%I #10 Feb 22 2020 02:32:20
%S 0,827,0,2744337540964913,49135832685980009261,0,
%T 16194277163870641658137516777216169745931717351217373979,
%U 64348566539203664467267512696859000696787170778887189057,0
%N Smallest prime formed by concatenation of n consecutive cubes, 0 if no such prime exists.
%C From _Robert Israel_, Feb 20 2020: (Start)
%C a(n)=0 if n is divisible by 3, as the sum of three consecutive cubes is divisible by 3.
%C a(22) has 2132 digits, too large for a b-file: it is the concatenation of 99999999999999999999999999999998^3 to 100000000000000000000000000000019^3. (End)
%H Robert Israel, <a href="/A104375/b104375.txt">Table of n, a(n) for n = 1..21</a>
%e a(2)=827 because 827 is the smallest prime formed from concatenation of 2 consecutive cubes i.e. 8 and 27.
%p ccat:= proc(L)
%p local t,i;
%p t:= L[1];
%p for i from 2 to nops(L) do
%p t:= t*10^(ilog10(L[i])+1)+L[i]
%p od;
%p t
%p end proc:
%p f:= proc(n) local k,p;
%p if n mod 3 = 0 then return 0 fi;
%p for k from floor(n/2)*2+1 by 2 do
%p p:= ccat([seq((k-i)^3,i=n-1..0,-1)]);
%p if isprime(p) then return p fi
%p od
%p end proc:
%p f(1):= 0:
%p map(f, [$1..10]); # _Robert Israel_, Feb 20 2020
%K base,nonn
%O 1,2
%A _Shyam Sunder Gupta_, Apr 17 2005