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a(0)=1; a(n) = the smallest prime dividing (n+a(n-1)), for n>=1.
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%I #11 Sep 16 2015 13:59:24

%S 1,2,2,5,3,2,2,3,11,2,2,13,5,2,2,17,3,2,2,3,23,2,2,5,29,2,2,29,3,2,2,

%T 3,5,2,2,37,73,2,2,41,3,2,2,3,47,2,2,7,5,2,2,53,3,2,2,3,59,2,2,61,11,

%U 2,2,5,3,2,2,3,71,2,2,73,5,2,2,7,83,2,2,3,83,2,2,5,89,2,2,89,3,2,2,3,5,2,2

%N a(0)=1; a(n) = the smallest prime dividing (n+a(n-1)), for n>=1.

%C a(4n+1) = a(4n+2) = 2, for all n >= 0. a(4n) and a(4n+3) are odd primes, for all n >= 0.

%H Harvey P. Dale, <a href="/A132850/b132850.txt">Table of n, a(n) for n = 0..1000</a>

%e a(8) + 9 = 11 + 9 = 20. The smallest prime divisor of 20 is 2. So a(9) = 2.

%t a = {1}; Do[AppendTo[a, FactorInteger[n + a[[ -1]]][[1, 1]]], {n, 1, 100}]; a (* _Stefan Steinerberger_, Nov 25 2007 *)

%t nxt[{n_,a_}]:={n+1,FactorInteger[n+1+a][[1,1]]}; Transpose[NestList[nxt,{0,1},100]][[2]] (* _Harvey P. Dale_, Jan 21 2015 *)

%Y Cf. A076561.

%K nonn

%O 0,2

%A _Leroy Quet_, Nov 21 2007

%E More terms from _Stefan Steinerberger_, Nov 25 2007