A169598 - OEIS

%I #31 Feb 05 2017 14:18:56

%S 3,18,26,41,49,64,72,87,95,110,118,133,141,156,164,179,187,202,210,

%T 225,233,248,256,271,279,294,302,317,325,340,348,363,371,386,394,409,

%U 417,432,440,455,463,478,486,501,509,524,532,547,555,570,578,593,601,616,624

%N Numbers that are congruent to {3,18} mod 23.

%C Conjecture: For no n 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="/A169598/b169598.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) = (46*n + 7*(-1)^n - 27)/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*(3+15*x+5*x^2) / ( (1+x)*(x-1)^2 ). - _R. J. Mathar_, Jul 07 2015

%p A169598:=n->(46*n + 7*(-1)^n - 27)/4: seq(A169598(n), n=1..100); # _Wesley Ivan Hurt_, Feb 05 2017

%t Select[Range[624],MemberQ[{3,18},Mod[#,23]]&] (* _Ray Chandler_, Jul 08 2015 *)

%t LinearRecurrence[{1,1,-1},{3,18,26},55] (* _Ray Chandler_, Jul 08 2015 *)

%t Rest[CoefficientList[Series[x*(3+15*x+5*x^2)/((1+x)*(x-1)^2),{x,0,55}],x]] (* _Ray Chandler_, Jul 08 2015 *)

%Y Cf. A169599.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Dec 03 2009

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