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%I #19 Aug 08 2020 14:35:53
%S 1,2,3,12,13,52,53,131,271,811,1601,2711,8111,13997,34589,74551,
%T 147773,310567,621227,1230343,2627759,4921373,10741931,24965191,
%U 45887291,111477631,183638843,394195667,788380493,1576798931
%N Smallest number requiring n steps to reach 0 when iterating the function: f(n)=abs(lpd(n)-Lpf(n)), where lpd(n) is the largest proper divisor of n and Lpf(n) is the largest prime factor of n.
%C 2*10^9 < a(31) <= 2938669883. a(32) <= 7511549827. a(33) <= 11754740251. a(34) <= 30050593523. - _Donovan Johnson_, Dec 22 2010
%H Jason Earls, <a href="https://www.semanticscholar.org/paper/Smarandache-iterations-of-the-first-kind-on-and-Earls/3685dfcee58bf0d2e15a15198e4005ecbadc0c8a">Smarandache iterations of the first kind on functions involving divisors and prime factors</a>, in Smarandache Notions Journal (2004), Vol. 14.1, page 259.
%o (PARI) {m=25; z=11000000; v=listcreate(m); for(i=1,m,listinsert(v,-1,i)); for(n=1,z,c=1; b=1; k=n; while(b&&c<=m,d=divisors(k); i=matsize(d)[2]-1; p=if(i>0,d[i],1); q=if(k==1,1,vecmax(component(factor(k),1))); a=abs(p-q); if(a==0,b=0,k=a; c++)); if(a==0,if(v[c]<0,v[c]=n; print1([c,n])))); print(); for(i=1,m,print1(v[i],","))}
%Y Cf. A075660.
%K more,nonn
%O 1,2
%A _Jason Earls_, Sep 23 2002
%E Four more terms from _Klaus Brockhaus_, Oct 01 2002
%E a(24)-a(30) from _Donovan Johnson_, Dec 22 2010
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