A329593 a(n) = (2^(A003558(n)) - A332433(n))/(2*n+1), for n >= 0.

1, 1, 1, 1, 1, 3, 5, 1, 1, 27, 3, 89, 41, 19, 565, 1, 1, 117, 7085, 105, 25, 3, 91, 178481, 42799, 5, 1266205, 19065, 9, 9099507, 17602325, 1, 1, 128207979, 60787, 483939977, 7, 13981, 13944699, 6958934353, 1657009, 26494256091, 3, 3085465, 23, 45, 11

Offset: 0

Author

Wolfdieter Lang, Feb 17 2020

Keywords

• nonn

Comments

Only for n=0 with A003558(0) = 1 only the minimal solution for a(0), namely 1, for sign A332433(0) = +1 is recorded here.

For n >= 1 only one sign qualifies in A003558(n).

A comment on the iteration of f(x) = x^2 - 2 (called R(2, x) in A127672) with seed rho(n) := 2*cos(Pi/n) for odd n >= 3, used in the comment from Gary W. Adamson, Sep 06 2011 in A065941. The proof that the cycle length coincides with A003558(n) is done by using the known formulas, for integers k and m: (i) R(k, m*x) = R(k*m, x), (ii) R(-k, x) = R(k, x), and the periodicity formula (iii) R(j, rho(n)) = R(+/-(j + k*2*n), rho(n)), j >= 0, k integer, n odd >= 3. The iterations are then R(2^q, rho(n)), for q >= 1. The primitive period length P(n) is obtained from R(2^(P(n)+1), rho(n)) = R(2^1, rho(n)) = R(+-(2 + k*2*n), rho(n)), that is 2^P(n) = +-(1 + k*n) or 2^P(n) == +-1 (mod n) with the least P(n), hence P(n) = A003558(n).

Examples

			a(3) = 1 because 2^3 - 1 = 1*7,
a(4) = 1 because 2^3 + 1 = 1*9,
a(5) = 3 because 2^5 + 1 = 3*11,
a(9) = 27 because 2^9 + 1 = 27*19.
		

Links

• Wolfdieter Lang, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A003558, A065941, A127672, A332433.

Programs

Mathematica

Suborder[a_, n_] := If[n > 1 && GCD[a, n] == 1,
   Min[MultiplicativeOrder[a, n, {-1, 1}]], 0];
A003558[n_] := If[n == 1, 1, Suborder[2, 2n+1]];
A332433[n_] := If[n == 0, 1, If[PowerMod[2, A003558[n], 2n+1]-1 == 0, 1, -1]];
a[n_] := If[n == 0, 1, (2^(A003558[n]) - A332433[n])/(2n+1)];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Oct 17 2024 *)

Formula

a(n) = (2^(A003558(n)) - A332433(n))/(2*n+1), for n >= 0.