Sequences of "periodic factors"; Jacobsthal-Lucas numbers

[creigh at o2online.de]

Observe the sequence:

a(0) = 2
a(2) = 7
a(3) = 11
a(4) = 25 = 5*5
a(5) = 47 
a(6) = 97
a(7) = 191
a(8) = 385 = 5*7*11
a(9) = 767 = 13*59
a(10) = 1537 = 29*53
a(11) = 3071 = 37*83
a(12) = 6145 = 5*1229
a(13) = 12287 = 11*1117
a(14) = 24577 = 7*3511
a(15) = 49151 = 23*2137
a(16) = 98305 = 5*19661
a(17) = 196607 = 421*467
a(18) = 393217 = 11*35747
a(19) = 786431 
a(20) = 1572865 = 5*7*44939
a(21) = 3145727 = 13*241979
a(22) = 6291457 = 347*18131

One could assume (law of small numbers) that the following holds: 
for m > n 
( s | a(n) ) and ( s | a(m) ) -> ( s | a(2m - n) )  

For ex. ( 7 divides a(2) = 7 ) and ( 7 divides a(8) = 385 = 5*7*11 )  
and ( 7 divides a(2*8 - 2) = a(14) = 24577 = 7*3511 ) 
and ( 7 divides a(2*14 - 8) = a(20) = 1572865 = 5*7*44939 )

[ If this is true, then 13 should divide a(21*2 - 9) =  a(33) ]

Do these types of sequences have a name (if so, what would be the best way 
to prove it...)? In any case, the above sequence (unlisted in OEIS) is connected 
with the Jacobsthal-Lucas numbers 
http://www.research.att.com/projects/OEIS?Anum=A014551 via

 c(n) +  b(n) + A014551(n+1) = 4*a(n) 
 
(b(n)) =  (5, 19, 29, 67, 125, 259, 509, 1027, 2045, 4099, 8189, ) 
; unlisted
(c(n)) = (2, 4, 8, 16, 32, 64, 128, )

Apparently, the above "law of small numbers" holds for A014551 as well,
ex.  A014551(5) = 31 and A014551(15) = 7*31*151 and 
A014551(25) = 31*601*1801
It probably also holds for (b(n)) [and trivially for (c(n)) ]

Now, for (Fib(n)), one has ( s | Fib(n) ) -> ( s | Fib(kn) ) which is 
readily seen to be a stronger condition than the above statement. 

Proof:  
( s | Fib(n) ) and ( s | Fib(m) ) ->   ( s | Fib(kn) ) and ( s | Fib(k'm) 
) 

A simple calculation shows:

F(2m - n) = | 2F(2m - n + 2) - F(2m - n + 3) | 
                = | -3F(2m - n + 3) + 2F(2m - n + 4) |
                = | 5F(2m - n + 4) - 3F(2m - n + 5) |
                = | F(n) F(2m-1) - F(n-1)F(2m) |

Thus, setting k' = 2 and k = 1, it follows that s |  F(2m - n),   q.e.d. 


Thanks/Sincerely, 
Creighton