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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
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
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
KEYWORD
nonn,base
AUTHOR
Robert Israel, Aug 14 2019
STATUS
approved