OFFSET
1,1
COMMENTS
Conjecture: Numbers n>6 such that 36*n^2+72*n+35 = (6*n+5)*(6*n+7) is not 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) = 41*k-8 and a(2k+1) = 41*k+6, therefore 6*a(2k)+7 = 41*(6*k-1) and 6*a(2k+1)+5 = 41*(6*k+1). [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
G.f.: x*(6 + 27*x + 8*x^2)/((1 + x)*(1 - x)^2). - Vincenzo Librandi, Jan 05 2013
a(n) = (82*n + 13*(-1)^n - 45)/4. - Vincenzo Librandi, Jan 05 2013
a(n) = a(n-1) + a(n-2) - a(n-3). - Vincenzo Librandi, Jan 05 2013
MATHEMATICA
CoefficientList[Series[(6 + 27*x + 8*x^2)/((1 + x)*(1 - x)^2), {x, 0, 60}], x] (* Vincenzo Librandi, Jan 05 2013 *)
(* By definition: *) Flatten[#+{6, 33}&/@(41*Range[0, 26])] (* Bruno Berselli, Jan 05 2013 *)
LinearRecurrence[{1, 1, -1}, {6, 33, 47}, 60] (* Harvey P. Dale, Aug 05 2023 *)
PROG
(Magma) [(82*n+13*(-1)^n-45)/4: n in [1..60]]; // Vincenzo Librandi, Jan 05 2013
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Dec 04 2009
STATUS
approved