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.

3, 4, 2, 33, 17, 319, 639, 1279, 2559, 5119, 10239, 20479, 40959, 81919, 163839, 327679, 655359, 1310719, 2621439, 5242879, 10485759, 20971519, 41943039, 83886079, 167772159, 335544319, 671088639

Offset: 1

Author

Derek Orr, Jun 26 2014

Keywords

• nonn
• base
• hard
• more

Comments

It is unknown if a(n) ends in a 9 for all n > 5.

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

Examples

			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.
		

Crossrefs

Cf. A243977.

Programs

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))))))
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)}
n=1;while(n<100,for(x=2,10^3,if(hup(x)==n,print1(x,", ");break));n++)

Formula

Conjecture for n>5: a(n) = 10*2^(n-1)-1, k = (2r+1)*5^(n-2), r>=0. - Lars Blomberg, Jul 08 2014

Extensions

a(9)-a(27) from Lars Blomberg, Jul 08 2014

Definition and example edited by Robert Israel, Jul 10 2014