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a(n+1) = (3*a(n) + 2*a(n-1)) mod 37; a(0) = 0, a(1) = 1.
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%I #13 Jan 31 2015 16:55:05

%S 0,1,3,11,2,28,14,24,26,15,23,25,10,6,1,15,10,23,15,17,7,18,31,18,5,

%T 14,15,36,27,5,32,32,12,26,28,25,20,36,0,35,31,15,33,18,9,26,22,7,28,

%U 24,17,25,35,7,17,28,7,3,23,1,12,1,27,9,7,2,20,27,10,10,13,22,18,24,34,2,0,4

%N a(n+1) = (3*a(n) + 2*a(n-1)) mod 37; a(0) = 0, a(1) = 1.

%C Has period 1368 = (37 + 1) * (37 - 1), indicating that all possible consecutive pairs other than 0,0 appear.

%H Michel Marcus, <a href="/A133369/b133369.txt">Table of n, a(n) for n = 0..1400</a>

%H Gupta, Rockstroh and Su, <a href="http://arxiv.org/abs/0909.0362">Splitting fields and Periods of Fibonacci Sequences Mod Primes</a>, arXiv:0909.0362v1; Sep 02 2009.

%t RecurrenceTable[{a[0]==0,a[1]==1,a[n]==Mod[3a[n-1]+2a[n-2],37]},a,{n,80}] (* _Harvey P. Dale_, Apr 11 2014 *)

%K nonn

%O 0,3

%A _Gary W. Adamson_, Sep 04 2009

%E More terms from _Max Alekseyev_, Apr 18 2010

%E Definition clarified by _Harvey P. Dale_, Apr 11 2014