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a(n) is the least positive integer k such that k^n starts with 2.
1

%I #29 Sep 08 2022 08:46:22

%S 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,

%T 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,

%U 14,5,6,4,2,7,3,2,3,13,3,5,15,4,8,6,9,2,4,11,6,5,22,4

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

%C For n > 1, take integer d > -n log_10(3^(1/n)-2^(1/n)).

%C Then 10^(d/n) > 1/(3^(1/n) - 2^(1/n))

%C so (3*10^d)^(1/n) - (2*10^d)^(1/n) > 1

%C and therefore a(n) <= ceiling((2*10^d)^(1/n)).

%C In particular, a(n) always exists.

%H Robert Israel, <a href="/A309735/b309735.txt">Table of n, a(n) for n = 1..10000</a>

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

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

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

%p f:= proc(n) local x,y;

%p for x from 2 do

%p y:= x^n;

%p if floor(y/10^ilog10(y)) = 2 then return x fi

%p od

%p end proc:

%p map(f, [$1..100]);

%o (PARI) a(n) = for(k=1, oo, if(digits(k^n)[1]==2, return(k))) \\ _Felix Fröhlich_, Aug 14 2019

%o (Python)

%o n = 1

%o while n < 100:

%o k, s = 2, str(2**n)

%o while s[0] != "2":

%o k = k+1

%o s = str(k**n)

%o print(n,k)

%o n = n+1 # _A.H.M. Smeets_, Aug 14 2019

%o (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

%Y Cf. A000030, A067443, A309707.

%K nonn,base

%O 1,1

%A _Robert Israel_, Aug 14 2019