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A367228 Products of two consecutive Fermat numbers: a(n) = A000215(n) * A000215(n+1).

%I A367228 #10 Nov 11 2023 10:39:49
%S A367228 15,85,4369,16843009,281479271743489,79228162532711081671548469249,
%T A367228 6277101735386680764176071790128604879584176795969512275969
%N A367228 Products of two consecutive Fermat numbers: a(n) = A000215(n) * A000215(n+1).
%C A367228 a(7) has 116 digits and is too large to include in the data section.
%C A367228 Szymiczek (1966) proved that a(n) is a super-Poulet number (A050217) for all n >= 2. All the composite Fermat numbers (A281576) are also super-Poulet numbers.
%D A367228 Michal Krížek, Florian Luca and Lawrence Somer, 17 Lectures on Fermat Numbers, Springer-Verlag, N.Y., 2001, p. 142.
%H A367228 Amiram Eldar, <a href="/A367228/b367228.txt">Table of n, a(n) for n = 0..10</a>
%H A367228 Andrzej Rotkiewicz, <a href="https://eudml.org/doc/35153">On pseudoprimes having special forms and a solution of K. Szymiczek's problem</a>, Acta Mathematica Universitatis Ostraviensis, Vol. 13, No. 1 (2005), pp. 57-71.
%H A367228 Kazimierz Szymiczek, <a href="https://www.e-periodica.ch/digbib/view?pid=edm-001%3A1966%3A21%3A%3A4&amp;referrer=search#65">Note on Fermat numbers</a>, Elemente der Mathematik, Vol. 21, No. 3 (1966), p. 59.
%F A367228 a(n) = (2^(2^n) + 1) * (2^(2^(n+1)) + 1).
%t A367228 f[n_] := 2^(2^n) + 1; a[n_] := f[n] * f[n + 1]; Array[a, 7, 0]
%o A367228 (PARI) f(n) = 2^(2^n) + 1;
%o A367228 a(n) = f(n) * f(n+1);
%Y A367228 Cf. A000215, A050217, A281576.
%K A367228 nonn,easy
%O A367228 0,1
%A A367228 _Amiram Eldar_, Nov 11 2023