Proof that 9555…5559 is Composite
                                                         By Cino Hilliard
                                                            07/16/2005

Theorem 1:
There are no prime numbers in the sequence 959,9559,95559,..955..559) 
The generating function for this sequence is  S(x)=9*10^(x+1)+9 + 50*(10^x-1)/9.

Proof:
We complete our proof by eliminating the 6 possible cases that the value of x can take.

Case 0. S(x) is div by   3 if x = 6k
Case 1. S(x) is div by   7 if x = 6k+1
Case 2. S(x) is div by 11 if x = 6k+2
Case 3. S(x) is div by   3 if x = 6k+3
Case 4. S(x) is div by 11 if x = 6k+4
Case 5. S(x) is div by 13 if x = 6k+5
 k = 0, 1, 2. .

Some expansions from algebra.

1. a^n – b^n = (a-b)(a^(n-1)  + a^(n-2)b + . . . + b^(n-1))
2. a^n – b^n = (a+b)(a^(n-1)  - a^(n-2)b + . . .  - b^(n-1))  if n is even.
3. a^n + b^n = (a+b)(a^(n-1) - a^(n-2)b + . . . + b^(n-1))  if n is odd.

Case 0.
S(6k) = 9*10^(6k+1) + 9 + 50*(10^6k-1)/9. Obviously, 3 divides the first 2 terms. Now
(10^6k-1) = (1000^2k-1) = (999)h from 1. for some integer h. then 999/9=111 = 37*3. So 3
divides S(6k).  <o:p></o:p>

Case 1.
S(6k+1) = 9*10^(6k+2) + 9 + 50*(10^(6k+1)-1)/9. Subtract 6 from 9*10^(6k+2) + 9 to get
9*(10^(6k+2) +3. Subtract 7*12=84 to get 9*(10^(6k+2) - 9). Now we test 10^(6k+2) – 9 by
subtracting 91 to get  10^(6k+2) – 100 = 100*(10^(6k) -1) = 100*(1000^2k-1) =100*1001h
divisible by 7 according to 2. So 9*(10^(6k+2) +3 is divisible by 7. Since we subtracted 6 from
9*10^(6k+2) + 9, we now add 6 to  50*(10^(6k+1)-1)/9 to get (5*10^(6k+2)-50)/9 +15 =
5*(10^(6k+2)+17)/9 . We subtract 10^(6k+2)+17 – 9*13 = 100*(1000^2k-1). So from 2. we
 50*(10^(6k+1)-1)/9 +6 = 50*1001h = 0 mod 7 . Therefore 7 divides S(6k+1)<o:p></o:p>

Case 2.
S(6k+2) = 9*10^(6k+3) + 9 + 50*(10^(6k+2)-1)/9. From 3. 9*(1000^(2k+1)+1)  =1001h for
some h. so 11 divides  9*10^(6k+3) + 9. similarly, from 2, 10^(6k+2)-1) = 1001h for some h.
Therefore 11 divides S(6k+2).

Case 3.
S(6k+3) = 9*10^(6k+4) + 9 + 50*(10^(6k+3)-1)/9. Obviously, 3 divides the first 2 terms.
 Now according to 1., (10^(6k+3)-1) = (1000^(2k+1) -1) = (999)h for some integer h. then
999/9=111 = 37*3. So 3 divides S(6k+3).  <o:p></o:p>

Case 4.
S(6k+4) = 9*10^(6k+5) + 9 + 50*(10^(6k+4)-1)/9. From 3. (1000^(6k+5)+1)  =1001h for
some h. so 11 divides  9*10^(6k+5) + 9. Similarly, from 2., 10^(6k+4)-1) = 1001h for some h.
Therefore 11 divides S(6k+4).

Case 5.
S(6k+6) = 9*10^(6k+6) + 9 + 50*(10^(6k+5)-1)/9. Subtract 5 from 9*10^(6k+6) + 9 to get
9*10^(6k+6) + 4. Now subtract 13 to get 9*10^(6k+6)  -9 = 9*(10^(6k+6)-1). Then by 2.,
(1000^(2k+2) – 1) =1001h for some h. So 7 divides = 9*10^(6k+6) + 9- 5 Since we
subtracted 5 from the first 2 terms, we now add 5 to 50*(10^(6k+5)-1)/9 to get
(5*10^(6k+6)–40)/9 + 5 = (5*10^(6k+6)–50 + 45)/9 = 5*(10^(6k+6)–1)/9 and by 2. we
have 5*(1001^(2k+2) -1)/9. So 7 divides 50*(10^(6k+5)-1)/9 + 5. Therefore, 7 divides
S(6k+6).

We have shown in all possible cases of x that S(x) = 9*10^(x+1) + 9 + 50*(10^x-1)/9 +9 
is composite and divisible by the primes 3,7,11 or 13.<o:p></o:p>

Cino Hilliard