%I
%S 2,5,8,11,13,16,19,22,25,27,30,33,36,38,41,44,47,50,52,55,58,61,63,66,
%T 69,72,75,77,80,83,86,88,91,94,97,100,102,105,108,111,114,116,119,122,
%U 125,127,130,133,136,139,141,144,147,150,152,155,158,161,164,166,169
%N Upper Beatty sequence for e^G, G = Euler's gamma constant.
%C Determine the continued fraction convergents to e^(G) = .561459484...; the first few being 1/1, 1/2, 4/7, 5/9, 9/16, 32/57...(check: 32/57 = .561403508...). Pick a convergent, a/b say 5/9. Then through (a+b) = n = 14, 5 of those integers are in the upper Beatty pair set: 2, 5, 8, 11, 13; while 9 terms are in the lower Beatty pair set, being 1, 3, 4, 6, 7, 9, 10, 12, 14. Since the upper Beatty pair set is derived from (k+1) and the lower from (k+1)/k, the ratio of upper to lower converges to k = 1.789107241...= e^G.
%F a(n) = floor (n*(k+1)) where k = 1.781072417...= e^G, G = Euler's gamma constant, .577215664901...
%e a(7) = 19 since floor(n*2.7810724...) = 19.
%t Table[ Floor[ n*(E^EulerGamma + 1)], {n, 65}]
%Y Beatty complement is A093610.
%K nonn
%O 1,1
%A _Gary W. Adamson_, Apr 04 2004
%E More terms from _Robert G. Wilson v_, Apr 05 2004
