A378872 -id:A378872 - cp's OEIS frontend

A378873 Squarefree part of A378872(n) (the discriminant of the minimal polynomial of a number whose continued fraction expansion has periodic part given by the n-th composition (in standard order)).

5, 2, 5, 13, 3, 3, 5, 5, 21, 2, 10, 21, 10, 10, 5, 29, 2, 15, 17, 15, 85, 85, 6, 2, 17, 85, 6, 17, 6, 6, 5, 10, 5, 6, 26, 13, 37, 37, 165, 6, 37, 2, 221, 37, 3, 221, 65, 5, 26, 37, 165, 37, 221, 3, 65, 26, 165, 221, 65, 165, 65, 65, 5, 53, 15, 35, 37, 3, 229

Offset: 1

Pontus von Brömssen, Dec 10 2024

nonn

Any number x whose continued fraction expansion is eventually periodic can be written uniquely as x = (c+f*sqrt(d))/b, where b, c, f, d are integers, b > 0, d > 0 is squarefree, and GCD(b,c,f) = 1. a(n) is equal to d when the periodic part of the continued fraction of x is given by the n-th composition. If two numbers have eventually periodic continued fraction expansions with the same periodic part, their respective values of d are the same.

			For n = 6, the 5th composition is (1,2). The value of the continued fraction 1+1/(2+1/(1+1/(2+...))) is (1+sqrt(3))/2, so a(6) = 3.

Wikipedia, Discriminant of an algebraic number field.

Cf. A007913, A066099 (compositions in standard order), A246904, A246922, A259911, A259912, A305311, A378872, A378874.

a(n) = A007913(A378872(n)) = A378872(n)/A378874(n)^2.
a(2^n) = A259912(n+1) if a(2^n) == 1 (mod 4), a(2^n) = A259912(n+1)/4 otherwise.
For n > k >= 0, a(2^n+2^k) = A259911(n,k+1) if a(2^n+2^k) == 1 (mod 4), a(2^n+2^k) = A259911(n,k+1)/4 otherwise.

A378874 a(n) = sqrt(A378872(n)/A378873(n)).

Original entry on oeis.org

1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 4, 2, 1, 2, 1, 1, 4, 4, 1, 1, 4, 1, 4, 4, 1, 2, 3, 2, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 1, 2, 3, 2, 2, 1, 2, 1, 2, 2, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 1, 8, 1, 1, 2, 8, 1, 1, 6, 1, 8, 6, 1, 2, 1, 1, 6, 1, 4, 4, 1

Offset: 1