A072016 Numbers k such that gcd(k, reverse(k)) = 27 = 3^3, where reverse(x) = A004086(x).

2889, 3699, 3888, 3969, 4779, 4887, 5589, 5697, 5778, 5859, 5886, 5967, 6399, 6669, 6777, 6885, 6939, 7398, 7479, 7587, 7668, 7695, 7749, 7776, 7857, 7884, 7938, 7965, 8289, 8397, 8559, 8667, 8775, 8829, 8883, 8937, 9099, 9288, 9369, 9396, 9477, 9558, 9585

Offset: 1

Author

Labos Elemer, Jun 05 2002

Keywords

• nonn
• base

Comments

Solutions to gcd(k, reverse(k)) = 1,3,9 (lower powers of 3) are trivial (see A072005).

Examples

			2889 = 107*3*3*3, 9889 = 3*3*3*3*2*61.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1030

Crossrefs

Cf. A004086, A055483, A069554, A071686, A072005.

Programs

Magma

[n: n in [1..10^4] | Gcd(n,Seqint(Reverse(Intseq(n)))) eq 27]; // Vincenzo Librandi, Jul 11 2018

Mathematica

Select[Range[10^4], GCD[#, FromDigits[Reverse[IntegerDigits[#]]]] == 27 &] (* Vincenzo Librandi, Jul 11 2018 *)

PARI

isok(n) = gcd(n, fromdigits(Vecrev(digits(n)))) == 27; \\ Michel Marcus, Jul 11 2018