login
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
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
1,2
COMMENTS
2*10^9 < a(31) <= 2938669883. a(32) <= 7511549827. a(33) <= 11754740251. a(34) <= 30050593523. - Donovan Johnson, Dec 22 2010
LINKS
PROG
(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], ", "))}
CROSSREFS
Cf. A075660.
Sequence in context: A102660 A081347 A309176 * A102034 A102150 A039588
KEYWORD
more,nonn
AUTHOR
Jason Earls, Sep 23 2002
EXTENSIONS
Four more terms from Klaus Brockhaus, Oct 01 2002
a(24)-a(30) from Donovan Johnson, Dec 22 2010
STATUS
approved