A076423 Number of iterations of the mapping k -> abs(reverse(lpd(k))-reverse(gpf(k))) to reach zero, or -1 if zero is never reached. lpd(k) is the largest proper divisor and gpf(k) is the greatest prime factor of k.
1, 2, 3, 1, 2, 1, 2, 3, 1, 1, 2, 4, 3, 1, 1, 2, 3, 2, 3, 2, 1, 1, 6, 3, 1, 1, 2, 2, 2, 2, 5, 3, 1, 1, 1, 3, 5, 1, 1, 4, 4, 3, 2, 3, 2, 1, 5, 2, 1, 6, 1, 6, 2, 2, 1, 7, 1, 1, 2, 3, 2, 1, 3, 2, 1, 2, 7, 3, 1, 2, 3, 4, 4, 1, 6, 4, 1, 2, 4, 2, 2, 1, 6, 4, 1, 1, 1, 2, 4, 2, 1, 4, 1, 1, 1, 3, 3, 2, 2, 1, 2, 8, 3, 2, 2
Offset: 1
Examples
For n = 13: lpd(13) = 1, gpf(13)=13, abs(reverse(1)-reverse(13)) = 30; lpd(30) = 15, gpf(30) = 5, abs(reverse(15)-reverse(5)) = 46; lpd(46) = 23, gpf(46)=23, abs(reverse(23)-reverse(23)) = 0. Three iterations to reach zero, so a(13) = 3.
Programs
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PARI
{stop=20; for(n=1,105,c=1; b=1; k=n; while(b&&c<=stop,w=divisors(k); s=matsize(w)[2]-1; z=if(s>0,w[s],1); p=0; while(z>0,d=divrem(z,10); z=d[1]; p=10*p+d[2]); z=if(k==1,1,vecmax(component(factor(k),1))); q=0; while(z>0,d=divrem(z,10); z=d[1]; q=10*q+d[2]); k=abs(p-q); if(k>0,c++,b=0)); print1(if(c>stop,-1,c),","))}
Comments