A248802 Smallest prime factor of 2^(2^n+2) + 3.

11, 19, 67, 13, 262147, 13, 1669, 13, 255127, 13, 2383, 13, 67, 13, 32544331, 13, 271, 13, 4057, 13

Offset

0,1

Comments

These numbers do not occur in A023394 (prime factors of Fermat numbers A000215).

From Chai Wah Wu, Oct 21 2019: (Start)

a(22) = 67, a(26) = 1399, a(28) = 10957, a(30) = 117127, a(32) = 67, a(36) = 12781849, a(38) = 262147, a(42) = 67, a(48) = 6391117, a(50) = 1265347, a(52) = 67, a(54) = 2383, a(58) = 26833, a(62) = 67, a(64) = 517261, a(68) = 2251, a(72) = 67, a(74) = 137077, a(78) = 562273, a(82) = 67, a(84) = 1399, a(86) = 3253, a(88) = 271, a(92) = 67, a(94) = 2203, a(96) = 329347, a(98) = 2383, a(100) = 5323, a(110) = 2759137, a(114) = 122653, a(116) = 659941, a(126) = 48337, a(130) = 2403229, a(134) = 2534659, a(140) = 41257.

Theorem: a(n) >= 13 for n > 0.

Proof. 2^(2^n+2) + 3 is odd and not a multiple of 3, so a(n) > 3. For all primes 3 < p < 14, p-3 is a power of 2. For p = 5, 2^4 == 1 mod 5, so for n = 1, 2^(2^n+2) + 3 == 4 mod 5 and for n > 1, 2^(2^n+2) + 3 == 7 == 2 mod 5. For p = 7, 2^3 == 1 mod 7. Since 2^n+2 <> 2 mod 3, 2^(2^n+2) <> 4 mod 7 and thus 2^(2^n+2) + 3 <> 0 mod 7.

For p = 11, 2^10 == 1 mod 11. Since 2^n+2 is even for n > 0, 2^n+2 <> 3 mod 10 and thus 2^(2^n+2) <> 2^3 mod 11 and 2^(2^n+2) + 3 <> 0 mod 11. End of proof.

Theorem: a(2n+1) = 13 for n >= 1.

Proof by induction. a(3) = 13 since 2^(2^3+2) + 3 = 1027 = 13*79.

Suppose a(2n+1) = 13, this implies that 2^(2^(2n+1)+2) == 10 mod 13.

Then 2^(2^(2n+3)+2) = 2^(3*2^(2n+1)) * 2^(2^(2n+1)+2). For n >= 1, 2^(2n+1) is a multiple of 4, and thus 2^(3*2^(2n+1)) == 2^12 == 1 mod 13.

This implies that 2^(2^(2n+3)+2) == 2^(2^(2n+1)+2) == 10 mod 13 and thus a(2n+3) <= 13. By the first result above, a(2n+3) = 13.

End of proof.

Conjecture 1: a(10n+2) = 67 for n >= 0.

Conjecture 2: a(36n+16) = 271 for n >= 0 and n <> 1 mod 5.

Conjecture 3: a(84n+22) = 523 for n >= 0 and n <> 0 mod 5.

Conjecture 4: a(58n+26) = 1399 for n >= 0 and when it is not covered by Conjectures 1-3.

Conjecture 5: a(138n+6) = 1669 for n >= 0 and n <> 2 mod 5.

Conjecture 6: a(44n+10) = 2383 for n >= 0 and when it is not covered by Conjectures 1-5.

(End)

Links

Table of n, a(n) for n=0..19.

Formula

Smallest prime factor of 4*A000215(n) - 1, with the Fermat numbers A000215. - Wolfdieter Lang, Nov 05 2014

Mathematica

PrimeFactors[n_]:= Flatten[Table[#[[1]], {1}]&/@FactorInteger[n]]; Table[PrimeFactors[2^(2^n + 2) + 3] [[1]], {n, 0, 7}] (* Vincenzo Librandi, Oct 15 2014 *)

Prog

(PARI) a(n) = factor(2^(2^n+2) + 3)[1, 1]; \\ Michel Marcus, Oct 15 2014
(PARI) for(n=1, 19, my(x=2^(2^n+2)+3); forprime(k=3, oo, if(x%k==0, print1(k, ", "); break))) \\ Hugo Pfoertner, Aug 08 2019

Crossrefs

Cf. A000215, A023394, A057733.

Sequence in context: A057770 A080788 A363023 * A290078 A089032 A344042

Adjacent sequences: A248799 A248800 A248801 * A248803 A248804 A248805

Keyword

nonn,hard,more

Author

Arkadiusz Wesolowski, Oct 14 2014

Status

approved