cp's OEIS Frontend

A103477 Positive integers k for which 1 + 3*2^(k+2) divides the Fermat number 1 + 2^(2^k).

%I A103477 #37 Nov 01 2024 16:57:44
%S A103477 207,157167,213319,382447,2145351
%N A103477 Positive integers k for which 1 + 3*2^(k+2) divides the Fermat number 1 + 2^(2^k).
%C A103477 The smallest prime factor any Fermat number 2^(2^k)+1 can have is 3*2^(k+2)+1 (for k >= 3). - _Jeppe Stig Nielsen_, Dec 03 2018
%C A103477 On Keller's linked page, to find the terms, you run through the tables and find all rows with k = 3 and with n exactly 2 greater than m, then that m belongs to this sequence. - _Jeppe Stig Nielsen_, Dec 04 2018
%C A103477 All terms in this sequence are odd. Proof: by a theorem of Fermat, each prime p that is 1 mod 3 is expressible uniquely in the form a^2 + 3b^2, where a and b are positive integers. Gauss proved that 2 is a cubic residue iff 3|b. However, for even k and p = 1 + 3*2^(k+2), we have b = 2^(k/2+1) which is not divisible by 3. - _William Hu_, Oct 14 2024
%H A103477 Wilfrid Keller, <a href="http://www.prothsearch.com/fermat.html">Prime factors k*2^n + 1 of Fermat numbers F_m</a>
%e A103477 a(1)=207 because 207 is the smallest positive integer k for which 1 + 3*2^(k+2) divides the Fermat number 1 + 2^(2^k).
%t A103477 aQ[n_] := PowerMod[2, 2^n, 1 + 3*2^(n+2)] == 3*2^(n+2); Select[Range[100000], aQ] (* _Amiram Eldar_, Dec 04 2018 *)
%o A103477 (PARI) isOK(n) = Mod(2, 1+3*2^(n+2))^(2^n) + 1 == 0 \\ _Jeppe Stig Nielsen_, Dec 03 2018
%Y A103477 Cf. A103478, A103479.
%K A103477 nonn,hard,more
%O A103477 1,1
%A A103477 Serhat Sevki Dincer (mesti_mudam(AT)yahoo.com), Feb 07 2005
%E A103477 Sequence name trimmed by _Jeppe Stig Nielsen_, Dec 03 2018