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Two consecutive odd numbers separated by multiples of four, repeated twice, between them, written in increasing order.
5

%I #44 Aug 29 2024 23:33:31

%S 1,3,4,4,5,7,8,8,9,11,12,12,13,15,16,16,17,19,20,20,21,23,24,24,25,27,

%T 28,28,29,31,32,32,33,35,36,36,37,39,40,40,41,43,44,44,45,47,48,48,49,

%U 51,52,52,53,55,56,56,57,59,60,60,61,63,64,64,65,67,68,68,69,71,72,72

%N Two consecutive odd numbers separated by multiples of four, repeated twice, between them, written in increasing order.

%C Essentially partial sums of A007877.

%C a(n) is the number of odd coefficients of the q-binomial coefficient [n+2 choose 2]. (Easy to prove.) - _Richard Stanley_, Oct 12 2016

%H G. C. Greubel, <a href="/A098181/b098181.txt">Table of n, a(n) for n = 0..10000</a>

%H P. Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Barry4/barry142.html">On a Generalization of the Narayana Triangle</a>, J. Int. Seq. 14 (2011) # 11.4.5.

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

%F G.f.: (1+x)/((1-x)^2*(1+x^2)).

%F a(n) = ( (2*n+3) - cos(Pi*n/2) + sin(Pi*n/2) )/2.

%F a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - a(n-4).

%F a(n) = floor(C(n+3, 2)/2)-floor(C(n+1, 2)/2). - _Paul Barry_, Jan 01 2005

%F a(4*n) = 4*n+1, a(4*n+1) = 4*n+3, a(4*n+2) = a(4*n+3) = 4*n+4. - _Philippe Deléham_, Apr 06 2007

%F Euler transform of length 4 sequence [ 3, -2, 0, 1]. - _Michael Somos_, Sep 11 2014

%F a(-3-n) = -a(n) for all n in Z. - _Michael Somos_, Sep 11 2014

%F a(n) = log_2(|A174882(n+2)|). [Barry] - _R. J. Mathar_, Aug 18 2017

%F a(n) = (2*n+3 - (-1)^ceiling(n/2))/2. - _Wesley Ivan Hurt_, Sep 29 2017

%e G.f. = 1 + 3*x + 4*x^2 + 4*x^3 + 5*x^4 + 7*x^5 + 8*x^6 + 8*x^7 + 9*x^8 + ...

%p A:=seq((2*n+3 - cos(Pi*n/2) + sin(Pi*n/2))/2, n=0..50); \\ _Bernard Schott_, Jun 07 2019

%t Table[Floor[Binomial[n+3, 2]/2] -Floor[Binomial[n+1, 2]/2], {n, 0, 80}] (* or *) CoefficientList[Series[(1+x)/((1-x)^2*(1+x^2)), {x, 0, 80}], x] (* _Michael De Vlieger_, Oct 12 2016 *)

%o (PARI) {a(n) = n\4*4 + [1, 3, 4, 4][n%4+1]}; /* _Michael Somos_, Sep 11 2014 */

%o (Magma) R<x>:=PowerSeriesRing(Integers(), 80); Coefficients(R!( (1+x)/((1-x)^2*(1+x^2)) )); // _G. C. Greubel_, May 22 2019

%o (Sage) ((1+x)/((1-x)^2*(1+x^2))).series(x, 80).coefficients(x, sparse=False) # _G. C. Greubel_, May 22 2019

%o (GAP) a:=[1,3,4,4];; for n in [5..80] do a[n]:=2*a[n-1]-2*a[n-2]+2*a[n-3] -a[n-4]; od; a; # _G. C. Greubel_, May 22 2019

%Y Cf. A098180.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Aug 30 2004

%E Name edited by _G. C. Greubel_, Jun 06 2019