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A169599 Numbers that are congruent to {4, 23} mod 29. 6
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 (list; graph; refs; listen; history; text; internal format)
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

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Last modified May 28 15:04 EDT 2022. Contains 354115 sequences. (Running on oeis4.)