following Rainer Rosenthal

[benoit]

In an other hand, one can modify the parameters in the quadratic 
recursion instead of modifying the initial conditions  .

For ex. fixing :

a(1)=1,  a(2)=1 :

and defining :

a(n+1)*a(n-1)=a(n)^2+A*a(n)+B

sequence (a(n))_{n>0}  satisfies the 3-order linear recurrence :

a(1)=1, a(2)=1, a(3)= A+B+1 :

a(n)=(2*A+B+3)*a(n-1)-(2*A+B+3)*a(n-2)+a(n-3)

Therefore, for any (A,B) in Z^2  (a(n)) is an integer sequence.


(except offset)
(A,B)=(-1,2)-->A000124
(A,B)=(-1,3)--> A069403
(A,B)=(-2,4) -->A002061
(A,B)=(-3,6)-->A005448
....

Note (a(1),a(2))=(1,1) is a special case, since a(1)=1 and a(2)=2 don't 
produce integer sequences for all (A,B) in Z^2. The only pairs 
(a(1),a(2)) in Z^2 with this property are :

(a(1),a(2))=(1,1)
(a(1),a(2))=(-1,1)
(a(1),a(2))=(1,-1)
(a(1),a(2))=(-1,-1)

disregarding the case a(1) or a(2) = 0.

Benoit
-------------- next part --------------
A non-text attachment was scrubbed...
Name: not available
Type: text/enriched
Size: 948 bytes
Desc: not available
URL: <http://list.seqfan.eu/pipermail/seqfan/attachments/20040314/6f92775d/attachment-0001.bin>