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 A169597 Numbers that are congruent to {2, 15} mod 19. 1
 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 (list; graph; refs; listen; history; text; internal format)
 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, 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

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Last modified September 26 17:49 EDT 2023. Contains 365666 sequences. (Running on oeis4.)