A189325 Number of nondecreasing arrangements of n+2 numbers in 0..8 with the last equal to 8 and each after the second equal to the sum of one or two of the preceding four.
13, 24, 49, 95, 179, 321, 548, 866, 1267, 1733, 2248, 2806, 3408, 4056, 4752, 5498, 6296, 7148, 8056, 9022, 10048, 11136, 12288, 13506, 14792, 16148, 17576, 19078, 20656, 22312, 24048, 25866, 27768, 29756, 31832, 33998, 36256, 38608, 41056, 43602
Offset: 1
Keywords
Examples
Some solutions for n=3: ..2....0....1....1....2....7....2....4....3....1....0....4....1....3....3....2 ..4....4....6....3....6....8....3....4....5....4....4....4....7....5....4....4 ..6....4....7....4....8....8....3....4....5....4....4....4....7....5....4....4 ..8....4....7....4....8....8....5....8....8....5....8....4....8....5....4....6 ..8....8....8....8....8....8....8....8....8....8....8....8....8....8....8....8
Links
- R. H. Hardin, Table of n, a(n) for n = 1..200
Crossrefs
Cf. A189326.
Formula
Empirical: a(n) = (1/3)*n^3 + 10*n^2 + (587/3)*n - 1558 for n>10.
Empirical g.f.: x*(13 - 28*x + 31*x^2 - 9*x^3 + 10*x^4 + 3*x^5 + 7*x^6 - 21*x^7 - 14*x^8 - 10*x^9 + 2*x^10 + 10*x^11 + 7*x^12 + x^13) / (1 - x)^4. - Colin Barker, May 02 2018
Comments