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%I #15 Sep 08 2022 08:45:06
%S 2889,3699,3888,3969,4779,4887,5589,5697,5778,5859,5886,5967,6399,
%T 6669,6777,6885,6939,7398,7479,7587,7668,7695,7749,7776,7857,7884,
%U 7938,7965,8289,8397,8559,8667,8775,8829,8883,8937,9099,9288,9369,9396,9477,9558,9585
%N Numbers k such that gcd(k, reverse(k)) = 27 = 3^3, where reverse(x) = A004086(x).
%C Solutions to gcd(k, reverse(k)) = 1,3,9 (lower powers of 3) are trivial (see A072005).
%H Vincenzo Librandi, <a href="/A072016/b072016.txt">Table of n, a(n) for n = 1..1030</a>
%e 2889 = 107*3*3*3, 9889 = 3*3*3*3*2*61.
%t Select[Range[10^4], GCD[#, FromDigits[Reverse[IntegerDigits[#]]]] == 27 &] (* _Vincenzo Librandi_, Jul 11 2018 *)
%o (PARI) isok(n) = gcd(n, fromdigits(Vecrev(digits(n)))) == 27; \\ _Michel Marcus_, Jul 11 2018
%o (Magma) [n: n in [1..10^4] | Gcd(n,Seqint(Reverse(Intseq(n)))) eq 27]; // _Vincenzo Librandi_, Jul 11 2018
%Y Cf. A004086, A055483, A069554, A071686, A072005.
%K nonn,base
%O 1,1
%A _Labos Elemer_, Jun 05 2002