OFFSET
0,8
COMMENTS
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. - Frank M Jackson, Jun 29 2012
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 (1,1,-1,1,-1,0,0,-1,1,-1,1,1,-1).
FORMULA
G.f.: x^4*(1+x+x^5)/((1-x^2)*(1-x^3)*(1-x^4)*(1-x^6)).
a(2*n+6) = A026810(2*n+6) - A000601(n), a(2*n+7) = A026810(2*n+7) - A000601(n) for n >= 0. - Seiichi Manyama, Jun 08 2017
From Wesley Ivan Hurt, Jan 01 2021: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-4) - a(n-5) - a(n-8) + a(n-9) - a(n-10) + a(n-11) + a(n-12) - a(n-13).
a(n) = Sum_{k=1..floor(n/4)} Sum_{j=k..floor((n-k)/3)} Sum_{i=j..floor((n-j-k)/2)} sign(floor((i+j+k)/(n-i-j-k+1))). (End)
EXAMPLE
a(7) = 2 as the two partitions are (1,2,2,2), (1,1,2,3) and in each sum of any three is more than the fourth.
MATHEMATICA
CoefficientList[Series[x^4*(1+x+x^5)/((1-x^2)*(1-x^3)*(1-x^4)*(1-x^6)), {x, 0, 60}], x] (* Frank M Jackson, Jun 09 2017 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Amarnath Murthy, Jun 29 2001
EXTENSIONS
More terms from Vladeta Jovovic and Dean Hickerson, Jul 01 2001
STATUS
approved