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A206636
a(n) = 2^^(n+2) modulo 10^n, where ^^ denotes a power tower (see A133612).
4
6, 36, 736, 8736, 48736, 948736, 2948736, 32948736, 432948736, 3432948736, 53432948736, 353432948736, 5353432948736, 75353432948736, 75353432948736, 5075353432948736, 15075353432948736, 615075353432948736, 8615075353432948736, 98615075353432948736, 98615075353432948736, 8098615075353432948736
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
Marco Ripà, La strana coda della serie n^n^...^n, Trento, UNI Service, Nov 2011. 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
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 *)
(* this hard-codes the needed Chinese Remainder step using a modular inverse *) {b, M} = {2, 5}; Prepend[Rest[NestList[(M *= 5; b *= 2; b Mod[PowerMod[2, #, M] ModularInverse[b, M], M]) & , 36, 10]], 6] (* Stan Wagon, Oct 23 2025 *)
KEYWORD
nonn
AUTHOR
Marco Ripà, Feb 10 2012
STATUS
approved