A062890 Number of quadrilaterals that can be formed with perimeter n. In other words, number of partitions of n into four parts such that the sum of any three is more than the fourth.
0, 0, 0, 0, 1, 1, 1, 2, 3, 4, 5, 7, 8, 11, 12, 16, 18, 23, 24, 31, 33, 41, 43, 53, 55, 67, 69, 83, 86, 102, 104, 123, 126, 147, 150, 174, 177, 204, 207, 237, 241, 274, 277, 314, 318, 358, 362, 406, 410, 458, 462, 514, 519, 575, 579, 640, 645, 710
Offset: 0
Examples
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.
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).
Crossrefs
Programs
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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 *)
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)
Extensions
More terms from Vladeta Jovovic and Dean Hickerson, Jul 01 2001
Comments