
solutions to the recurrence relation, and it is up to the user
to work out what their Linear() function is in terms of them.
The 'primitive' solutions have factorization properties, while
the general Linear() solution does not.
If the Linear() function fails to fit the data, the expression
will not evaluate.

I am not sure about how it is done!

The manual goes on to say:

C.3.4 lucasV
The Lucas sequences are defined via a standard recurrence
relation:

U(0)=0 U(1)=1 U(n) = pU(n-1)-qU(n-2)
V(0)=2 V(1)=p V(n) = pV(n-1)-qV(n-2)

These definitions allow the functions U and V to possess a
standard set of identities, no matter the values of p and q.
However, even values of p will produce a V-sequence that is
always even - searchers should either remove this factor 2,
or use the primitive part.

So if your sequence cannot have <0,1> nor <2,p>, then I guess you are stuck with Linear. If you are running Linux you can redirect the noise to /dev/null

[CRGreathouse]

You know, I think I just miscalculated the first time around, because it looks like it is indeed a linear combination of the relevant Lucas sequences, which in this case are the Lucas sequence and the Fibonacci sequence:

(20425*L($a) - 4039*F($a))/2

Sorry for the trouble.