OFFSET
1,1
COMMENTS
That s(n) > 0 for n >=1 follows from the chain 1 < log 2 < 3/4 < 2 log 3/2 < 5/6 < 3 log 4/3 < 7/8 < 4 log 5/4 < ... ; i.e., n log((n+1)/n) - (2n-1)/(2n) > 0 and (2n+1)/(2n+2) - n log((n+1)/n) > 0. For the first, closeness to 0 is indicated by A227719 and A227720, and for the second, by A227721 and a sequence which possibly equals A094159. Conjecture: the four sequences are linearly recurrent.
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..1000
FORMULA
a(n) = -2*a(n-1) - a(n-2) + a(n-4) - 2*a(n-5) + a(n-6) (conjectured).
G.f.: (-5 - 6 x - 7 x^2 - 5 x^3 - x^4)/((-1 + x)^3 (1 + x + x^2 + x^3)) (conjectured).
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jul 22 2013
STATUS
approved