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%I #10 May 21 2018 18:32:09
%S 1,2,3,5,7,10,13,17,22,27,33,39,46,54,62,71,80,90,100,111,123,135,148,
%T 161,175,189,204,220,236,253,270,288,306,325,345,365,386,407,429,451,
%U 474,498,522,547,572,598,625,652,680,708,737,766,796,827,858,890,922
%N a(1)=1. a(n+1) = a(n) + (number of terms of this sequence that are <= (1/n)sum{k=1 to n} a(k)).
%H Ivan Neretin, <a href="/A136413/b136413.txt">Table of n, a(n) for n = 1..10000</a>
%e The sum of the first 7 terms of this sequence is 1+2+3+5+7+10+13 = 41. So the arithmetic average of the first 7 terms is 41/7. The terms of this sequence that are <= 41/7 (= 5 +6/7) are 1,2,3,5. There are therefore 4 such terms <= 41/7. So a(8) = a(7) + 4 = 13 + 4 = 17.
%t cnt = Function[a, Length@Select[a, # <= Total[a]/Length[a] &]]; Nest[Append[#, #[[-1]] + cnt[#]] &, {1}, 56] (* _Ivan Neretin_, May 21 2018*)
%K nonn
%O 1,2
%A _Leroy Quet_, Mar 31 2008
%E More terms from _Sean A. Irvine_, May 03 2010
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