OFFSET
0,8
COMMENTS
From Frank M Jackson, Jul 10 2012: (Start)
I recently commented on A062890 that:
"Partition sets of n into four parts (sides) such that the sum of any three is more than the fourth do not uniquely define a quadrilateral, even if it is further constrained to be cyclic. This is because the order of adjacent sides is important. E.g. the partition set [1,1,2,2] for a perimeter n=6 can be reordered to generate two non-congruent cyclic quadrilaterals, [1,2,1,2] and [1,1,2,2], where the first is a rectangle and the second a kite."
This comment applies to all integer polygons (other than triangles) that are generated from a perimeter of length n. Not sure how best to correct for the above observation but my suggestion would be to change the definition of the present sequence to read:
"The number of cyclic integer pentagons differing only in circumradius that can be generated from an integer perimeter n." (End)
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 0..1000 from T. D. Noe)
G. E. Andrews, P. Paule and A. Riese, MacMahon's partition analysis III. The Omega package, p. 19.
G. E. Andrews, P. Paule and A. Riese, MacMahon's Partition Analysis IX: k-gon partitions, Bull. Austral Math. Soc., 64 (2001), 321-329.
Index entries for linear recurrences with constant coefficients, signature (0, 1, 0, 1, 1, 0, -1, 0, -1, -2, 0, 0, 0, 0, 2, 1, 0, 1, 0, -1, -1, 0, -1, 0, 1).
FORMULA
G.f.: x^5*(1-x^11)/((1-x)*(1-x^2)*(1-x^4)*(1-x^5)*(1-x^6)*(1-x^8)).
a(2*n+8) = A026811(2*n+8) - A002621(n), a(2*n+9) = A026811(2*n+9) - A002621(n) for n >= 0. - Seiichi Manyama, Jun 08 2017
MATHEMATICA
CoefficientList[Series[x^5(1-x^11)/((1-x)(1-x^2)(1-x^4)(1-x^5)(1-x^6) (1-x^8)), {x, 0, 60}], x] (* Harvey P. Dale, Dec 16 2011 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, May 05, 2002
STATUS
approved