A169597 Numbers that are congruent to {2, 15} mod 19.

2, 15, 21, 34, 40, 53, 59, 72, 78, 91, 97, 110, 116, 129, 135, 148, 154, 167, 173, 186, 192, 205, 211, 224, 230, 243, 249, 262, 268, 281, 287, 300, 306, 319, 325, 338, 344, 357, 363, 376, 382, 395, 401, 414, 420, 433, 439, 452, 458, 471, 477, 490, 496, 509, 515, 528

Offset

1,1

Comments

Conjecture: For no term n>2 of 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) = 19k-4 and a(2k+1) = 19*k+2, therefore 6*a(2k)+5 = 19*(6*k-1) and 6*a(2k+1)+7 = 19*(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) = (38*n + 7*(-1)^n -23)/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*(2+13*x+4*x^2) / ( (1+x)*(x-1)^2 ). - R. J. Mathar, Jul 07 2015

Maple

seq(seq(19*i+j, j=[2, 15]), i=0..100); # Robert Israel, Jul 07 2015

Mathematica

Select[Range[528], MemberQ[{2, 15}, Mod[#, 19]]&] (* Ray Chandler, Jul 08 2015 *)
LinearRecurrence[{1, 1, -1}, {2, 15, 21}, 56] (* Ray Chandler, Jul 08 2015 *)
Rest[CoefficientList[Series[x*(2+13*x+4*x^2)/((1+x)*(x-1)^2), {x, 0, 56}], x]] (* Ray Chandler, Jul 08 2015 *)

Crossrefs

Sequence in context: A198391 A278889 A075722 * A280288 A153712 A153711

Adjacent sequences: A169594 A169595 A169596 * A169598 A169599 A169600

Keyword

nonn,easy

Author

Vincenzo Librandi, Dec 03 2009

Extensions

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

Status

approved