A169598 Numbers that are congruent to {3,18} mod 23.

3, 18, 26, 41, 49, 64, 72, 87, 95, 110, 118, 133, 141, 156, 164, 179, 187, 202, 210, 225, 233, 248, 256, 271, 279, 294, 302, 317, 325, 340, 348, 363, 371, 386, 394, 409, 417, 432, 440, 455, 463, 478, 486, 501, 509, 524, 532, 547, 555, 570, 578, 593, 601, 616, 624

Offset

1,1

Comments

Conjecture: For no n 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) = (46*n + 7*(-1)^n - 27)/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*(3+15*x+5*x^2) / ( (1+x)*(x-1)^2 ). - R. J. Mathar, Jul 07 2015

Maple

A169598:=n->(46*n + 7*(-1)^n - 27)/4: seq(A169598(n), n=1..100); # Wesley Ivan Hurt, Feb 05 2017

Mathematica

Select[Range[624], MemberQ[{3, 18}, Mod[#, 23]]&] (* Ray Chandler, Jul 08 2015 *)
LinearRecurrence[{1, 1, -1}, {3, 18, 26}, 55] (* Ray Chandler, Jul 08 2015 *)
Rest[CoefficientList[Series[x*(3+15*x+5*x^2)/((1+x)*(x-1)^2), {x, 0, 55}], x]] (* Ray Chandler, Jul 08 2015 *)

Crossrefs

Cf. A169599.

Sequence in context: A263578 A048080 A306279 * A202359 A118474 A163242

Adjacent sequences: A169595 A169596 A169597 * A169599 A169600 A169601

Keyword

nonn,easy

Author

Vincenzo Librandi, Dec 03 2009

Extensions

Added leading terms. Clarified comment. - R. J. Mathar, Jul 07 2015

Status

approved