%I
%S 1,10,14,23,27,36,40,49,53,62,66,75,79,88,92,101,105,114,118,127,131,
%T 140,144,153,157,166,170,179,183,192,196,205,209,218,222,231,235,244,
%U 248,257,261,270,274,283,287,296,300,309,313,322,326,335,339,348,352,361,365,374
%N Numbers that are congruent to {1, 10} mod 13.
%C Conjecture: For no n>1 in the sequence 36*n^2+72*n+35 is equal to p*(p+2), where p, p+2 are twin primes.
%C The conjecture is evident, it can be proved as in A169599. [_Bruno Berselli_, Jan 07 2013]
%H Vincenzo Librandi, <a href="/A168671/b168671.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,1).
%F From _Vincenzo Librandi_, Jul 11 2012, modified Jul 07 2015: (Start)
%F G.f.: x*(1+9*x+3*x^2)/((1+x)*(1x)^2).
%F a(n) = (26*n+5*(1)^n17)/4.
%F a(n) = a(n2) +13 = a(n1) +a(n2) a(n3). (End)
%t Select[Range[374],MemberQ[{1,10},Mod[#,13]]&] (* _Ray Chandler_, Jul 08 2015 *)
%t LinearRecurrence[{1,1,1},{1,10,14},58] (* _Ray Chandler_, Jul 08 2015 *)
%t Rest[CoefficientList[Series[x*(1+9*x+3*x^2)/((1+x)*(1x)^2),{x,0,58}],x]] (* _Ray Chandler_, Jul 08 2015 *)
%K nonn,easy
%O 1,2
%A _Vincenzo Librandi_, Dec 02 2009
%E 4 leading terms added. Conjecture clarified.  _R. J. Mathar_, Jul 07 2015
