A173836 Natural numbers n such that the concatenation 1331//n^3 is a prime number.

21, 27, 29, 41, 101, 119, 141, 171, 173, 177, 191, 197, 219, 243, 267, 291, 309, 327, 333, 369, 371, 383, 411, 417, 1019, 1049, 1059, 1091, 1157, 1163, 1211, 1311, 1337, 1343, 1359, 1371, 1379, 1409, 1461, 1473, 1481, 1503, 1521, 1593, 1599, 1613, 1637

Offset

1,1

Comments

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.

The number k of digits that 1331=11^3 is shifted is not a multiple of 3,

because the form a^3+b^3 = (a^2+a*b+b^2) * (a - b) cannot construct a prime.

References

K. Haase, P. Mauksch: Spass mit Mathe, Urania-Verlag Leipzig, Verlag Dausien Hanau, 2. Auflage 1985

Links

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Example

21 is in the sequence because 21^3=9261, and the concatenation is 13319261=prime(868687).
27 is in the sequence because 27^3=19683, and the concatenation is 133119683=prime(7545064).

Mathematica

Select[Range[2000], PrimeQ[FromDigits[Join[{1, 3, 3, 1}, IntegerDigits[ #^3]]]]&] (* Harvey P. Dale, Oct 14 2011 *)

Crossrefs

Cf. A102006, A167535, A168147, A168219, A168274, A173579, A173733

Sequence in context: A344978 A299594 A318544 * A064507 A249729 A280107

Adjacent sequences: A173833 A173834 A173835 * A173837 A173838 A173839

Keyword

base,nonn

Author

Eva-Maria Zschorn (e-m.zschorn(AT)zaschendorf.km3.de), Feb 26 2010

Extensions

Comments sligthly rephrased - R. J. Mathar, Mar 05 2010

Status

approved