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a(n) = [ a(n-1)/a(1) ] + [ a(n-1)/a(2) ] + ... + [ a(n-1)/a(n-1) ] for n >= 3.
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%I #12 May 22 2019 04:02:20

%S 2,1,1,2,6,19,60,194,628,2035,6597,21393,69382,225031,729865,2367255,

%T 7678002,24902998,80770936,261974262,849693202,2755914018,8938593456,

%U 28991634898,94032120191,304985891949,989198096368

%N a(n) = [ a(n-1)/a(1) ] + [ a(n-1)/a(2) ] + ... + [ a(n-1)/a(n-1) ] for n >= 3.

%F a(n) ~ c * d^n, where d = 3.243422474587042277724011259031636390127037162673948731178241793178427249639..., c = 0.015782950162188139059124323554918048406538374877378493529663483700401917... - _Vaclav Kotesovec_, May 22 2019

%t Clear[a]; a[n_] := a[n] = If[n==1, 2, Sum[Floor[a[n-1]/a[k]], {k, 1, n-1}]]; Table[a[n], {n, 1, 30}] (* _Vaclav Kotesovec_, May 22 2019 *)

%Y Cf. A022855, A022858.

%K nonn

%O 1,1

%A _Clark Kimberling_

%E a(26) corrected by _Sean A. Irvine_, May 22 2019