login

Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.

a(n) = numerator(r(n)) where r(n) = (((1/2)*(sqrt(3) + 1))^n - ((1/2)*(sqrt(3) - 1))^n * cos(Pi*n))/sqrt(3).
3

%I #32 Dec 06 2018 11:09:47

%S 0,1,1,3,2,11,15,41,7,153,209,571,195,2131,2911,7953,679,29681,40545,

%T 110771,37829,413403,564719,1542841,263445,5757961,7865521,21489003,

%U 7338631,80198051,109552575,299303201,12776743,1117014753,1525870529,4168755811,1423656585

%N a(n) = numerator(r(n)) where r(n) = (((1/2)*(sqrt(3) + 1))^n - ((1/2)*(sqrt(3) - 1))^n * cos(Pi*n))/sqrt(3).

%C Let f(x, y) = ((y+1)^x - (y-1)^x * cos(Pi*x))/(y * 2^x). Then f(n, sqrt(3)) are the rational numbers a(n)/A060723(n) and f(n, sqrt(5)) the Fibonacci numbers A000045(n).

%C From _Paul Curtz_, Dec 05 2018: (Start)

%C The binomial inverse of the rational sequence r(n) starts 0, 1, -1, 3/2, -2, 11/4, -15/4, 41/8, -7, 153/16, -209/16, ... and is up to signs equal to r(n). The difference table starts:

%C 0, 1, 1, 3/2, 2, 11/4, 15/4, 41/8, ...

%C 1, 0, 1/2, 1/2, 3/4, 1, 11/8, 15/8, ...

%C -1, 1/2, 0, 1/4, 1/4, 3/8, 1/2, 11/16, ...

%C 3/2, -1/2, 1/4, 0, 1/8, 1/8, 3/16, 1/4, ...

%C ...

%C Let s(n) = 2*r(n+1) - r(n) then s(n) = 1, 2, 5/2, 7/2, 19/4, 13/2, ... = A173299(n)/A173300(n) for n >= 1. (End)

%F A recurrence for r(n) is given in A060723.

%t Table[Numerator[Simplify[((1/2 (Sqrt[3] + 1))^x - (1/2 (Sqrt[3] - 1))^x Cos[Pi x])/Sqrt[3]]], {x, 0, 36}]

%Y Cf. A060723 (denominators), A060755, A000045, A305492.

%Y Cf. A173299/A173300.

%K nonn

%O 0,4

%A _Peter Luschny_, Jun 02 2018