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%I #7 Nov 21 2013 12:50:00
%S 21,27,29,41,101,119,141,171,173,177,191,197,219,243,267,291,309,327,
%T 333,369,371,383,411,417,1019,1049,1059,1091,1157,1163,1211,1311,1337,
%U 1343,1359,1371,1379,1409,1461,1473,1481,1503,1521,1593,1599,1613,1637
%N Natural numbers n such that the concatenation 1331//n^3 is a prime number.
%C Given the cube n^3 with k = A111393(n) decimal digits, we have to check whether the concatenation, 11^3 * 10^k + n^3, is a prime.
%C The number k of digits that 1331=11^3 is shifted is not a multiple of 3,
%C because the form a^3+b^3 = (a^2+a*b+b^2) * (a - b) cannot construct a prime.
%D K. Haase, P. Mauksch: Spass mit Mathe, Urania-Verlag Leipzig, Verlag Dausien Hanau, 2. Auflage 1985
%H Harvey P. Dale, <a href="/A173836/b173836.txt">Table of n, a(n) for n = 1..1000</a>
%e 21 is in the sequence because 21^3=9261, and the concatenation is 13319261=prime(868687).
%e 27 is in the sequence because 27^3=19683, and the concatenation is 133119683=prime(7545064).
%t Select[Range[2000],PrimeQ[FromDigits[Join[{1,3,3,1}, IntegerDigits[ #^3]]]]&] (* _Harvey P. Dale_, Oct 14 2011 *)
%Y Cf. A102006, A167535, A168147, A168219, A168274, A173579, A173733
%K base,nonn
%O 1,1
%A Eva-Maria Zschorn (e-m.zschorn(AT)zaschendorf.km3.de), Feb 26 2010
%E Comments sligthly rephrased - _R. J. Mathar_, Mar 05 2010
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