OFFSET
1,1
COMMENTS
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.
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]
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).
FORMULA
a(n) = (38*n + 7*(-1)^n -23)/4. - Vincenzo Librandi, Jan 06 2013, modified Jul 07 2015
a(n) = a(n-1) + a(n-2) - a(n-3). - Vincenzo Librandi, Jan 06 2013
G.f.: x*(2+13*x+4*x^2) / ( (1+x)*(x-1)^2 ). - R. J. Mathar, Jul 07 2015
MAPLE
seq(seq(19*i+j, j=[2, 15]), i=0..100); # Robert Israel, Jul 07 2015
MATHEMATICA
Select[Range[528], MemberQ[{2, 15}, Mod[#, 19]]&] (* Ray Chandler, Jul 08 2015 *)
LinearRecurrence[{1, 1, -1}, {2, 15, 21}, 56] (* Ray Chandler, Jul 08 2015 *)
Rest[CoefficientList[Series[x*(2+13*x+4*x^2)/((1+x)*(x-1)^2), {x, 0, 56}], x]] (* Ray Chandler, Jul 08 2015 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Dec 03 2009
EXTENSIONS
Added missing leading terms. Clarified the comment. - R. J. Mathar, Jul 07 2015
STATUS
approved