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)).

%I A378873 #8 Dec 12 2024 09:25:15
%S A378873 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,
%T A378873 17,6,6,5,10,5,6,26,13,37,37,165,6,37,2,221,37,3,221,65,5,26,37,165,
%U A378873 37,221,3,65,26,165,221,65,165,65,65,5,53,15,35,37,3,229
%N 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)).
%C A378873 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.
%H A378873 Wikipedia, <a href="https://en.wikipedia.org/wiki/Discriminant_of_an_algebraic_number_field">Discriminant of an algebraic number field</a>.
%F A378873 a(n) = A007913(A378872(n)) = A378872(n)/A378874(n)^2.
%F A378873 a(2^n) = A259912(n+1) if a(2^n) == 1 (mod 4), a(2^n) = A259912(n+1)/4 otherwise.
%F A378873 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.
%e A378873 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.
%Y A378873 Cf. A007913, A066099 (compositions in standard order), A246904, A246922, A259911, A259912, A305311, A378872, A378874.
%K A378873 nonn
%O A378873 1,1
%A A378873 _Pontus von Brömssen_, Dec 10 2024