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A058187
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Expansion of (1+x)/(1-x^2)^4: duplicated tetrahedral numbers.
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29
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1, 1, 4, 4, 10, 10, 20, 20, 35, 35, 56, 56, 84, 84, 120, 120, 165, 165, 220, 220, 286, 286, 364, 364, 455, 455, 560, 560, 680, 680, 816, 816, 969, 969, 1140, 1140, 1330, 1330, 1540, 1540, 1771, 1771, 2024, 2024, 2300, 2300, 2600, 2600, 2925, 2925, 3276, 3276
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OFFSET
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0,3
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COMMENTS
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a(n) = A108299(n-3,n)*(-1)^floor(n/2) for n>2. - Reinhard Zumkeller, Jun 01 2005
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
From Johannes W. Meijer, May 20 2011: (Start)
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).
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)
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
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REFERENCES
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H. Gupta, Enumeration of incongruent cyclic k-gons, Indian J. Pure and Appl. Math., 10 (1979), no.8, 964-999.
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LINKS
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Vincenzo Librandi, Table of n, a(n) for n = 0..880
V. Shevelev, A problem of enumeration of two-color bracelets with several variations
Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1).
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FORMULA
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a(n) = A006918(n+1) - a(n-1).
a(2*n) = a(2*n+1) = A000292(n) = (n+1)*(n+2)*(n+3)/6.
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
Euler transform of finite sequence [1, 3]. - Michael Somos, Jun 07 2005
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.
a(n) = binomial(floor(n/2) + 3, 3). - Vladimir Shevelev, May 03 2011
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MAPLE
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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
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MATHEMATICA
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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 *)
With[{tetra=Binomial[Range[30]+2, 3]}, Riffle[tetra, tetra]] (* Harvey P. Dale, Mar 22 2015 *)
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PROG
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(PARI) {a(n) = binomial(n\2+3, 3)}; /* Michael Somos, Jun 07 2005 */
(Haskell)
a058187 n = a058187_list !! n
a058187_list = 1 : f 1 1 [1] where
f x y zs = z : f (x + y) (1 - y) (z:zs) where
z = sum $ zipWith (*) [1..x] [x, x-1..1]
-- Reinhard Zumkeller, Dec 21 2011
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CROSSREFS
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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.
Cf. A096338, A005513, A032280
Cf. A000292, A190717, A190718. - Johannes W. Meijer, May 20 2011
Sequence in context: A168326 A101256 A116569 * A188271 A219939 A219471
Adjacent sequences: A058184 A058185 A058186 * A058188 A058189 A058190
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KEYWORD
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easy,nonn
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AUTHOR
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Henry Bottomley, Nov 20 2000
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STATUS
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approved
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