cp's OEIS Frontend

A244364 a(n) is the least integer m > 1 such that n is the largest number of identical digits that can end m^k for positive integer k.

%I A244364 #21 Jul 13 2014 10:18:49
%S A244364 3,4,2,33,17,319,639,1279,2559,5119,10239,20479,40959,81919,163839,
%T A244364 327679,655359,1310719,2621439,5242879,10485759,20971519,41943039,
%U A244364 83886079,167772159,335544319,671088639
%N A244364 a(n) is the least integer m > 1 such that n is the largest number of identical digits that can end m^k for positive integer k.
%C A244364 It is unknown if a(n) ends in a 9 for all n > 5.
%C A244364 a(n) <= 5*2^n-1 because (5*2^n-1)^(5^(n-1)) = -1 (mod 10^n), (5*2^n-1)^k != -1 (mod 10^(n+1)), and (5*2^n-1)^k != 11 (mod 100). - _Hiroaki Yamanouchi_, Jul 11 2014
%F A244364 Conjecture for n>5: a(n) = 10*2^(n-1)-1, k = (2r+1)*5^(n-2), r>=0. - _Lars Blomberg_, Jul 08 2014
%e A244364 33^k ends in 4 identical digits (let k == 187 mod 500) and never ends in more than 4 identical digits. Since this is not true for 2 <= m <= 32, a(4) = 33.
%o A244364 (PARI) a(n,p)=for(c=0,10^p,st=Str(n^c);if(#st>p,jo=(eval(st)%(10^p));if(jo==1,return(c));if(ispower(jo,,&k),if(ispower(n)!=0,if(jo^(1/(ispower(jo,,&k)))==(n^(1/(ispower(n)))),return(c)));if(ispower(n)==0,if(jo==n^(ispower(jo,,&k)),return(c))))))
%o A244364 hup(x)={m=1;for(i=2,100,f=0;for(j=m,a(x,i),dt=(x^j)%(10^i);b="";for(w=1,i,b=concat(b,"1"));if(dt%eval(b)==0,f++;r=j;break));if(f==0,return(i-1));m=r)}
%o A244364 n=1;while(n<100,for(x=2,10^3,if(hup(x)==n,print1(x,", ");break));n++)
%Y A244364 Cf. A243977.
%K A244364 nonn,base,hard,more
%O A244364 1,1
%A A244364 _Derek Orr_, Jun 26 2014
%E A244364 a(9)-a(27) from _Lars Blomberg_, Jul 08 2014
%E A244364 Definition and example edited by _Robert Israel_, Jul 10 2014