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A146535
Numerator of (2*n-1)/3.
9
1, 1, 5, 7, 3, 11, 13, 5, 17, 19, 7, 23, 25, 9, 29, 31, 11, 35, 37, 13, 41, 43, 15, 47, 49, 17, 53, 55, 19, 59, 61, 21, 65, 67, 23, 71, 73, 25, 77, 79, 27, 83, 85, 29, 89, 91, 31, 95, 97, 33, 101, 103, 35, 107, 109, 37, 113, 115, 39, 119, 121, 41, 125, 127, 43, 131, 133, 45
OFFSET
1,3
COMMENTS
From Jaroslav Krizek, May 28 2010: (Start)
a(n+1) = numerators of antiharmonic mean of the first n positive integers for n >= 1.
See A169609(n-1) - denominators of antiharmonic mean of the first n positive integers for n >= 1. (End)
FORMULA
From R. J. Mathar, Nov 21 2008: (Start)
a(n) = 2*a(n-3) - a(n-6).
G.f.: x(1+x)(1+5x^2+x^4)/((1-x)^2*(1+x+x^2)^2). (End)
Sum_{k=1..n} a(k) ~ (7/9) * n^2. - Amiram Eldar, Apr 04 2024
EXAMPLE
Fractions begin with 1/6, 1/2, 5/6, 7/6, 3/2, 11/6, 13/6, 5/2, 17/6, 19/6, 7/2, 23/6, ...
MATHEMATICA
Table[Numerator[(2 n - 1)/6], {n, 1, 100}]
LinearRecurrence[{0, 0, 2, 0, 0, -1}, {1, 1, 5, 7, 3, 11}, 100] (* Harvey P. Dale, Feb 24 2015 *)
PROG
(PARI) a(n) = numerator((2*n-1)/3); \\ Altug Alkan, Apr 13 2018
CROSSREFS
KEYWORD
nonn,easy,frac
AUTHOR
Artur Jasinski, Oct 31 2008
EXTENSIONS
Name edited by Altug Alkan, Apr 13 2018
STATUS
approved