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a(n) = n + floor[sum_k{k<n}a(k)/2] with a(0)=0.
2

%I #6 Aug 01 2021 19:31:10

%S 0,1,2,4,7,12,19,29,45,68,103,156,235,353,531,797,1197,1796,2695,4044,

%T 6067,9101,13653,20480,30721,46083,69125,103689,155534,233302,349954,

%U 524932,787399,1181100,1771651,2657477,3986217,5979326,8968990,13453486

%N a(n) = n + floor[sum_k{k<n}a(k)/2] with a(0)=0.

%H Harvey P. Dale, <a href="/A079719/b079719.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n)=A073941(n+4)-2 (and appears to be A005428(n+1)) =round[1.82505431574536323...*1.5^n - 2]

%e a(4) = 4+floor[(0+1+2+4)/2] = 4+floor[7/2] = 4+3 = 7

%t nxt[{n_,t_,a_}]:=Module[{c=n+1+Floor[t/2]},{n+1,t+c,c}]; NestList[nxt,{0,0,0},40][[All,3]] (* _Harvey P. Dale_, Aug 01 2021 *)

%K nonn

%O 0,3

%A _Henry Bottomley_, Feb 17 2003