A309735 a(n) is the least positive integer k such that k^n starts with 2.

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

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: A030660 A253720 A371353 * A275726 A355505 A146096

Adjacent sequences: A309732 A309733 A309734 * A309736 A309737 A309738

Keyword

nonn,base

Author

Robert Israel, Aug 14 2019

Status

approved