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%I #31 Feb 26 2024 12:54:20
%S 1,1,5,1,1,1,5,1,1,1,5,1,1,1,5,1,1,1,5,1,1,1,5,1,1,1,5,1,1,1,5,1,1,1,
%T 5,1,1,1,5,1,1,1,5,1,1,1,5,1,1,1,5,1,1,1,5,1,1,1,5,1,1,1,5,1,1,1,5,1,
%U 1,1,5,1,1,1,5,1,1,1,5,1,1,1,5,1,1,1,5
%N a(n) = gcd(3^n-2,2^n-3).
%C The first time the inequality a(n) > 5 occurs for n = A196628(2) = 3783 with a(3783) = 26665 = 5*5333 = A196627(1)*A196627(2). - _Max Alekseyev_, Oct 04 2011
%H Antti Karttunen, <a href="/A109768/b109768.txt">Table of n, a(n) for n = 1..10000</a>
%H Suggested by Max Alekseyev in <a href="http://list.seqfan.eu/pipermail/seqfan/2005-August/005982.html">a Seqfan memo</a> Aug 09 2005.
%H Anatoly S. Izotov, <a href="http://www.fq.math.ca/Papers1/43-2/paper43-2-6.pdf">On prime divisors of GCD(3^n-2,2^n-3)</a>, Fibonacci Quarterly 43, May 2005, pp. 130-131.
%t Table[GCD[3^n - 2, 2^n - 3], {n, 120}] (* _Michael De Vlieger_, Mar 10 2016 *)
%o (PARI) a(n) = gcd(3^n-2, 2^n-3); \\ _Michel Marcus_, Mar 10 2016
%Y Cf. A196627, A196628.
%K nonn
%O 1,3
%A _John W. Layman_, Aug 13 2005
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