A308964 -id:A308964 - cp's OEIS frontend

A308972 Least k > 0 such that A114561(k) == A114561(k+1) mod n.

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

Offset: 1

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Jinyuan Wang, Aug 30 2019

nonn

A114561(k+1) - A114561(k) is the largest n such that a(n) = k.

			4, 4^4, 4^4^4, ... mod 8 equal 4, 0, 0, ..., so A114561(k) mod 8 = 0 for all k >= 2, hence a(8) = 2.

Cf. A114561, A308964, A322418.

a(n) = {c=0; k=1; x=0; d=n; while(k==1, z=x++; y=0; b=1; while(z>0, while(y++

a(n) <= A003434(n).

a(n) <= a(A000010(n)) + 1. Proof: a(n) <= a(eulerphi(n)) + 1. Proof: If A114561(i) == b(i) mod eulerphi(n), 0 < b(i) <= eulerphi(n), then a(n) is the least k > 0 such that 2^b(k-1) == 2^b(k) mod n. Since A114561(a(eulerphi(n))) == A114561(a(eulerphi(n)) + 1), k <= a(A000010(n)) + 1.

cp's OEIS Frontend

A308972 Least k > 0 such that A114561(k) == A114561(k+1) mod n.

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