



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
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OFFSET

0,6


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).


LINKS

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


FORMULA

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


EXAMPLE

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.


CROSSREFS

Cf. A003558, A065941, A127672, A332433.
Sequence in context: A016610 A305470 A141707 * A263490 A190180 A190178
Adjacent sequences: A329590 A329591 A329592 * A329594 A329595 A329596


KEYWORD

nonn


AUTHOR

Wolfdieter Lang, Feb 17 2020


STATUS

approved



