cp's OEIS Frontend

a(n) is also the number of non-congruent solutions to x^2 - y^2 == 1 (mod n). - Yuval Dekel (dekelyuval(AT)hotmail.com), Sep 21 2003

a(n) is the size of a square companion matrix of the minimal cyclotomic polynomial of (-1)^(1/n). - Eric Desbiaux, Dec 08 2015

a(n) is the degree of the (2n)-th cyclotomic field Q(zeta_(2n)). Note that Q(zeta_n) = Q(zeta_(2n)) for odd n. - Jianing Song, May 17 2021

The number of integers k from 1 to n such that gcd(n,k) is a power of 2. - Amiram Eldar, May 18 2025

Given the quasi-order and coach theorems of Hilton and Pedersen (cf. A003558): phi(b) = 2 * k * c with b odd. Dividing through by 2 we obtain phi(b)/2 = k * c which is the same as a(n) = A003558(n-1) * A135303(n-1). Here k refers to the "least exponent" (cf. A003558), while c is the number of coaches for odd b (cf. A135303). - Gary W. Adamson, Sep 25 2012

After the initial term, this is the totient function phi(2n-1)/2 (A037225(n-1)/2). Also a(n) is the number of partitions of the odd number (2n+1) into two ordered relatively prime parts. If p and q are such parts where p > q and p+q = 2n+1 then they can generate primitive Pythagorean triples of the form (p^2 - q^2, 2*p*q, p^2 + q^2). - Frank M Jackson, Oct 30 2012

Number of primitive irreducible factors of x^(2n+1) - 1 over integers mod 2. There are no primitive irreducible factors for x^(2n)-1 because it always has the same factors as x^n-1. Considering that A000374 also counts the cycles in the map f(x) = 2x mod n, a(n) is also the number of primitive cycles of that mapping. - T. D. Noe, Aug 01 2003

Equals number of irreducible factors of the cyclotomic polynomial Phi(2n+1,x) over Z/2Z. All factors have the same degree. - T. D. Noe, Mar 01 2008

The n-th term is the number of notes of the (2n-1)-limit tonality diamond. This is a term from music theory and means the scale consisting of the rational numbers r, 1 <= r < 2, such that the odd part of both the numerator and the denominator of r, when reduced to lowest terms, is less than or equal to the fixed odd number 2n-1. - Gene Ward Smith, Mar 27 2006

(1/4)*Number of distinct angular positions under which an observer positioned at the center of a square of a square lattice can see the (2n) X (2n) points symmetrically surrounding his position.

(1/8)*number of distinct angular positions under which an observer positioned at a lattice point of a square lattice can see the (2n+1)X(2n+1) points symmetrically surrounding his position gives A002088.

(1/2)*number of distinct angular positions under which an observer positioned at the center of an edge of a square lattice can see the (2n)X(2n-1) points symmetrically surrounding his position gives A099958.

Moebius transform of A181983.

The length of row n is A037225(n), for n >= 0.

The doubling sequence modulo N = 2*n+1, for n >= 0, has entries DS(N, s(N,i), j) = s(N,i)*2^j (mod N), with j >= 0, and certain positive odd integer seeds s(N, i), for i = 1, 2, ..., S(N) = A037226((N-1)/2), where gcd(s(N, i), N) = 1 (restricted seeds modulo N). These doubling sequences are periodic with period length P(N) = A002326((N-1)/2) (order of 2 modulo N). Only the periods (cycles) {DS(N, s(N, i), j)}_{j=0..P(N)-1}, for i = 1, 2, ..., S(N), are listed.

N = 1 (n=0) is special: one takes here the restricted residue system modulo N not as [0] but as [1]. The order of 2 modulo 1 is 1, because 2^1 == 1 (mod 1) (== 0 (mod 1)).

In order to obtain the complete system of doubling sequences one starts with seed s(N, 1) = 1, and if all numbers from the smallest positive reduced residue system modulo N (called RRS(N), given in row N of A038566) are obtained, i.e., if P(N) = #RRS(N) = phi(N) = A000010(N), then the system is complete. Otherwise the smallest missing number from RRS(N) is taken as new seed s(N, 2), etc. until the system is complete. This means that the number of seeds needed is S(N) = phi(N)/P(N) = A037226((N-1)/2)).

The irregular subtriangle where only seed s(N, 1) = 1 has been used is given in A201908. But there 0 (not 1) for N = 1 has been used.

From Gary W. Adamson and Wolfdieter Lang, Dec 15 2020: (Start)

The cycles in row n, for N = 2*n + 1, of period length P(N) = A002326((N-1)/2) give the periods of the iterated doubling function D(x) = frac(2*x) with seeds x = s(N, i)/N, for i = 1, 2, ..., S(N) = A037226((N-1)/2), after multiplication with N. This is the doubling function used in the Devaney reference, pp. 24-25, 27, 125. 132, 171,289.

Each cycle in row n can also be used to find from the base 2 version of its first entry (the seed s = s(N, i)) divided by N the other entries by repeated application of a cyclic left shift by one step (called sigma operation) to the period of the base 2 expression of s/N. E.g., n = 7, N = 15, P(N) = 4, s = 1: (1/15){10->2} = .repeat(0001), then (.repeat(0010)){2->10} = 2/10, (.repeat(0100)){2->10} = 4/10 and (.repeat(1000)){2->10} = 8/15. Similarly for s = 7: from (7/15)_{10->2} = .repeat(0111) one obtains by repeated sigma operations 14/15, 13/15 and 11/15. The proof uses the elementary formulas for the conversion from base 10 to base 2, and the reverse one, from base 2 to base 10. See also a comment on the period length P(N) given in A002326. (End)