A309707 a(n) is the least integer k > 1 such that k^n starts with 1.

10, 4, 5, 2, 4, 5, 2, 6, 3, 2, 3, 4, 3, 2, 3, 5, 2, 6, 3, 2, 3, 4, 5, 2, 4, 5, 2, 8, 5, 2, 6, 3, 5, 2, 4, 3, 2, 3, 5, 2, 8, 3, 5, 2, 4, 5, 2, 10, 5, 2, 7, 10, 3, 2, 3, 5, 2, 6, 3, 2, 3, 6, 3, 2, 3, 5, 2, 10, 5, 2, 6, 6, 5, 2, 4, 3, 2, 3, 5, 2, 6, 3, 5, 2, 4, 3, 2, 10, 5, 2, 8

Offset: 1

Author

Robert Israel, Aug 13 2019

Keywords

• nonn
• base

Comments

1 <= a(n) <= 10.

Examples

			a(3)=5 because 5^3=125 starts with 1, and none of 2^3, 3^3, 4^3 do.
		

Links

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

Crossrefs

Cf. A067442.

Programs

Magma

m:=1; sol:=[]; for n in [1..100] do k:=2; while Reverse(Intseq(k^n))[1] ne 1 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 d,x,y;
  for x from 2 to 10 do
    y:= x^n;
    if floor(y/10^ilog10(y)) = 1 then return x fi
od
end proc:
map(f, [$1..100]);

Mathematica

lik[n_]:=Module[{k=2},While[IntegerDigits[k^n][[1]]!=1,k++];k]; Array[ lik,100] (* Harvey P. Dale, Dec 06 2019 *)

PARI

a(n) = {my(k=2); while(digits(k^n)[1] != 1, k++); k;} \\ Michel Marcus, Aug 15 2019

Python

print(1,10)
n = 1
while n < 100:
    n, p = n+1, 2
    s = str(p**n)
    while s[0] != "1":
        p = p+1
        s = str(p**n)
    print(n,p) # A.H.M. Smeets, Aug 16 2019

Formula

a(n) = A067442(n)^(1/n) for n >= 2.