A337714 Euler totient function phi(N) divided by the multiplicative order of 3 modulo N, with N = N(n) = floor((3*n-1)/2), for n >= 1.

1, 1, 1, 1, 1, 2, 1, 2, 4, 1, 2, 1, 1, 2, 2, 2, 1, 4, 2, 1, 1, 2, 1, 2, 2, 1, 4, 5, 1, 2, 2, 2, 1, 1, 4, 1, 2, 4, 1, 2, 6, 1, 2, 4, 3, 2, 2, 2, 6, 2, 2, 2, 1, 8, 5, 2, 4, 1, 4, 1, 12, 2, 2, 2, 2, 1, 2, 1, 3, 8, 1, 2, 4, 2, 4, 1

Offset: 1

Author

Wolfdieter Lang, Oct 22 2020

Keywords

• nonn
• easy

Comments

For the multiplicative order of 3 modulo N = N(n), with N(n) = floor((3*n-1)/2) = A001651(n), see A053446(n), for n >= 1.

For n >= 2 this sequence gives also the number of seeds s(N(n), i) needed to cover all numbers of the smallest positive restricted residue system (called RRS(N(n))) from the cycles obtained from s(N(n), i)*3^k (mod(N(n)), for k = 0..(P(N(n))-1), and certain s(N(n), i) chosen from RRS(N(n)). See A337936 for the choice of these seeds s(N, i). The cycles have period length P(N(n)) = A053446(n). For n = 1, N = 1, RRS(1) = [1] (not [0])

For the complete system of tripling sequences modulo N(n), for n >= 1, see A337936.

Examples

			The pairs [N(n),a(n)] begin, for n >= 1:
[1, 1], [2, 1], [4, 1], [5, 1], [7, 1], [8, 2], [10, 1], [11, 2], [13, 4], [14, 1], [16, 2], [17, 1], [19, 1], [20, 2], [22, 2], [23, 2], [25, 1], [26, 4], [28, 2], [29, 1], [31, 1], [32, 2], [34, 1], [35, 2], [37, 2], [38, 1], [40, 4], [41, 5], [43, 1], [44, 2], ...
The pairs [N(n)= floor((3*n-1)/2), P(N(n)) = A053446(n)] begin, for n >= 1:
[1, 1], [2, 1], [4, 2], [5, 4], [7, 6], [8, 2], [10, 4], [11, 5], [13, 3], [14, 6], [16, 4], [17, 16], [19, 18], [20, 4], [22, 5], [23, 11], [25, 20], [26, 3], [28, 6], [29, 28], [31, 30], [32, 8], [34, 16], [35, 12], [37, 18], [38, 18], [40, 4], [41, 8], [43, 42], [44, 10], ...
		

Crossrefs

Cf. A000010, A001651, A053446, A337936.

Programs

Mathematica

a[n_] := EulerPhi[(f = Floor[(3*n - 1)/2])] / MultiplicativeOrder[3, f]; Array[a, 100] (* Amiram Eldar, Oct 22 2020 *)

PARI

a(n) = my(N=(3*n-1)\2); eulerphi(N)/znorder(Mod(3, N)); \\ Michel Marcus, Oct 22 2020

Formula

Bisection: a(2*k+1) = phi(3*k+1)/A053446(2*k+1), a(2*k+2) = phi(3*k+2)/A053446(2*k+2), for k >= 0, where phi = A000010.