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a(1) = 1; a(n) = smallest number such that the concatenation a(1)a(2)...a(n) is an n-th power.
8

%I #15 Oct 06 2020 08:18:48

%S 1,6,6375,34623551127976881,

%T 18860302374385155610185422853070042488899966126368559233360607121925651097253827765970857

%N a(1) = 1; a(n) = smallest number such that the concatenation a(1)a(2)...a(n) is an n-th power.

%C Is this sequence infinite? - _Charles R Greathouse IV_, Sep 19 2012

%C From _Robert Israel_, Oct 05 2020:

%C If 10^m > ((x+1)^(1/n)-(x+1/10)^(1/n))^(-n), where x is the concatenation a(1)...a(n-1), then a(n) < 10^m.

%C In particular, the sequence is infinite.

%C a(6) has 558 digits, a(7) has 4014 digits, and a(8) has 32783 digits. (End)

%D Amarnath Murthy, Exploring some new ideas on Smarandache type sets, functions and sequences, Smarandache Notions Journal Vol. 11, No. 1-2-3, Spring 2000.

%e a(1) = 1, a(1)a(2) = 16 = 4^2, a(1)a(2)a(3) = 166375 = 55^3, a(1)a(2)a(3)a(4) = 16637534623551127976881 = 359147^4.

%p ncat:= (a,b) -> a*10^(1+ilog10(b))+b:

%p f:= proc(n,x)

%p local z,d;

%p for d from 1 do

%p z:= ceil(((x+1/10)*10^d)^(1/n));

%p if z^n < (x+1)*10^d then return z^n - x*10^d fi

%p od

%p end proc:

%p R[1]:= 1: C:= 1:

%p for n from 2 to 6 do

%p R[n]:= f(n,C);

%p C:= ncat(C, R[n]);

%p od:

%p seq(R[i],i=1..6); # _Robert Israel_, Oct 05 2020

%K base,nonn

%O 1,2

%A _Amarnath Murthy_, Apr 20 2001

%E Corrected and extended by Ulrich Schimke, Feb 08 2002

%E Offset corrected by _Robert Israel_, Oct 05 2020