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%I #31 Nov 27 2019 02:57:35
%S 0,0,0,0,1,0,1,0,0,6,5,0,1,8,6,0,1,0,1,16,15,16,2,0,6,14,0,8,23,6,1,0,
%T 27,18,1,0,1,20,27,16,18,36,1,16,36,2,37,0,43,6,18,40,44,0,16,8,39,52,
%U 5,36,9,32,36,0,1,60,14,52,48,36,6,0,1,38,6,20,71
%N a(n) = 6^(6^6) mod n.
%C The indices of zeros in this sequence, i.e., divisors of 6^(6^6), are all numbers of the form 2^i * 3^j, with 0 <= i, j <= 6^6. [Edited by _M. F. Hasler_, Feb 25 2018]
%C If c and N are any positive integers, and p^k is the largest prime power divisor of c, then the divisors of c^N less than p^(k*N+1) are precisely those numbers in that range whose prime factorization includes only primes that divide c. This is the case c = 6, N = 6^6, so p^k = 2^1 = 2; so the first difference in the divisor list from A003586 is for A003586(n) = 2^(6^6+1). _Franklin T. Adams-Watters_, Jul 12 2012
%C Eventually constant: see formula. - _M. F. Hasler_, Feb 24 2018
%C If n > 1 is coprime to 6 and A000010(n) divides 6^6 then a(n)=1. - _Robert Israel_, Nov 27 2019
%H Robert Israel, <a href="/A214128/b214128.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Con#constant">Index entries for eventually constant sequences</a>.
%F a(n) = 0 if and only if n = 2^i 3^j, 0 <= i, j <= 6^6; after the last of these zeros at n = 6^6^6, a(n) = 6^6^6 for all n > 6^6^6 ~ 2.659*10^36305. - _M. F. Hasler_, Feb 24 2018
%e a(1) = 6^(6^6) mod 1 = 0.
%e a(2) = 6^(6^6) mod 2 = 0.
%e a(3) = 6^(6^6) mod 3 = 0.
%e a(4) = 6^(6^6) mod 4 = 0.
%p seq(6 &^ (6^6) mod n, n=1..100); # _Robert Israel_, Nov 27 2019
%t Table[PowerMod[6, 6^6, n], {n, 100}]
%o (PARI) a(n)=lift(Mod(6,n)^6^6) \\ _Charles R Greathouse IV_, Jul 05 2012
%Y Cf. A129810 (9^9^9 mod n), A003586.
%K nonn
%O 1,10
%A _Marvin Ray Burns_, Jul 04 2012
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