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

1,1

Comments

1 <= a(n) <= 10.

Links

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

Formula

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

Example

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

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 *)

Prog

(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
(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

Crossrefs

Cf. A067442.

Sequence in context: A063566 A153690 A018811 * A100844 A076587 A266999

Adjacent sequences: A309704 A309705 A309706 * A309708 A309709 A309710

Keyword

nonn,base

Author

Robert Israel, Aug 13 2019

Status

approved