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A309735 a(n) is the least positive integer k such that k^n starts with 2.

%I A309735 #29 Sep 08 2022 08:46:22
%S A309735 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 A309735 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 A309735 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 A309735 a(n) is the least positive integer k such that k^n starts with 2.
%C A309735 For n > 1, take integer d > -n log_10(3^(1/n)-2^(1/n)).
%C A309735 Then 10^(d/n) > 1/(3^(1/n) - 2^(1/n))
%C A309735 so (3*10^d)^(1/n) - (2*10^d)^(1/n) > 1
%C A309735 and therefore a(n) <= ceiling((2*10^d)^(1/n)).
%C A309735 In particular, a(n) always exists.
%H A309735 Robert Israel, <a href="/A309735/b309735.txt">Table of n, a(n) for n = 1..10000</a>
%F A309735 a(n) = A067443(n)^(1/n).
%F A309735 A000030(a(n)^n)=2.
%e A309735 a(5) = 3 because 3^5 = 243 starts with 2, while 1^5=1 and 2^5=32 do not start with 2.
%p A309735 f:= proc(n) local x,y;
%p A309735   for x from 2  do
%p A309735     y:= x^n;
%p A309735       if floor(y/10^ilog10(y)) = 2 then return x fi
%p A309735   od
%p A309735 end proc:
%p A309735 map(f, [$1..100]);
%o A309735 (PARI) a(n) = for(k=1, oo, if(digits(k^n)[1]==2, return(k))) \\ _Felix Fröhlich_, Aug 14 2019
%o A309735 (Python)
%o A309735 n = 1
%o A309735 while n < 100:
%o A309735     k, s = 2, str(2**n)
%o A309735     while s[0] != "2":
%o A309735         k = k+1
%o A309735         s = str(k**n)
%o A309735     print(n,k)
%o A309735     n = n+1 # _A.H.M. Smeets_, Aug 14 2019
%o A309735 (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 A309735 Cf. A000030, A067443, A309707.
%K A309735 nonn,base
%O A309735 1,1
%A A309735 _Robert Israel_, Aug 14 2019