A224392 Number of 3 X n 0..3 arrays with diagonals and antidiagonals unimodal and rows nondecreasing.
64, 1000, 6094, 27790, 102232, 319769, 881519, 2196522, 5038720, 10788462, 21789398, 41858498, 76994510, 136338455, 233448747, 387963222, 627730762, 991506308, 1532314870, 2321602662, 3454306716, 5054988261, 7285189791
Offset: 1
Keywords
Examples
Some solutions for n=3: ..0..1..1....0..1..1....1..1..2....0..0..1....0..1..1....1..1..3....0..2..2 ..2..2..2....0..0..1....2..3..3....2..2..2....2..2..2....1..2..3....1..2..3 ..0..2..2....0..0..0....1..1..2....1..3..3....0..0..1....1..3..3....0..3..3
Links
- R. H. Hardin, Table of n, a(n) for n = 1..210
Crossrefs
Cf. A224391.
Formula
Empirical: a(n) = (353/181440)*n^9 + (17/560)*n^8 + (9731/30240)*n^7 + (1283/720)*n^6 + (52457/8640)*n^5 + (776/45)*n^4 + (294499/11340)*n^3 + (28837/2520)*n^2 + (8207/126)*n - 18 for n>1.
Conjectures from Colin Barker, Aug 30 2018: (Start)
G.f.: x*(64 + 360*x - 1026*x^2 + 4170*x^3 - 7998*x^4 + 10591*x^5 - 9351*x^6 + 5629*x^7 - 2165*x^8 + 478*x^9 - 46*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