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

Author

Robert Israel, Aug 14 2019

Keywords

• nonn
• base

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.

Examples

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

Links

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

Crossrefs

Cf. A000030, A067443, A309707.

Programs

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

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]);

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

Formula

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

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