

A058187


Expansion of (1+x)/(1x^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
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OFFSET

0,3


COMMENTS

a(n) = A108299(n3,n)*(1)^floor(n/2) for n>2.  Reinhard Zumkeller, Jun 01 2005
For n>=i, i=6,7, a(ni) is the number of incongruent twocolor 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 180degree 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 ConnellPol triangle A159797 are linear sums of shifted versions of the duplicated tetrahedral numbers, e.g., Fi1(n) = a(n1) + 5*a(n2) + a(n3) + 5*a(n4).
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 kgons, Indian J. Pure and Appl. Math., 10 (1979), no.8, 964999.


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..880
V. Shevelev, A problem of enumeration of twocolor 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(n1). 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(7n) 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, x1..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



