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Smallest number that requires n steps to reach 0 when iterating the mapping k -> abs(reverse(lpd(k))-reverse(Lpf(k))). lpd(k) is the largest proper divisor and Lpf(k) is the largest prime factor of k.
1

%I #3 Mar 30 2012 17:27:39

%S 1,2,3,12,31,23,56,102,193,257,570,1129,4970,3229,11551,11969,24232,

%T 20094,24103,35996,100090,222284,116269,231488,388768,1751753,2046872,

%U 1140163,1149979,2156214,3199384,2971734,7018074,10163234,13135933

%N Smallest number that requires n steps to reach 0 when iterating the mapping k -> abs(reverse(lpd(k))-reverse(Lpf(k))). lpd(k) is the largest proper divisor and Lpf(k) is the largest prime factor of k.

%e a(5) =31 since 31 requires 5 steps, but no m < 31 does. Although 23 < 31, 23 requires 6 steps.

%o (PARI) {m=36; z=19200000; 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; z=if(i>0,d[i],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]); 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. A074347, A076423.

%K base,nonn

%O 1,2

%A _Klaus Brockhaus_, Oct 11 2002