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## Certain product sequences and their factors

define the sequence $$P_n$$ as follows:

$$P_{0} = 1$$ ; $$P_{1} = a$$ and $$P_{n} = 6P_{(n-1)}-P_{(n-2)} + 2a^2-8a+4$$

Then each term is a product of two numbers as follows
$$P_{n}= {1*1,1*a,a*b,b*c,c*d,d*e,\dots}$$
where
b = 2a-1
c = 4b-a
d = 2c-b
e = 4d-c
f = 2e-d
...
...

Has anyone come across this before

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Blog Entries: 2
 Quote by ramsey2879 define the sequence $$P_n$$ as follows: $$P_{0} = 1$$ ; $$P_{1} = a$$ and $$P_{n} = 6P_{(n-1)}-P_{(n-2)} + 2a^2-8a+4$$ Then each term is a product of two numbers as follows $$P_{n}= {1*1,1*a,a*b,b*c,c*d,d*e,\dots}$$ where b = 2a-1 c = 4b-a d = 2c-b e = 4d-c f = 2e-d ... ... Has anyone come across this before
Let a and b be integers
define the sequence $$P_n$$ as follows:

$$P_{0} = ab$$ ; $$P_{1} = b^{2}-ab$$ and $$P_{n} = 6P_{(n-1)}-P_{(n-2)} + 2a^{2}-8ab+4b^{2}$$

Then each term is a product of two numbers as follows
$$P_{n}= \{a*b,b*c,c*d,d*e,\dots\}$$
where
c = 4b-a
d = 2c-b
e = 4d-c
f = 2e-d
...
I am thinking of at looking at this further in various finite number systems $$Z_n$$ but right now I dont know if any of my sequences consist of complete residue sets or if they are just subsets thereof. The constant that is added in the recursive formula can also be written as $$4(b-a)^{2}-2a^2$$ and the sets (c,d) (e,f) (g,h) can also be interchanged for the a and b thereof.