%I #19 Sep 08 2022 08:45:51

%S 1,0,3,1,5,2,7,3,9,4,11,5,13,6,15,7,17,8,19,9,21,10,23,11,25,12,27,13,

%T 29,14,31,15,33,16,35,17,37,18,39,19,41,20,43,21,45,22,47,23,49,24,51,

%U 25,53,26,55,27,57,28,59,29,61,30,63,31,65,32,67,33,69,34,71,35,73,36,75,37,77,38,79,39,81

%N a(n) = (3*n + 1 + (-1)^n*(n+3))/4.

%C Obtained from A026741 by swapping pairs of consecutive entries.

%C The main diagonal of an array with this sequence in the top row and further rows defined by the first differences of their previous row is essentially 1 followed by 3*A045623(.):

%C 1, 0, 3, 1, 5, 2, 7, 3, 9, 4, 11, 5, 13, 6, 15, 7, 17, 8, ...

%C -1, 3, -2, 4, -3, 5, -4, 6, -5, 7, -6, 8, -7, 9, -8, 10, -9, ...

%C 4, -5, 6, -7, 8, -9, 10, -11, 12, -13, 14, -15, 16, -17, ...

%C -9, 11, -13, 15, -17, 19, -21, 23, -25, 27, -29, 31, ...

%C 20, -24, 28, -32, 36, -40, 44, -48, 52, -56, 60, -64, ...

%C -44, 52, -60, 68, -76, 84, -92, 100, -108, 116, -124, 132, ...

%C 96, -112, 128, -144, 160, -176, 192, -208, 224, -240, ...

%C Also, numerator of (Nimsum n+1)/2 = A004442(n)/2. - _Wesley Ivan Hurt_, Mar 21 2015

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,2,0,-1).

%F a(2n) = 2n+1; a(2n+1) = n.

%F a(n) = 2*a(n-2) - a(n-4).

%F a(2n+1) - 2*a(2n) = -A016789(n+1).

%F a(2n+2) - 2*a(2n+1) = 3.

%F G.f.: ( 1+x^2+x^3 ) / ( (x-1)^2*(1+x)^2 ). - _R. J. Mathar_, Feb 07 2011

%p A174239:=n->(3*n+1+(-1)^n*(n+3))/4: seq(A174239(n), n=0..100); # _Wesley Ivan Hurt_, Mar 21 2015

%t Table[(3 n + 1 + (-1)^n*(n + 3))/4, {n, 0, 100}] (* _Wesley Ivan Hurt_, Mar 21 2015 *)

%t LinearRecurrence[{0,2,0,-1},{1,0,3,1},90] (* _Harvey P. Dale_, Jul 16 2018 *)

%o (Magma) [(3*n+1 +(-1)^n*(n+3))/4: n in [0..80]]; // _Vincenzo Librandi_, Feb 08 2011

%Y Cf. A004442.

%K nonn,easy

%O 0,3

%A _Paul Curtz_, Mar 13 2010