A305492 -id:A305492 - cp's OEIS frontend

A305491 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).

0, 1, 1, 3, 2, 11, 15, 41, 7, 153, 209, 571, 195, 2131, 2911, 7953, 679, 29681, 40545, 110771, 37829, 413403, 564719, 1542841, 263445, 5757961, 7865521, 21489003, 7338631, 80198051, 109552575, 299303201, 12776743, 1117014753, 1525870529, 4168755811, 1423656585

Offset: 0

Peter Luschny, Jun 02 2018

nonn

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).
From Paul Curtz, Dec 05 2018: (Start)
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:
      0,     1,    1,  3/2,    2, 11/4,  15/4,   41/8, ...
      1,     0,  1/2,  1/2,  3/4,    1,  11/8,   15/8, ...
     -1,   1/2,    0,  1/4,  1/4,  3/8,   1/2,  11/16, ...
    3/2,  -1/2,  1/4,    0,  1/8,  1/8,  3/16,    1/4, ...
    ...
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)

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

Cf. A173299/A173300.

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

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

cp's OEIS Frontend

A305491 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).

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cp's OEIS Frontend

A305491 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).

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

Formula

A305491 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).