A329593 - OEIS

%I #23 Oct 17 2024 16:40:12

%S 1,1,1,1,1,3,5,1,1,27,3,89,41,19,565,1,1,117,7085,105,25,3,91,178481,

%T 42799,5,1266205,19065,9,9099507,17602325,1,1,128207979,60787,

%U 483939977,7,13981,13944699,6958934353,1657009,26494256091,3,3085465,23,45,11

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

%C 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.

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

%C 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).

%H Wolfdieter Lang, <a href="/A329593/b329593.txt">Table of n, a(n) for n = 0..1000</a>

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

%e a(3) = 1 because 2^3 - 1 = 1*7,

%e a(4) = 1 because 2^3 + 1 = 1*9,

%e a(5) = 3 because 2^5 + 1 = 3*11,

%e a(9) = 27 because 2^9 + 1 = 27*19.

%t Suborder[a_, n_] := If[n > 1 && GCD[a, n] == 1,

%t    Min[MultiplicativeOrder[a, n, {-1, 1}]], 0];

%t A003558[n_] := If[n == 1, 1, Suborder[2, 2n+1]];

%t A332433[n_] := If[n == 0, 1, If[PowerMod[2, A003558[n], 2n+1]-1 == 0, 1, -1]];

%t a[n_] := If[n == 0, 1, (2^(A003558[n]) - A332433[n])/(2n+1)];

%t Table[a[n], {n, 0, 100}] (* _Jean-François Alcover_, Oct 17 2024 *)

%Y Cf. A003558, A065941, A127672, A332433.

%K nonn

%O 0,6

%A _Wolfdieter Lang_, Feb 17 2020