A244188 - OEIS

%I #6 Jun 24 2014 00:10:52

%S 2,12,22,33,37,42,52,62,72,73,77,92,102,112,113,117,122,142,152,153,

%T 162,172,192,197,202,212,222,233,237,242,252,262,272,273,277,292,302,

%U 312,313,317,322,342,352,353,362,372,392,397,402,412,422,433,437,442,452,462,472

%N Numbers n such that the digit that repeats the most at the end of n^k for some k is not the last digit of n.

%C a(n) ends in either a 2, 3, or 7.

%C If a(n) ends in a 2, the digit that repeats itself the most at the end of a(n)^k is 8.

%C If a(n) ends in a 3, the digit that repeats itself the most at the end of a(n)^k is 7.

%C If a(n) ends in a 7, the digit that repeats itself the most at the end of a(n)^k is 3.

%C The numbers that end in 2 are congruent to {2, 12, 22, 42} mod 50.

%C The numbers that end in 3 are congruent to {33, 73, 113, 153} mod 200.

%C The numbers that end in 7 are congruent to {37, 77, 117, 197} mod 200.

%e 2^k ends in 2 of the same digit for k = 18 mod 20 (last digit is 4) and 19 mod 20 (last digit is 8). 2^k ends in 3 of the same digit for k = 39 mod 100 (last digit is 8). Since 8 is the only possibility, 8 must remain the only possibility for any larger run of identical digits. Since 8 is not the last digit of 2, then 2 is a member of this sequence.

%e 33^k ends in 2 of the same digit for k = 1 mod 20 (last digit is 3) and 7 mod 20 (last digit is 7). 33^k ends in 3 of the same digit for k = 87 mod 100 (last digit is 7). Since 7 is the only possibility, 7 must remain the only possibility for any larger run of identical digits. Since 7 is not the last digit of 33, 33 is a member of this sequence.

%o (PARI) seq(n)=for(m=2,6,cc=0;for(i=10^(m-1),10^m,st1=(n^i)%10^m;b="";for(j=1,m,b=concat(b,"1"));if(st1%eval(b)==0,for(d=i+1,10^m,sb1=(n^d)%10^m;if(sb1%eval(b)==0,if(sb1%10==st1%10,return(st1%10));if(sb1%10!=st1%10,cc++;break)))));if(cc==0,return(n%10)))

%o n=1;while(n<1000,if(seq(n)!=n%10,print1(n,", "));n++)

%Y Cf. A243977, A244187, A243911, A243912.

%K nonn,base

%O 1,1

%A _Derek Orr_, Jun 22 2014