For example, the Dana Scott sequence:

*f*(n) * *f*(n-4) = *f*(n-1) * *f*(n-3) +
*f*(n-2)

Where *f*(1) = *f*(2) = *f*(3) = *f*(4)
= 1.

#begin Maple Code interface(prettyprint=false): f := proc(n) option remember ; if n<=4 then 1 else (f(n-1) * f(n-3) + f(n-2))/f(n-4) fi end: seq(f(n),n=1..38); #end Maple Code 1, 1, 1, 1, 2, 3, 5, 13, 22, 41, 111, 191, 361, 982, 1693, 3205, 8723, 15042, 28481, 77521, 133681, 253121, 688962, 1188083, 2249605, 6123133, 10559062, 19993321, 54419231, 93843471, 177690281, 483649942, 834032173, 1579219205, 4298430243, 7412446082, 14035282561, 38202222241

This can be generalized to the recurrance:

F(n,k) * F(n-4,k-2) = F(n-1,k-1) * F(n-3,k-1) + F(n-2,k-1)

*n* is a positive integer and *k* is an integer.

F(n,k) = x_{n,k} for *n* < 5

Then F(n,k) is always a Laurent polynomial in the x_{n,k} variables
with all coefficents equal to 1. (a *faithful polynomial*)

Define the sequence {*g*}

*g*(n) = 10 *g*(n-3) -10 *g*(n-6) + *g*(n-9)

where *g*(n) = *f*(n) for 1 < n < 9

*g*(n) = *f*(n) for all *n* up to 6000.

If we can prove that *g*(n) = *f*(n) for all
*n*, then we have shown that {*f*} is always an
integer.

Update: I just did that.