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%I #23 Jul 08 2015 10:16:59
%S 4,23,33,52,62,81,91,110,120,139,149,168,178,197,207,226,236,255,265,
%T 284,294,313,323,342,352,371,381,400,410,429,439,458,468,487,497,516,
%U 526,545,555,574,584,603,613,632,642,661,671,690,700,719,729,748,758,777,787,806,816
%N Numbers that are congruent to {4, 23} mod 29.
%C Conjecture: For no number n>4 in 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) = 29*k-6 and a(2k+1) = 29*k+4, therefore 6*a(2k)+7 = 29*(6*k-1) and 6*a(2k+1)+5 = 29*(6*k+1). [_Bruno Berselli_, Jan 07 2013, modified Jul 07 2015]
%H Vincenzo Librandi, <a href="/A169599/b169599.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) = (58*n + 9*(-1)^n -33)/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*(4+19*x+6*x^2) / ( (1+x)*(x-1)^2 ). - _R. J. Mathar_, Jul 07 2015
%t Select[Range[816],MemberQ[{4,23},Mod[#,29]]&] (* _Ray Chandler_, Jul 08 2015 *)
%t LinearRecurrence[{1,1,-1},{4,23,33},57] (* _Ray Chandler_, Jul 08 2015 *)
%t Rest[CoefficientList[Series[x*(4+19*x+6*x^2)/((1+x)*(x-1)^2),{x,0,57}],x]] (* _Ray Chandler_, Jul 08 2015 *)
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Dec 03 2009
%E Missing leading terms added. Conjecture clarified. - _R. J. Mathar_, Jul 07 2015
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