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a(n) = floor(concatenation(1,2,3,...,n-1,n)^(1/n)).
2

%I #12 Nov 18 2018 10:33:35

%S 1,3,4,5,6,7,7,7,7,10,12,14,17,19,21,24,26,28,30,31,33,35,37,38,40,41,

%T 42,44,45,46,47,49,50,51,52,53,53,54,55,56,57,58,58,59,60,60,61,62,62,

%U 63

%N a(n) = floor(concatenation(1,2,3,...,n-1,n)^(1/n)).

%C Does lim_{n->infinity} a(n)/n converge? - _Ryan Propper_, Aug 29 2005

%H Harvey P. Dale, <a href="/A095254/b095254.txt">Table of n, a(n) for n = 1..1000</a>

%e a(4) = floor(1234^(1/4)) = floor(5.926...) = 5.

%t s = ""; Do[s = s <> ToString[n]; k = ToExpression[s]; Print[Floor[k^(1/n)]], {n, 1, 50}] (* _Ryan Propper_, Aug 29 2005 *)

%t Table[Floor[Surd[FromDigits[Flatten[IntegerDigits/@Range[n]]],n]],{n,60}] (* _Harvey P. Dale_, Sep 26 2016 *)

%Y Cf. A068995, A095253.

%K base,nonn

%O 1,2

%A _Amarnath Murthy_, Jun 17 2004

%E More terms from _Ryan Propper_, Aug 29 2005