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Largest prime factor of the n-th Fermat number F(n) = 2^(2^n) + 1.
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%I #24 Oct 25 2024 07:32:02

%S 3,5,17,257,65537,6700417,67280421310721,5704689200685129054721,

%T 93461639715357977769163558199606896584051237541638188580280321

%N Largest prime factor of the n-th Fermat number F(n) = 2^(2^n) + 1.

%H Eric M. Schmidt, <a href="/A070592/b070592.txt">Table of n, a(n) for n = 0..11</a>

%H C. L. Stewart, <a href="https://doi.org/10.1112/plms/s3-35.3.425">On Divisors of Fermat, Fibonacci, Lucas, and Lehmer Numbers</a>, Proceedings of the London Mathematical Society, Vol. s3-35, No. 3 (1977), pp. 425-447. See p. 430.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FermatNumber.html">Fermat Number</a>.

%F From _Amiram Eldar_, Oct 25 2024: (Start)

%F a(n) = A006530(A000215(n)).

%F a(n) > c * n * 2^n for n >= 1, where c is a positive absolute constant (Stewart, 1977). (End)

%o (PARI) a(n) = vecmax(factor(2^(2^n) + 1)[,1]); \\ _Michel Marcus_, Jul 05 2017

%Y Cf. A000215, A006530, A050922, A093179.

%K nonn,hard,changed

%O 0,1

%A _Benoit Cloitre_, May 12 2002

%E Offset changed by _Arkadiusz Wesolowski_, Jul 09 2011