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Numbers that are congruent to {2,13} mod 17.
1

%I #27 Jul 08 2015 09:59:36

%S 2,13,19,30,36,47,53,64,70,81,87,98,104,115,121,132,138,149,155,166,

%T 172,183,189,200,206,217,223,234,240,251,257,268,274,285,291,302,308,

%U 319,325,336,342,353,359,370,376,387,393,404,410,421,427,438,444,455,461,472,478,489

%N Numbers that are congruent to {2,13} mod 17.

%C Conjecture: For no term n>2 in the sequence 36*n^2+72*n+35 is equal to p*(p+2), where p, p+2 are twin primes.

%C The conjecture is evident, it can be proved as in A169599. [_Bruno Berselli_, Jan 07 2013]

%H Vincenzo Librandi, <a href="/A168672/b168672.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) = a(n-1) +a(n-2) -a(n-3). - _Vincenzo Librandi_, Jul 11 2012

%F a(n) = (34*n +5*(-1)^n -21)/4. - _Vincenzo Librandi_, Jan 06 2013, modified Jul 07 2015

%F G.f.: x*(2+11*x+4*x^2) / ( (1+x)*(x-1)^2 ). - _R. J. Mathar_, Jul 07 2015

%t Select[Range[489],MemberQ[{2,13},Mod[#,17]]&] (* _Ray Chandler_, Jul 08 2015 *)

%t LinearRecurrence[{1,1,-1},{2,13,19},58] (* _Ray Chandler_, Jul 08 2015 *)

%t Rest[CoefficientList[Series[x*(2+11*x+4*x^2)/((1+x)*(x-1)^2),{x,0,58}],x]] (* _Ray Chandler_, Jul 08 2015 *)

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Dec 02 2009

%E 5 leading terms added. Conjecture clarified. - _R. J. Mathar_, Jul 07 2015