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A309735 a(n) is the least positive integer k such that k^n starts with 2. 1
2, 5, 3, 4, 3, 8, 3, 2, 4, 7, 2, 5, 9, 4, 9, 6, 7, 2, 4, 21, 2, 5, 7, 3, 5, 3, 8, 2, 4, 3, 2, 5, 11, 4, 5, 7, 8, 2, 6, 23, 2, 5, 6, 14, 3, 16, 3, 2, 3, 14, 2, 4, 15, 17, 5, 7, 4, 2, 11, 18, 2, 4, 47, 14, 5, 6, 4, 2, 7, 3, 2, 3, 13, 3, 5, 15, 4, 8, 6, 9, 2, 4, 11, 6, 5, 22, 4 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

For n > 1, take integer d > -n log_10(3^(1/n)-2^(1/n)).

Then 10^(d/n) > 1/(3^(1/n) - 2^(1/n))

so (3*10^d)^(1/n) - (2*10^d)^(1/n) > 1

and therefore a(n) <= ceiling((2*10^d)^(1/n)).

In particular, a(n) always exists.

LINKS

Robert Israel, Table of n, a(n) for n = 1..10000

FORMULA

a(n) = A067443(n)^(1/n).

A000030(a(n)^n)=2.

EXAMPLE

a(5) = 3 because 3^5 = 243 starts with 2, while 1^5=1 and 2^5=32 do not start with 2.

MAPLE

f:= proc(n) local x, y;

  for x from 2  do

    y:= x^n;

      if floor(y/10^ilog10(y)) = 2 then return x fi

  od

end proc:

map(f, [$1..100]);

PROG

(PARI) a(n) = for(k=1, oo, if(digits(k^n)[1]==2, return(k))) \\ Felix Fröhlich, Aug 14 2019

(Python)

n = 1

while n < 100:

    k, s = 2, str(2**n)

    while s[0] != "2":

        k = k+1

        s = str(k**n)

    print(n, k)

    n = n+1 # A.H.M. Smeets, Aug 14 2019

(MAGMA) m:=1; sol:=[]; for n in [1..100] do k:=2; while Reverse(Intseq(k^n))[1] ne 2 do k:=k+1; end while; sol[m]:=k; m:=m+1; end for; sol; // Marius A. Burtea, Aug 15 2019

CROSSREFS

Cf. A000030, A067443, A309707.

Sequence in context: A077057 A030660 A253720 * A275726 A146096 A011310

Adjacent sequences:  A309732 A309733 A309734 * A309736 A309737 A309738

KEYWORD

nonn,base

AUTHOR

Robert Israel, Aug 14 2019

STATUS

approved

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Last modified July 11 20:03 EDT 2020. Contains 335652 sequences. (Running on oeis4.)