%I #27 Jul 08 2015 10:11:12

%S 2,15,21,34,40,53,59,72,78,91,97,110,116,129,135,148,154,167,173,186,

%T 192,205,211,224,230,243,249,262,268,281,287,300,306,319,325,338,344,

%U 357,363,376,382,395,401,414,420,433,439,452,458,471,477,490,496,509,515,528

%N Numbers that are congruent to {2, 15} mod 19.

%C Conjecture: For no term n>2 of the sequence 36*n^2+72*n+35 = (6*n+5)*(6*n+7) is of the form p*(p+2), where p and p+2 are primes.

%C This conjecture is evident: in fact, it is sufficient to observe that a(2k) = 19k-4 and a(2k+1) = 19*k+2, therefore 6*a(2k)+5 = 19*(6*k-1) and 6*a(2k+1)+7 = 19*(6*k+1). [_Bruno Berselli_, Jan 07 2013, modified Jul 07 2015]

%H Vincenzo Librandi, <a href="/A169597/b169597.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) = (38*n + 7*(-1)^n -23)/4. - _Vincenzo Librandi_, Jan 06 2013, modified Jul 07 2015

%F a(n) = a(n-1) + a(n-2) - a(n-3). - _Vincenzo Librandi_, Jan 06 2013

%F G.f.: x*(2+13*x+4*x^2) / ( (1+x)*(x-1)^2 ). - _R. J. Mathar_, Jul 07 2015

%p seq(seq(19*i+j,j=[2,15]),i=0..100); # _Robert Israel_, Jul 07 2015

%t Select[Range[528],MemberQ[{2,15},Mod[#,19]]&] (* _Ray Chandler_, Jul 08 2015 *)

%t LinearRecurrence[{1,1,-1},{2,15,21},56] (* _Ray Chandler_, Jul 08 2015 *)

%t Rest[CoefficientList[Series[x*(2+13*x+4*x^2)/((1+x)*(x-1)^2),{x,0,56}],x]] (* _Ray Chandler_, Jul 08 2015 *)

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Dec 03 2009

%E Added missing leading terms. Clarified the comment. - _R. J. Mathar_, Jul 07 2015