A308964 Least k > 0 such that A014222(k) == A014222(k+1) (mod n).

1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 3, 1, 2, 2, 2, 2, 3, 2, 3, 2, 2, 3, 4, 1, 3, 2, 2, 2, 3, 2, 3, 2, 3, 3, 2, 2, 3, 3, 2, 2, 2, 2, 3, 3, 2, 4, 5, 2, 3, 3, 3, 2, 3, 2, 3, 2, 3, 3, 4, 2, 3, 3, 2, 3, 2, 3, 4, 3, 4, 2, 3, 2, 2, 3, 3, 3, 3, 2, 3, 2, 3, 2, 3, 2, 3, 3, 3

Offset

1,5

Comments

a(n) is the least positive integer k such that the sequence {A014222(i) mod n} for i >= k is constant. Proof: if g(i) = A014222(i), then n divides 3^g(i-1)*(3^(g(i)-g(i-1)) - 1) when g(i) == g(i+1) (mod n). g(j)-g(j-1) divides g(j+1)-g(j) for any j > 0, therefore, 3^(g(i)-g(i-1)) - 1 divides 3^(g(i+1)-g(i)) - 1. Then n divides 3^g(i)*(3^(g(i+1)-g(i)) - 1) = g(i+2) - g(i+1), that is, g(i) == g(i+1) == g(i+2) (mod n). Since a(n) is the least positive integer k such that g(k) == g(k+1) (mod n), the sequence {A014222(i) mod n} for i >= k is constant.

A014222(k+1) - A014222(k) is the largest n such that a(n) = k.

Links

Table of n, a(n) for n=1..87.

Formula

a(n) <= A003434(n).

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

Example

3, 27, 7625597484987, ... mod 5 equal 3, 2, 2, ..., so A014222(k) mod 5 = 2 for all k >= 2, hence a(5) = 2.

Prog

(PARI) a(n) = {my(c=0, k=1, x=0, d=n); while(k==1, z=x++; y=0; b=1; while(z>0, while(y++<z, d=eulerphi(d)); b=3^b-floor((3^b-1)/d)*d; z=z-1; y=0; d=n); if(c==b, k=0); c=b); x-1; }

Crossrefs

Cf. A014222, A308972, A322418.

Sequence in context: A324117 A106698 A097848 * A124830 A291598 A287820

Adjacent sequences: A308961 A308962 A308963 * A308965 A308966 A308967

Keyword

nonn

Author

Jinyuan Wang, Aug 30 2019

Status

approved