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A128863 a(0)=1. For n >= 1, a(n) = number of positive divisors of (n+a(n-1)).

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%I A128863 #13 Mar 12 2019 20:22:42
%S A128863 1,2,3,4,4,3,3,4,6,4,4,4,5,6,6,4,6,2,6,3,2,2,8,2,4,2,6,4,6,4,4,4,9,8,
%T A128863 8,2,4,2,8,2,8,3,6,3,2,2,10,4,6,4,8,2,8,2,8,6,4,2,12,2,4,4,8,2,8,2,6,
%U A128863 2,8,4,4,6,8,5,2,4,10,4,4,2,4,4,4,4,8,4,12,6,4,4,4,4,12,8,8,2,6,2,9,12,10,4
%N A128863 a(0)=1. For n >= 1, a(n) = number of positive divisors of (n+a(n-1)).
%C A128863 Starting the sequence at a(1)=1 instead and using the recursion to get all terms after this, gets the sequence beginning: 1,2,2,4,3,3,4,6,4,4,4,5,... The sequences are the same from term a(4) on, whether starting with a(0)=1 or with a(1)=1.
%H A128863 Harvey P. Dale, <a href="/A128863/b128863.txt">Table of n, a(n) for n = 0..1000</a>
%e A128863 a(11)+12 = 16. So a(12) is the number of positive divisors of 16, which is 5.
%p A128863 with(numtheory): a:=proc(n) if n=0 then 1 else tau(n+a(n-1)) fi end: seq(a(n),n=0..130); # _Emeric Deutsch_, Apr 26 2007
%t A128863 nxt[{n_,a_}]:={n+1,DivisorSigma[0,a+n+1]}; NestList[nxt,{0,1},110][[All,2]] (* _Harvey P. Dale_, Mar 12 2019 *)
%K A128863 nonn
%O A128863 0,2
%A A128863 _Leroy Quet_, Apr 17 2007
%E A128863 More terms from _Emeric Deutsch_, Apr 26 2007