%I #27 Jul 08 2015 10:20:10
%S 4,25,35,56,66,87,97,118,128,149,159,180,190,211,221,242,252,273,283,
%T 304,314,335,345,366,376,397,407,428,438,459,469,490,500,521,531,552,
%U 562,583,593,614,624,645,655,676,686,707,717,738,748,769,779,800,810,831,841,862,872,893,903,924,934
%N Numbers that are congruent to {4, 25} mod 31.
%C Conjecture: For no 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 The conjecture is evident, it can be proved as in A169599. [_Bruno Berselli_, Jan 07 2013]
%H Vincenzo Librandi, <a href="/A169600/b169600.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) = (62*n + 11*(-1)^n - 35)/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+21*x+6*x^2) / ( (1+x)*(x-1)^2 ). - _R. J. Mathar_, Jul 07 2015
%t Select[Range[934],MemberQ[{4,25},Mod[#,31]]&] (* _Ray Chandler_, Jul 08 2015 *)
%t LinearRecurrence[{1,1,-1},{4,25,35},61] (* _Ray Chandler_, Jul 08 2015 *)
%t Rest[CoefficientList[Series[x*(4+21*x+6*x^2)/((1+x)*(x-1)^2),{x,0,61}],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
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