

A217368


Smallest number having a power that in decimal has exactly n copies of all ten digits.


3



32043, 69636, 643905, 421359, 320127, 3976581, 47745831, 15763347, 31064268, 44626422, 248967789, 85810806, 458764971, 500282265, 2068553967, 711974055, 2652652791, 901992825, 175536645, 3048377607, 3322858521, 1427472867, 3730866429, 9793730157
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OFFSET

1,1


COMMENTS

The exponents that produce the number with a fixed number of copies of each digit are listed in sequence A217378. See there for further comments.
Since we allow A217378(n)=1, the sequence is well defined, with the upper bound a(n) <= 100...99 ~ 10^(10n1) (n copies of each digit, sorted in increasing order, except for one "1" permuted to the first position).  M. F. Hasler, Oct 05 2012
What is the minimum value of a(n)? Can it be proved that a(n) > 2 for all n?  Charles R Greathouse IV, Oct 16 2012


LINKS



EXAMPLE

The third term raised to the fifth power (A217378(3)=5), 643905^5 = 110690152879433875483274690625, has three copies of each digit (in its decimal representation), and no number smaller than 643905 has a power with this feature.


MATHEMATICA

f[n_] := Block[{k = 2, t = Table[n, {10}], r = Range[0, 9]}, While[c = Count[ IntegerDigits[k^Floor[ Log[k, 10^(10 n)]]], #] & /@ r; c != t, k++]; k] (* Robert G. Wilson v, Nov 28 2012 *)


PROG

(PARI) is(n, k)=my(v); for(e=ceil((10*n1)*log(10)/log(k)), 10*n*log(10)/log(k), v=vecsort(digits(k^e)); for(i=1, 9, if(v[i*n]!=i1  v[i*n+1]!=i, return(0))); return(1)); 0


CROSSREFS



KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



