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 A103477 Positive integers k for which 1 + 3*2^(k+2) divides the Fermat number 1 + 2^2^k. 2
 207, 157167, 213319, 382447, 2145351 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS The smallest prime factor any Fermat number 2^(2^k)+1 can have is 3*2^(k+2)+1. - Jeppe Stig Nielsen, Dec 03 2018 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 LINKS Table of n, a(n) for n=1..5. Wilfrid Keller, Prime factors k*2^n + 1 of Fermat numbers F_m EXAMPLE 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. MATHEMATICA aQ[n_] := PowerMod[2, 2^n, 1 + 3*2^(n+2)] == 3*2^(n+2); Select[Range, aQ] (* Amiram Eldar, Dec 04 2018 *) PROG (PARI) isOK(n) = Mod(2, 1+3*2^(n+2))^(2^n) + 1 == 0 \\ Jeppe Stig Nielsen, Dec 03 2018 CROSSREFS Cf. A103478, A103479. Sequence in context: A223650 A232346 A052093 * A080531 A123136 A092153 Adjacent sequences: A103474 A103475 A103476 * A103478 A103479 A103480 KEYWORD nonn,hard,more AUTHOR Serhat Sevki Dincer (mesti_mudam(AT)yahoo.com), Feb 07 2005 EXTENSIONS Sequence name trimmed by Jeppe Stig Nielsen, Dec 03 2018 STATUS approved

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Last modified September 30 02:05 EDT 2023. Contains 365781 sequences. (Running on oeis4.)