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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. 0
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

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

LINKS

Table of n, a(n) for n=1..27.

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

EXAMPLE

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.

PROG

(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++)

CROSSREFS

Cf. A243977.

Sequence in context: A225475 A259334 A210488 * A218610 A260958 A260965

Adjacent sequences:  A244361 A244362 A244363 * A244365 A244366 A244367

KEYWORD

nonn,base,hard,more

AUTHOR

Derek Orr, Jun 26 2014

EXTENSIONS

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

Definition and example edited by Robert Israel, Jul 10 2014

STATUS

approved

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Last modified September 26 05:09 EDT 2017. Contains 292502 sequences.