A171141 Numbers that are congruent to {6,33} mod 41.

6, 33, 47, 74, 88, 115, 129, 156, 170, 197, 211, 238, 252, 279, 293, 320, 334, 361, 375, 402, 416, 443, 457, 484, 498, 525, 539, 566, 580, 607, 621, 648, 662, 689, 703, 730, 744, 771, 785, 812, 826, 853, 867, 894, 908, 935, 949, 976, 990, 1017, 1031, 1058

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

Sequence in context: A222749 A132548 A140521 * A069065 A073343 A157872

Adjacent sequences: A171138 A171139 A171140 * A171142 A171143 A171144

Keyword

nonn,easy

Author

Vincenzo Librandi, Dec 04 2009

Status

approved