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a(1) = 2, a(2) = 3; thereafter a(n) = a(n-1) + a(|n-a(T)|), where a(T) is the largest term in the sequence before a(n) such that 0 < |n-a(T)| < n.
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%I #32 Apr 09 2016 13:57:51

%S 2,3,5,7,10,17,22,25,50,72,89,161,322,411,483,533,558,580,597,607,614,

%T 619,622,624,629,1253,1875,2494,3108,3715,4312,4892,5450,5983,6466,

%U 6877,13343,19326,24776,29668,33980,37695,40803,43297,86594,127397,165092,199072,228740,253516

%N a(1) = 2, a(2) = 3; thereafter a(n) = a(n-1) + a(|n-a(T)|), where a(T) is the largest term in the sequence before a(n) such that 0 < |n-a(T)| < n.

%C When a(1) = 1, the formula generates the natural numbers.

%H Cody M. Haderlie, <a href="/A271063/b271063.txt">Table of n, a(n) for n = 1..1982</a>

%e a(11) = 89

%e a(T) = a(7) = 22

%e |12-22| = 10

%e a(10) = 72

%e 89 + 72 = 161

%e a(12) = 161

%t a = {2, 3}; Do[AppendTo[a, a[[n - 1]] + Part[a, #] &@ SelectFirst[ Reverse@ Abs[a - n], 0 < # < n &]], {n, 3, 50}]; a (* _Michael De Vlieger_, Apr 08 2016 *)

%o (PARI) lista(nn) = {va = vector(nn); print1(va[1] = 2, ", "); print1(va[2] = 3, ", "); for (n=3, nn, T = 0; forstep(k = n-1, 1, -1, vabs = abs(n - va[k]); if ((vabs < n) && (vabs > 0), T = k; break);); va[n] = va[n-1] + va[abs(n-va[T])]; print1(va[n], ", "););} \\ _Michel Marcus_, Apr 08 2016

%K nonn

%O 1,1

%A _Cody M. Haderlie_, Apr 05 2016