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%I #26 Jan 02 2023 12:30:46
%S 1,2,3,6,5,10,7,14,9,18,11,22,13,26,15,30,17,34,19,38,21,42,23,46,25,
%T 50,27,54,29,58,31,62,33,66,35,70,37,74,39,78,41,82,43,86,45,90,47,94,
%U 49,98,51,102,53,106,55,110,57,114,59,118,61,122,63,126,65,130,67,134
%N Odd numbers interspersed with double the previous odd number.
%C As pointed out by E. Angelini on the SeqFan list (cf. link), this is the lexicographically earliest sequence of positive integers without repetitions such that the sum of four consecutive terms is always a multiple of 4. - _M. F. Hasler_, Mar 22 2013
%H E. Angelini, <a href="http://list.seqfan.eu/oldermail/seqfan/2013-March/010956.html">k-chunks sum and division by k</a>, post to the SeqFan list, Mar 22 2013
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,2,0,-1).
%F a(n) = (2 - n) * (n - floor(n/2) * 2) + 2 * (n - 1).
%F G.f.: x*(1+2*x)*(1+x^2)/(1-x^2)^2. - _Ralf Stephan_, Jun 10 2003
%F a(2n-1) = 2n-1, a(2n) = 4n-2. - _M. F. Hasler_, Mar 22 2013
%F From _Wesley Ivan Hurt_, Nov 22 2015: (Start)
%F a(n) = 2*a(n-2) - a(n-4) for n>4.
%F a(n) = n*(-1)^n/2 - (-1)^n + 3*n/2 - 1. (End)
%e a(1)=1 because n is odd. a(2)=2 because a(1)*2=2.
%p A043547:=n->n*(-1)^n/2-(-1)^n+3*n/2-1: seq(A043547(n), n=1..100); # _Wesley Ivan Hurt_, Nov 22 2015
%t Flatten[Table[Accumulate[{2 n - 1, 2 n - 1}], {n, 40}]] (* _Wesley Ivan Hurt_, Nov 22 2015 *)
%t With[{o=Range[1,71,2]},Riffle[o,2o]] (* _Harvey P. Dale_, Sep 17 2019 *)
%o (PARI) A043547(n)=n+!bittest(n,0)*(n-2) \\ _M. F. Hasler_, Mar 22 2013
%o (Magma) [n*(-1)^n/2-(-1)^n+3*n/2-1 : n in [1..50]]; // _Wesley Ivan Hurt_, Nov 22 2015
%o (PARI) Vec(x*(1+2*x)*(1+x^2)/(1-x^2)^2 + O(x^100)) \\ _Altug Alkan_, Nov 22 2015
%Y For n>3, a(n) = A059029(n-3)+3. For n>1, a(n) = A022998(n-2)+2.
%K easy,nonn
%O 1,2
%A Jim Cook (jcook(AT)halcyon.com), Mar 01 2000
%E More terms from _James A. Sellers_, Mar 01 2000
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