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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.
1

%I #8 May 02 2018 09:22:57

%S 13,24,49,95,179,321,548,866,1267,1733,2248,2806,3408,4056,4752,5498,

%T 6296,7148,8056,9022,10048,11136,12288,13506,14792,16148,17576,19078,

%U 20656,22312,24048,25866,27768,29756,31832,33998,36256,38608,41056,43602

%N 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.

%C Column 8 of A189326.

%H R. H. Hardin, <a href="/A189325/b189325.txt">Table of n, a(n) for n = 1..200</a>

%F Empirical: a(n) = (1/3)*n^3 + 10*n^2 + (587/3)*n - 1558 for n>10.

%F 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

%e Some solutions for n=3:

%e ..2....0....1....1....2....7....2....4....3....1....0....4....1....3....3....2

%e ..4....4....6....3....6....8....3....4....5....4....4....4....7....5....4....4

%e ..6....4....7....4....8....8....3....4....5....4....4....4....7....5....4....4

%e ..8....4....7....4....8....8....5....8....8....5....8....4....8....5....4....6

%e ..8....8....8....8....8....8....8....8....8....8....8....8....8....8....8....8

%Y Cf. A189326.

%K nonn

%O 1,1

%A _R. H. Hardin_, Apr 20 2011