Start with the numbers x
1=9 and x
2=20. After that, each number x
n=2x
n-1+x
n-2.
x3=2x2+x1=2(20)+9=40+9=49
x4=2x3+x2=2(49)+20=98+20=118
x5=2x4+x3=2(118)+49=236+49=285
...
This gives the sequence 9, 20, 49, 118, 285, ... Prove that no number in this sequence is prime.
The relevant function is a(n) = 1/4 ((10 - 3 sqrt(2)) (1 + sqrt(2))^n + (10 + 3 sqrt(2)) (1 - sqrt(2))^n), or
a(n) = 1/4 ((10 - 3 sqrt(2)) (3 + 2 sqrt(2))^n + (10 + 3 sqrt(2)) (3 - 2 sqrt(2))^n) for even n, i.e. the odd-valued entries.
Strictly speaking 2 and 5 are prime and in the sequence, but they are specifically excluded by the wording of the problem.
Start with the idea that the function is cyclic mod every prime. Indeed this is true even for negative n.
Clearly every odd entry is divisible by 2: Mod2 n=(2k+1)=0
Then work out the cycles:
Mod3 {12202110}, n=2,6,10..,(4n+2)
Mod5 {310120240430}, n=3,6,9,..,(3n)
Mod7 {226065}, n=4,10,16,..,(6n+4)
Mod11{606186922631505(10)3529958(10)}, n=8,20,32,{12n+8}
Since 3,4,6,12 divide 12, consider these patterns mod 12.
1 2
2 3
3 2,3
4 7
5 2
6 3,5
7 2
8 11
9 2,5
10 3,7
11 2
12 5
thus completing the proof.
In like manner, it is also easily shown that no entry in the series is ever divisible by 13,17,29, or 41.
Edited on August 10, 2019, 11:19 pm
|
|
Posted by broll
on 2019-08-10 23:12:09 |
Possible solution
