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a(n) = floor(e^(n + 1/2)).
1

%I #13 Nov 01 2024 05:06:43

%S 1,4,12,33,90,244,665,1808,4914,13359,36315,98715,268337,729416,

%T 1982759,5389698,14650719,39824784,108254987,294267566,799902177,

%U 2174359553,5910522063,16066464720,43673179097,118716009132,322703570371

%N a(n) = floor(e^(n + 1/2)).

%C a(n) is the number k such that {log(k)} < 1/2 < {log(k+1)}, where { } = fractional part. Equivalently, the jump sequence of f(x) = log(x), in the sense that these are the positive integers k for which round(log(k)) < round(log(k+1)). For a guide to related sequences, see A219085.

%F a(n) = [e^(n + 1/2)].

%e log(1) = 0.000... ; log(2) = 0.693...

%e log(4) = 1.386... ; log(5) = 1.609...

%e log(12) = 2.484... ; log(13) = 2.564...

%t Table[Floor[E^(n + 1/2)], {n, 0, 100}]

%Y Cf. A001113, A219085.

%K nonn,easy,changed

%O 0,2

%A _Clark Kimberling_, Jan 01 2013