%I #28 Aug 21 2015 10:59:33
%S 1,8,12,19,23,30,34,41,45,52,56,63,67,74,78,85,89,96,100,107,111,118,
%T 122,129,133,140,144,151,155,162,166,173,177,184,188,195,199,206,210,
%U 217,221,228,232,239,243,250,254,261,265,272,276,283,287,294,298,305,309,316
%N Numbers that are congruent to {1, 8} mod 11.
%C Conjecture: For no term 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 This conjecture is evident: in fact, it is sufficient to observe that a(2k) = 11*k-3 and a(2k+1) = 11*k+1, therefore 6*a(2k)+7 = 11*(6*k-1) and 6*a(2k+1)+5 = 11*(6*k+1). [_Bruno Berselli_, Jan 07 2013, modified Jul 07 2015]
%H Vincenzo Librandi, <a href="/A168670/b168670.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 a(n) = (-15+3*(-1)^n+22*n)/4. a(n) = a(n-1)+a(n-2)-a(n-3). G.f.: x*(1+7*x+3*x^2)/((1-x)^2*(1+x)). [_Colin Barker_, May 15 2012, modified Jul 07 2015]
%t Select[Range[316],MemberQ[{1,8},Mod[#,11]]&] (* _Ray Chandler_, Jul 07 2015 *)
%t LinearRecurrence[{1,1,-1},{1,8,12},58] (* _Ray Chandler_, Jul 07 2015 *)
%t Rest[CoefficientList[Series[x*(1+7*x+3*x^2)/((1-x)^2*(1+x)),{x,0,58}],x]] (* _Ray Chandler_, Jul 07 2015 *)
%K nonn,easy
%O 1,2
%A _Vincenzo Librandi_, Dec 02 2009
%E 3 leading terms added. Conjecture clarified. - _R. J. Mathar_, Jul 07 2015