Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #8 Dec 12 2024 09:25:15
%S 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 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 37,221,3,65,26,165,221,65,165,65,65,5,53,15,35,37,3,229
%N 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 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 Wikipedia, <a href="https://en.wikipedia.org/wiki/Discriminant_of_an_algebraic_number_field">Discriminant of an algebraic number field</a>.
%F a(n) = A007913(A378872(n)) = A378872(n)/A378874(n)^2.
%F a(2^n) = A259912(n+1) if a(2^n) == 1 (mod 4), a(2^n) = A259912(n+1)/4 otherwise.
%F 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 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 Cf. A007913, A066099 (compositions in standard order), A246904, A246922, A259911, A259912, A305311, A378872, A378874.
%K nonn
%O 1,1
%A _Pontus von Brömssen_, Dec 10 2024