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%I #18 Mar 11 2020 10:06:40
%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.
%Y Cf. A003558, A065941, A127672, A332433.
%K nonn
%O 0,6
%A _Wolfdieter Lang_, Feb 17 2020
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