OFFSET
1,1
COMMENTS
Backward concatenation of A133612.
For all m>n+1, 2^^m == 2^^(n+2) (mod 10^n). Hence, each term represents the trailing decimal digits of 2^^m for every sufficiently large m.
REFERENCES
M. Ripà, La strana coda della serie n^n^...^n, Trento, UNI Service, Nov 2011. ISBN 978-88-6178-789-6.
LINKS
Robert G. Wilson v, Table of n, a(n) for n = 1..1000
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.
Robert G. Wilson v, Mathematica coding for "SuperPowerMod" from Vardi
FORMULA
a(n) = A014221(n+3) mod (10^n).
For n>1, a(n) = 2^a(n-1) mod 10^n.
MATHEMATICA
(* first load all lines of Super Power Mod by Ilan Vardi from the hyper-link, then *) $RecursionLimit = 2^14; a[n_] := SuperPowerMod[2, n +2, 10^n]; Array[a, 22] (* Robert G. Wilson v, Apr 20 2020 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Marco Ripà, Feb 10 2012
STATUS
approved