following Rainer Rosenthal

benoit abcloitre at
Sun Mar 14 20:24:29 CET 2004

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 :


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

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


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

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

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 :


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

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