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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.
4
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
1,2
COMMENTS
See A076424 for numbers such that zero is never reached, A076425 for the smallest numbers that need n iterations to reach zero, A076426 for fixed points of the mapping.
EXAMPLE
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.
PROG
(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), ", "))}
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Klaus Brockhaus, Oct 11 2002
STATUS
approved