A169599 Numbers that are congruent to {4, 23} mod 29.

4, 23, 33, 52, 62, 81, 91, 110, 120, 139, 149, 168, 178, 197, 207, 226, 236, 255, 265, 284, 294, 313, 323, 342, 352, 371, 381, 400, 410, 429, 439, 458, 468, 487, 497, 516, 526, 545, 555, 574, 584, 603, 613, 632, 642, 661, 671, 690, 700, 719, 729, 748, 758, 777, 787, 806, 816

Offset

1,1

Comments

Conjecture: For no number 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.

This conjecture is evident: in fact, it is sufficient to observe that a(2k) = 29*k-6 and a(2k+1) = 29*k+4, therefore 6*a(2k)+7 = 29*(6*k-1) and 6*a(2k+1)+5 = 29*(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) = (58*n + 9*(-1)^n -33)/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+19*x+6*x^2) / ( (1+x)*(x-1)^2 ). - R. J. Mathar, Jul 07 2015

Mathematica

Select[Range[816], MemberQ[{4, 23}, Mod[#, 29]]&] (* Ray Chandler, Jul 08 2015 *)
LinearRecurrence[{1, 1, -1}, {4, 23, 33}, 57] (* Ray Chandler, Jul 08 2015 *)
Rest[CoefficientList[Series[x*(4+19*x+6*x^2)/((1+x)*(x-1)^2), {x, 0, 57}], x]] (* Ray Chandler, Jul 08 2015 *)

Crossrefs

Sequence in context: A160613 A131545 A030716 * A106684 A239624 A179628

Adjacent sequences: A169596 A169597 A169598 * A169600 A169601 A169602

Keyword

nonn,easy

Author

Vincenzo Librandi, Dec 03 2009

Extensions

Missing leading terms added. Conjecture clarified. - R. J. Mathar, Jul 07 2015

Status

approved