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%I #82 Aug 18 2022 08:29:36
%S 1,1,4,4,10,10,20,20,35,35,56,56,84,84,120,120,165,165,220,220,286,
%T 286,364,364,455,455,560,560,680,680,816,816,969,969,1140,1140,1330,
%U 1330,1540,1540,1771,1771,2024,2024,2300,2300,2600,2600,2925,2925,3276,3276
%N Expansion of (1+x)/(1-x^2)^4: duplicated tetrahedral numbers.
%C For n >= i, i = 6,7, a(n - i) is the number of incongruent two-color bracelets of n beads, i of which are black (cf. A005513, A032280), having a diameter of symmetry. The latter means the following: if we imagine (0,1)-beads as points (with the corresponding labels) dividing a circumference of a bracelet into n identical parts, then a diameter of symmetry is a diameter (connecting two beads or not) such that a 180-degree turn of one of two sets of points around it (obtained by splitting the circumference by this diameter) leads to the coincidence of the two sets (including their labels). - _Vladimir Shevelev_, May 03 2011
%C From _Johannes W. Meijer_, May 20 2011: (Start)
%C The Kn11, Kn12, Kn13, Fi1 and Ze1 triangle sums, see A180662 for their definitions, of the Connell-Pol triangle A159797 are linear sums of shifted versions of the duplicated tetrahedral numbers, e.g., Fi1(n) = a(n-1) + 5*a(n-2) + a(n-3) + 5*a(n-4).
%C The Kn11, Kn12, Kn13, Kn21, Kn22, Kn23, Fi1, Fi2, Ze1 and Ze2 triangle sums of the Connell sequence A001614 as a triangle are also linear sums of shifted versions of the sequence given above. (End)
%C The number of quadruples of integers [x, u, v, w] that satisfy x > u > v > w >= 0, n + 5 = x + u. - _Michael Somos_, Feb 09 2015
%C Also, this sequence is the fourth column in the triangle of the coefficients of the sum of two consecutive Fibonacci polynomials F(n+1, x) and F(n, x) (n>=0) in ascending powers of x. - _Mohammad K. Azarian_, Jul 18 2018
%H Vincenzo Librandi, <a href="/A058187/b058187.txt">Table of n, a(n) for n = 0..880</a>
%H Hansraj Gupta, <a href="https://web.archive.org/web/20200806162943/https://www.insa.nic.in/writereaddata/UpLoadedFiles/IJPAM/20005a66_964.pdf">Enumeration of incongruent cyclic k-gons</a>, Indian J. Pure and Appl. Math., Vol. 10, No. 8 (1979), 964-999.
%H Vladimir Shevelev, <a href="https://arxiv.org/abs/0710.1370">A problem of enumeration of two-color bracelets with several variations</a>, arXiv:0710.1370 [math.CO], 2007-2011.
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-3,-3,3,1,-1).
%F a(n) = A006918(n+1) - a(n-1).
%F a(2*n) = a(2*n+1) = A000292(n) = (n+1)*(n+2)*(n+3)/6.
%F a(n) = (2*n^3 + 21*n^2 + 67*n + 63)/96 + (n^2 + 7*n + 11)(-1)^n/32. - _Paul Barry_, Aug 19 2003
%F a(n) = A108299(n-3,n)*(-1)^floor(n/2) for n > 2. - _Reinhard Zumkeller_, Jun 01 2005
%F Euler transform of finite sequence [1, 3]. - _Michael Somos_, Jun 07 2005
%F G.f.: 1 / ((1 - x) * (1 - x^2)^3) = 1 / ((1 + x)^3 * (1 - x)^4). a(n) = -a(-7-n) for all n in Z.
%F a(n) = binomial(floor(n/2) + 3, 3). - _Vladimir Shevelev_, May 03 2011
%F a(-n) = -a(n-7); a(n) = A000292(A008619(n)). - _Guenther Schrack_, Sep 13 2018
%F Sum_{n>=0} 1/a(n) = 3. - _Amiram Eldar_, Aug 18 2022
%p A058187:= proc(n) option remember; A058187(n):= binomial(floor(n/2)+3, 3) end: seq(A058187(n), n=0..51); # _Johannes W. Meijer_, May 20 2011
%t a[n_]:= Length @ FindInstance[{x>u, u>v, v>w, w>=0, x+u==n+5}, {x, u, v, w}, Integers, 10^9]; (* _Michael Somos_, Feb 09 2015 *)
%t With[{tetra=Binomial[Range[30]+2,3]},Riffle[tetra,tetra]] (* _Harvey P. Dale_, Mar 22 2015 *)
%o (PARI) {a(n) = binomial(n\2+3, 3)}; /* _Michael Somos_, Jun 07 2005 */
%o (Haskell)
%o a058187 n = a058187_list !! n
%o a058187_list = 1 : f 1 1 [1] where
%o f x y zs = z : f (x + y) (1 - y) (z:zs) where
%o z = sum $ zipWith (*) [1..x] [x,x-1..1]
%o -- _Reinhard Zumkeller_, Dec 21 2011
%o (Sage) [binomial((n//2)+3, 3) for n in (0..60)] # _G. C. Greubel_, Feb 18 2022
%Y Cf. A057884. Sum of 2 consecutive terms gives A006918, whose sum of 2 consecutive terms gives A002623, whose sum of 2 consecutive terms gives A000292, which is this sequence without the duplication. Continuing to sum 2 consecutive terms gives A000330, A005900, A001845, A008412 successively.
%Y Cf. A096338, A005513, A032280
%Y Cf. A000292, A190717, A190718. - _Johannes W. Meijer_, May 20 2011
%K easy,nonn
%O 0,3
%A _Henry Bottomley_, Nov 20 2000
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