# Sequences of "periodic factors"; Jacobsthal-Lucas numbers

creigh at o2online.de creigh at o2online.de
Mon Sep 13 20:29:32 CEST 2004

```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

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