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a(0)=1; a(n+1) = 1 + f(a(n))^2, where f(x) is the largest prime factor of x (A006530).
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%I #17 Feb 01 2022 18:57:32

%S 1,2,5,26,170,290,842,177242,160802,2810,78962,9223370,5033760602,

%T 2935496262242,2154284576409188208716642,

%U 1379590379356276893461978662419832989306970202,10320758390549056348725939119133160378521185060950774444682

%N a(0)=1; a(n+1) = 1 + f(a(n))^2, where f(x) is the largest prime factor of x (A006530).

%C Is the sequence bounded?

%C Essentially the same as A031439; a(n) = A031439(n-1)^2 + 1. - _Charles R Greathouse IV_, May 08 2009

%e Given a(5)=290: a(6) = 1 + lpf(a(5))^2 = 1 + lpf(290)^2 = 1 + 29^2 = 842.

%p with(numtheory): a[0]:=1: a[1]:=2: for n from 2 to 20 do b:=factorset(a[n-1]): a[n]:=1+op(nops(b),b)^2: od: seq(a[n],n=0..20); # _Emeric Deutsch_, Feb 05 2006

%t NestList[1+FactorInteger[#][[-1,1]]^2&,1,17] (* _Harvey P. Dale_, Feb 01 2022 *)

%Y Cf. A031439.

%K nonn

%O 0,2

%A _Reinhard Zumkeller_, Jul 08 2002

%E More terms from _Emeric Deutsch_, Feb 05 2006

%E a(16) corrected by _T. D. Noe_, Nov 26 2007