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%I #15 Apr 26 2020 05:47:34
%S 1,1,4,6,9,12,15,18,21,25,28,31,34,37,41,44,47,51,54,57,61,64,68,71,
%T 75,78,82,85,89,92,96,99,103,106,110,113,117,120,124,128,131,135,138,
%U 142,146,149,153,157,160,164,168,171,175,179,182,186,190,193,197,201,204
%N a(1)=1; a(n) = floor((n-1)*Sum_{k=1..n-1} 1/a(k)).
%H Vaclav Kotesovec, <a href="/A157124/b157124.txt">Table of n, a(n) for n = 1..10000</a>
%e For n = 5, (5-1)*(1 + 1 + 1/4 + 1/6) = 4*29/12 = 29/3. So a(5) = floor(29/3) = 9.
%p a[1] := 1: for n to 65 do a[n+1] := floor(n*(sum(1/a[k], k = 1 .. n))) end do: seq(a[n], n = 1 .. 65); # _Emeric Deutsch_, Mar 01 2009
%t a[1] = 1; a[n_] := a[n] = Floor[(n - 1)*Sum[1/a[k], {k, 1, n - 1}]]; Table[a[n], {n, 1, 100}] (* _Vaclav Kotesovec_, Apr 26 2020 *)
%K nonn
%O 1,3
%A _Leroy Quet_, Feb 23 2009
%E More terms from _Emeric Deutsch_, Mar 01 2009