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9^(9^9) mod n.
1

%I #15 Feb 24 2018 23:56:12

%S 0,1,0,1,4,3,1,1,0,9,5,9,1,1,9,9,9,9,1,9,15,5,8,9,14,1,0,1,9,9,4,9,27,

%T 9,29,9,1,1,27,9,9,15,11,5,9,31,34,9,15,39,9,1,9,27,49,1,39,9,28,9,34,

%U 35,36,9,14,27,22,9,54,29,12,9,72,1,39,1,71,27,21,9,0,9,61,57,9,11,9,49,42

%N 9^(9^9) mod n.

%C Eventually constant: a(n) = 9^9^9 for all n > 9^9^9 ~ 4.28*10^369693099. - _M. F. Hasler_, Feb 24 2018

%H Math Forum, Dr. Math <a href="http://www.mathforum.org/library/drmath/view/59172.html">9^387420489, [09/10/1997] </a>

%H Math Forum, Dr. Math <a href="http://www.mathforum.org/library/drmath/view/63676.html">Digits of 9^9^9, 06/11/2003</a>

%H <a href="/index/Con#constant">Index entries for eventually constant sequences</a>.

%F a(n) = 0 for all n = 3^k, 0 <= k <= 2*9^9; a(n) = 9^9^9 for all n > 9^9^9. - _M. F. Hasler_, Feb 24 2018

%e [9^(9^9)] mod 1 =0

%e [9^(9^9)] mod 2 =1

%e [9^(9^9)] mod 9 =0

%e [9^(9^9)] mod (3*387420489) =0

%t Table[PowerMod[9, 9^9, n], {n, 1, 100}]

%o (PARI) a(n)=lift(Mod(9,n)^9^9) \\ _Charles R Greathouse IV_, Jul 02 2012

%Y Cf. A081230, A214128 (6^6^6 mod n).

%K nonn,easy

%O 1,5

%A _Marvin Ray Burns_, May 19 2007