A224025 Number of 3 X n 0..3 arrays with rows nondecreasing and antidiagonals unimodal.
64, 1000, 6796, 32523, 122523, 387729, 1074167, 2679260, 6137666, 13104218, 26368076, 50439449, 92358199, 162782299, 277423483, 458907498, 739147146, 1162327788, 1788617172, 2698724343, 3999445995, 5830352933, 8371784327
Offset: 1
Keywords
Examples
Some solutions for n=3: ..2..2..2....1..2..3....0..0..1....2..3..3....1..1..3....3..3..3....2..2..2 ..1..2..2....3..3..3....1..1..3....1..1..1....2..2..2....1..1..2....2..3..3 ..3..3..3....0..2..2....2..2..2....1..1..3....0..0..0....1..1..2....0..1..1
Links
- R. H. Hardin, Table of n, a(n) for n = 1..210
Crossrefs
Cf. A224024.
Formula
Empirical: a(n) = (353/181440)*n^9 + (5/126)*n^8 + (12287/30240)*n^7 + (131/60)*n^6 + (64877/8640)*n^5 + (135/8)*n^4 + (998257/45360)*n^3 + (53933/2520)*n^2 + (3421/252)*n - 2 for n>1.
Conjectures from Colin Barker, Aug 26 2018: (Start)
G.f.: x*(64 + 360*x - 324*x^2 + 1883*x^3 - 3447*x^4 + 4386*x^5 - 3748*x^6 + 2193*x^7 - 825*x^8 + 182*x^9 - 18*x^10) / (1 - x)^10.
a(n) = 10*a(n-1) - 45*a(n-2) + 120*a(n-3) - 210*a(n-4) + 252*a(n-5) - 210*a(n-6) + 120*a(n-7) - 45*a(n-8) + 10*a(n-9) - a(n-10) for n>11.
(End)
Comments