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COMMENTS
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This sequence is finite and complete. Proof:
Suppose a(9) = p exists. Then, we obtain the sequence of 10 primes:
E = {4p + 3 -> 16p + 15 -> 64p + 63 ->...-> (2^20)p + (2^22 - 1)}.
The prime divisors or 2^(2*q) - 1 for q = 1,2,...,11 are
(2^2 - 1) -> {3};
(2^4 - 1) -> {3, 5};
(2^6 - 1) -> {3, 7};
(2^8 - 1) -> {3, 5, 17};
(2^10 - 1) -> {3, 11, 31};
(2^12 - 1) -> {3, 5, 7, 13};
(2^14 - 1) -> {3, 43, 127};
(2^16 - 1) -> {3, 5, 17, 257};
(2^18 - 1) -> {3, 7, 19, 73};
(2^20 - 1) -> {3, 5, 11, 31, 41};
(2^22 - 1) -> {3, 23, 89, 683}.
But p == r (mod 32) where r is element of the set {3, 7, 11, 15, 19, 23, 27, 31}, and one of the ten numbers of E is divisible by r. For example, 27 | (2^18)p + 2^18 - 1 if p == 27 (mod 32).
Remark: the map p -> 4p + 1 is not interesting because the corresponding sequence contains only two numbers: a(0) = 5 and a(1) = 13 if we consider only 2 iterations {4p + 1 -> 16p + 5 -> 64p + 21}: if p==0 (mod 3) => 64p + 21 is composite, if p==1 (mod 3) => 16p + 5 is composite and if p==2 (mod 3) => 4p + 1 is composite.
Proof of finiteness is incorrect. Flaw is last sentence: "For example, ...". Specifically, 27 does not divide quantity unless 27 | k where p = 32*k + 27.
No further terms < 10^11. (End)
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