cp's OEIS Frontend

A322418 Least k > 0 such that A014221(k) == A014221(k+1) mod n.

%I A322418 #67 Mar 08 2019 23:47:04
%S A322418 1,1,2,2,3,2,3,3,3,3,4,2,3,3,3,3,4,3,4,3,3,4,5,3,4,3,4,3,4,3,4,4,4,4,
%T A322418 3,3,4,4,3,3,4,3,4,4,3,5,6,3,4,4,4,3,4,4,4,3,4,4,5,3,4,4,3,4,3,4,5,4,
%U A322418 5,3,4,3,4,4,4,4,4,3,4,3,5,4,5,3,4,4,4,4,5,3
%N A322418 Least k > 0 such that A014221(k) == A014221(k+1) mod n.
%C A322418 For any fixed integer n > 0, the sequence 2 mod n, 2^2 mod n, 2^2^2 mod n, that is, the sequence {A014221(i) mod n} for i >= 1 is eventually constant. a(n) is the least index k such that A014221(k) mod n equals this constant.
%C A322418 A038081(k+1) is the largest n such that a(n) = k.
%H A322418 USAMO, <a href="https://artofproblemsolving.com/wiki/index.php/1991_USAMO_Problems/Problem_3">Problem 3</a>, 1991.
%F A322418 a(n) <= A003434(n).
%F A322418 a(n) <= a(A000010(n)) + 1.
%F A322418 If A014221(k) == b(k) mod eulerphi(n), 0 < b(k) <= eulerphi(n), then a(n) is the least m > 0 such that 2^b(m-1) == 2^b(m) mod n.
%e A322418 2, 4, 16, ... mod 6 equal 2, 4, 4, ..., so A014221(k) mod 6 = 4 for all k >= 2, hence a(6) = 2.
%o A322418 (PARI) a(n) = {c=0; k=1; x=1; d=n; while(k==1, z=x; y=1; b=1; while(z>0, while(y<z, d=eulerphi(d); y++); b=2^b-floor((2^b-1)/d)*d; z=z-1; y=1; d=n); if(c==b, k=0); c=b; x++); x-2;}
%Y A322418 Cf. A014221, A038081, A000010, A003434.
%K A322418 nonn
%O A322418 1,3
%A A322418 _Jinyuan Wang_, Dec 07 2018