%I #23 Feb 09 2018 09:09:47
%S 7,87,387,5387,95387,195387,4195387,64195387,464195387,2464195387,
%T 62464195387,262464195387,7262464195387,27262464195387,
%U 627262464195387,5627262464195387,75627262464195387,575627262464195387,4575627262464195387,4575627262464195387
%N a(n) = 3^^(n+1) modulo 10^n.
%C Backward concatenation of A133613.
%C For all m>n, 3^^m == 3^^(n+1) (mod 10^n). Hence, each term represents the tailing decimal digits of 3^^m for all sufficiently large m.
%D M. RipĂ , La strana coda della serie n^n^...^n, Trento, UNI Service, Nov 2011, p. 11-12, 69-78. ISBN 978-88-6178-789-6.
%H J. Jimenez Urroz and J. Luis A. Yebra, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL12/Yebra/yebra4.html">On the equation a^x == x (mod b^n)</a>, J. Int. Seq. 12 (2009) #09.8.8.
%F For n>1, a(n) = 3^a(n-1) mod 10^n.
%K nonn
%O 1,1
%A _Max Alekseyev_, Sep 08 2011