A074347 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.

1, 2, 3, 12, 13, 52, 53, 131, 271, 811, 1601, 2711, 8111, 13997, 34589, 74551, 147773, 310567, 621227, 1230343, 2627759, 4921373, 10741931, 24965191, 45887291, 111477631, 183638843, 394195667, 788380493, 1576798931

Offset: 1

Author

Jason Earls, Sep 23 2002

Keywords

• more
• nonn

Comments

2*10^9 < a(31) <= 2938669883. a(32) <= 7511549827. a(33) <= 11754740251. a(34) <= 30050593523. - Donovan Johnson, Dec 22 2010

Links

• Jason Earls, Smarandache iterations of the first kind on functions involving divisors and prime factors, in Smarandache Notions Journal (2004), Vol. 14.1, page 259.

Crossrefs

Cf. A075660.

Programs

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],","))}

Extensions

Four more terms from Klaus Brockhaus, Oct 01 2002

a(24)-a(30) from Donovan Johnson, Dec 22 2010