This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A248802 #18 Oct 22 2019 04:22:25
%S A248802 11,19,67,13,262147,13,1669,13,255127,13,2383,13,67,13,32544331,13,
%T A248802 271,13,4057,13
%N A248802 Smallest prime factor of 2^(2^n+2) + 3.
%C A248802 These numbers do not occur in A023394 (prime factors of Fermat numbers A000215).
%C A248802 From _Chai Wah Wu_, Oct 21 2019: (Start)
%C A248802 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.
%C A248802 Theorem: a(n) >= 13 for n > 0.
%C A248802 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.
%C A248802 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.
%C A248802 Theorem: a(2n+1) = 13 for n >= 1.
%C A248802 Proof by induction. a(3) = 13 since 2^(2^3+2) + 3 = 1027 = 13*79.
%C A248802 Suppose a(2n+1) = 13, this implies that 2^(2^(2n+1)+2) == 10 mod 13.
%C A248802 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.
%C A248802 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.
%C A248802 End of proof.
%C A248802 Conjecture 1: a(10n+2) = 67 for n >= 0.
%C A248802 Conjecture 2: a(36n+16) = 271 for n >= 0 and n <> 1 mod 5.
%C A248802 Conjecture 3: a(84n+22) = 523 for n >= 0 and n <> 0 mod 5.
%C A248802 Conjecture 4: a(58n+26) = 1399 for n >= 0 and when it is not covered by Conjectures 1-3.
%C A248802 Conjecture 5: a(138n+6) = 1669 for n >= 0 and n <> 2 mod 5.
%C A248802 Conjecture 6: a(44n+10) = 2383 for n >= 0 and when it is not covered by Conjectures 1-5.
%C A248802 (End)
%F A248802 Smallest prime factor of 4*A000215(n) - 1, with the Fermat numbers A000215. - _Wolfdieter Lang_, Nov 05 2014
%t A248802 PrimeFactors[n_]:= Flatten[Table[#[[1]], {1}]&/@FactorInteger[n]]; Table[PrimeFactors[2^(2^n + 2) + 3] [[1]], {n, 0, 7}] (* _Vincenzo Librandi_, Oct 15 2014 *)
%o A248802 (PARI) a(n) = factor(2^(2^n+2) + 3)[1, 1]; \\ _Michel Marcus_, Oct 15 2014
%o A248802 (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
%Y A248802 Cf. A000215, A023394, A057733.
%K A248802 nonn,hard,more
%O A248802 0,1
%A A248802 _Arkadiusz Wesolowski_, Oct 14 2014