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%I #22 Oct 31 2023 01:07:20
%S 1,1,4,6,9,11,14,16,19,22,24,27,31,33,36,38,42,44,48,51,54,56,60,62,
%T 67,69,71,75,79,81,84,87,91,95,97,99,105,107,111,113,116,118,123,125,
%U 131,134,136,138,145,147,149,152,155,157,163,166,171,174,176,178,183,185
%N a(0)=1; for n > 0, a(n) = Sum_{k=0..n-1} floor(n/a(k)).
%H Robin Visser, <a href="/A138812/b138812.txt">Table of n, a(n) for n = 0..10000</a>
%F Probably a(n) ~ sqrt(2) n log(n)^(1/2) as n -> oo. - _Robert Israel_, May 02 2008
%F From _Andrew V. Sutherland_, May 02 2008: (Start)
%F This is supported by the following data:
%F .
%F n a(n) a(n)/n a(n)/(n*sqrt(log(n)))
%F ------ ------ ------ ---------------------
%F 2 4 2.0000 2.4022
%F 4 9 2.2500 1.9110
%F 8 19 2.3750 1.6470
%F 16 42 2.6250 1.5765
%F 32 91 2.8438 1.5275
%F 64 196 3.0625 1.5017
%F 128 421 3.2891 1.4932
%F 256 896 3.5000 1.4863
%F 512 1892 3.6953 1.4795
%F 1024 3979 3.8857 1.4759
%F 2048 8335 4.0698 1.4739
%F 4096 17386 4.2446 1.4718
%F 8192 36146 4.4124 1.4699
%F 16384 74931 4.5734 1.4681
%F 32768 154964 4.7291 1.4666
%F 65536 319818 4.8800 1.4654
%F 131072 658761 5.0259 1.4641 (End)
%p a[0]:=1: for n to 65 do a[n]:=sum(floor(n/a[k]),k=0..n-1) end do: seq(a[n], n =0..65); # _Emeric Deutsch_, Apr 04 2008
%t a = {1}; Do[AppendTo[a, Sum[Floor[n/a[[k]]], {k, 1, n}]], {n, 1, 70}]; a (* _Stefan Steinerberger_, Apr 04 2008 *)
%Y Cf. A138813.
%K nonn
%O 0,3
%A _Leroy Quet_, Mar 31 2008
%E More terms from _Stefan Steinerberger_ and _Emeric Deutsch_, Apr 04 2008
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