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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 (list; graph; refs; listen; history; text; internal format)
 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 Robert Israel, Table of n, a(n) for n = 1..10000 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)==2, return(k))) \\ Felix Fröhlich, Aug 14 2019 (Python) n = 1 while n < 100:     k, s = 2, str(2**n)     while s != "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)) ne 2 do k:=k+1; end while; sol[m]:=k; m:=m+1; end for; sol; // Marius A. Burtea, Aug 15 2019 CROSSREFS Cf. A000030, A067443, A309707. Sequence in context: A077057 A030660 A253720 * A275726 A146096 A011310 Adjacent sequences:  A309732 A309733 A309734 * A309736 A309737 A309738 KEYWORD nonn,base AUTHOR Robert Israel, Aug 14 2019 STATUS approved

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Last modified July 11 20:03 EDT 2020. Contains 335652 sequences. (Running on oeis4.)