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A058187 Expansion of (1+x)/(1-x^2)^4: duplicated tetrahedral numbers. 29
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

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

REFERENCES

H. Gupta, Enumeration of incongruent cyclic k-gons, Indian J. Pure and Appl. Math., 10 (1979), no.8, 964-999.

LINKS

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).

FORMULA

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

MAPLE

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

MATHEMATICA

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 *)

PROG

(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

CROSSREFS

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

KEYWORD

easy,nonn

AUTHOR

Henry Bottomley, Nov 20 2000

STATUS

approved

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Last modified June 28 05:07 EDT 2017. Contains 288813 sequences.