OFFSET
1,1
COMMENTS
Backward concatenation of A133613.
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.
REFERENCES
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.
LINKS
J. Jimenez Urroz and J. Luis A. Yebra, On the equation a^x == x (mod b^n), J. Int. Seq. 12 (2009) #09.8.8.
FORMULA
For n>1, a(n) = 3^a(n-1) mod 10^n.
CROSSREFS
KEYWORD
nonn
AUTHOR
Max Alekseyev, Sep 08 2011
STATUS
approved