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Numbers that are congruent to {1, 3} mod 6.
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%I #115 Dec 11 2021 04:32:46

%S 1,3,7,9,13,15,19,21,25,27,31,33,37,39,43,45,49,51,55,57,61,63,67,69,

%T 73,75,79,81,85,87,91,93,97,99,103,105,109,111,115,117,121,123,127,

%U 129,133,135,139,141,145,147,151,153,157,159,163,165,169,171,175,177,181,183

%N Numbers that are congruent to {1, 3} mod 6.

%C Also the numbers k such that 10^p+k could possibly be prime. - _Roderick MacPhee_, Nov 20 2011 This statement can be written as follows. If 10^m + k = prime, for any m >= 1, then k is in this sequence. See the pink box comments by Roderick MacPhee from Dec 09 2014. - _Wolfdieter Lang_, Dec 09 2014

%C The odd-indexed terms are one more than the arithmetic mean of their neighbors; the even-indexed terms are one less than the arithmetic mean of their neighbors. - _Amarnath Murthy_, Jul 29 2003

%C Partial sums are A212959. - _Philippe Deléham_, Mar 16 2014

%C 12*a(n) is conjectured to be the length of the boundary after n iterations of the hexagon and square expansion shown in the link. The squares and hexagons have side length 1 in some units. The pattern is supposed to become the planar Archimedean net 4.6.12 when n -> infinity. - _Kival Ngaokrajang_, Nov 30 2014

%C Positive numbers k for which 1/2 + k/3 + k^2/6 is an integer. - _Bruno Berselli_, Apr 12 2018

%D L. Lovasz, J. Pelikan, K. Vesztergombi, Discrete Mathematics, Springer (2003); 14.4, p. 225.

%H Reinhard Zumkeller, <a href="/A047241/b047241.txt">Table of n, a(n) for n = 1..1000</a>

%H L. Lovász, J. Pelikán and K. Vesztergombi, <a href="https://doi.org/10.1007/b97469">Discrete Mathematics, Elementary and Beyond</a>, Springer (2003); 14.4, p. 225.

%H Kival Ngaokrajang, <a href="/A047241/a047241.pdf">Illustration of initial terms</a>.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).

%F From _Paul Barry_, Sep 04 2003: (Start)

%F O.g.f.: (1 + 2*x + 3*x^2)/((1 + x)*(1 - x)^2) = (1 + 2*x + 3*x^2)/((1 - x)*(1 - x^2)).

%F E.g.f.: (6*x + 1)*exp(x)/2 + exp(-x)/2;

%F a(n) = 3*n - 5/2 - (-1)^n/2. (End)

%F a(n) = 2*floor((n-1)/2) + 2*n - 1. - _Gary Detlefs_, Mar 18 2010

%F a(n) = 6*n - a(n-1) - 8 with n > 1, a(1)=1. - _Vincenzo Librandi_, Aug 05 2010

%F a(n) = 3*n - 2 - ((n+1) mod 2). - _Wesley Ivan Hurt_, Jun 29 2013

%F a(1)=1, a(2)=3, a(3)=7; for n>3, a(n) = a(n-1) + a(n-2) - a(n-3). - _Harvey P. Dale_, Oct 01 2013

%F From _Benedict W. J. Irwin_, Apr 13 2016: (Start)

%F A005408(a(n)+1) = A016813(A001651(n)),

%F A007310(a(n)) = A005408(A087444(n)-1),

%F A007310(A005408(a(n)+1)) = A017533(A001651(n)). (End)

%F Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(4*sqrt(3)) + log(3)/4. - _Amiram Eldar_, Dec 11 2021

%p seq(3*k-2-((k+1) mod 2), k=1..100); # _Wesley Ivan Hurt_, Sep 28 2013

%t Table[{2, 4}, {30}] // Flatten // Prepend[#, 1]& // Accumulate (* _Jean-François Alcover_, Jun 10 2013 *)

%t Select[Range[200], MemberQ[{1, 3}, Mod[#, 6]]&] (* or *) LinearRecurrence[{1, 1, -1}, {1, 3, 7}, 70] (* _Harvey P. Dale_, Oct 01 2013 *)

%o (Haskell)

%o a047241 n = a047241_list !! (n-1)

%o a047241_list = 1 : 3 : map (+ 6) a047241_list

%o -- _Reinhard Zumkeller_, Feb 19 2013

%o (PARI) a(n)=bitor(3*n-3,1) \\ _Charles R Greathouse IV_, Sep 28 2013

%o (Python) for n in range(1,10**5):print(3*n-2-((n+1)%2)) # _Soumil Mandal_, Apr 14 2016

%Y Cf. A047233, A056970, A007310, A047228, A047261, A047273, A212959.

%Y Subsequence of A186422.

%Y Union of A016921 and A016945. - _Wesley Ivan Hurt_, Sep 28 2013

%K nonn,easy

%O 1,2

%A _N. J. A. Sloane_

%E Formula corrected by _Bruno Berselli_, Jun 24 2010