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

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, 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, 5, 3, 4, 3, 4, 4, 4, 4, 4, 3, 4, 3, 5, 4, 5, 3, 4, 4, 4, 4, 5, 3

Offset: 1

Author

Jinyuan Wang, Dec 07 2018

Keywords

• nonn

Comments

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.

A038081(k+1) is the largest n such that a(n) = k.

Examples

			2, 4, 16, ... mod 6 equal 2, 4, 4, ..., so A014221(k) mod 6 = 4 for all k >= 2, hence a(6) = 2.
		

Links

• USAMO, Problem 3, 1991.

Crossrefs

Cf. A014221, A038081, A000010, A003434.

Programs

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
				

Formula

a(n) <= A003434(n).

a(n) <= a(A000010(n)) + 1.

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.