OFFSET
1,1
COMMENTS
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.
The conjecture is evident, it can be proved as in A169599. [Bruno Berselli, Jan 07 2013]
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) = (62*n + 11*(-1)^n - 35)/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*(4+21*x+6*x^2) / ( (1+x)*(x-1)^2 ). - R. J. Mathar, Jul 07 2015
MATHEMATICA
Select[Range[934], MemberQ[{4, 25}, Mod[#, 31]]&] (* Ray Chandler, Jul 08 2015 *)
LinearRecurrence[{1, 1, -1}, {4, 25, 35}, 61] (* Ray Chandler, Jul 08 2015 *)
Rest[CoefficientList[Series[x*(4+21*x+6*x^2)/((1+x)*(x-1)^2), {x, 0, 61}], 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