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Number of nondecreasing arrangements of n+2 numbers in 0..5 with the last equal to 5 and each after the second equal to the sum of one or two of the preceding four.
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%I #8 May 02 2018 09:22:33

%S 8,12,21,33,54,84,119,157,195,233,271,309,347,385,423,461,499,537,575,

%T 613,651,689,727,765,803,841,879,917,955,993,1031,1069,1107,1145,1183,

%U 1221,1259,1297,1335,1373,1411,1449,1487,1525,1563,1601,1639,1677,1715

%N Number of nondecreasing arrangements of n+2 numbers in 0..5 with the last equal to 5 and each after the second equal to the sum of one or two of the preceding four.

%C Column 5 of A189326.

%H R. H. Hardin, <a href="/A189322/b189322.txt">Table of n, a(n) for n = 1..200</a>

%F Empirical: a(n) = 38*n - 147 for n>6.

%F Empirical g.f.: x*(8 - 4*x + 5*x^2 + 3*x^3 + 9*x^4 + 9*x^5 + 5*x^6 + 3*x^7) / (1 - x)^2. - _Colin Barker_, May 02 2018

%e Some solutions for n=3:

%e ..1....1....2....5....1....1....1....1....1....4....1....2....3....2....1....0

%e ..2....2....3....5....3....2....2....1....3....5....3....5....5....3....4....5

%e ..2....3....3....5....3....2....3....2....4....5....4....5....5....3....4....5

%e ..4....3....5....5....4....3....4....3....5....5....4....5....5....3....4....5

%e ..5....5....5....5....5....5....5....5....5....5....5....5....5....5....5....5

%Y Cf. A189326.

%K nonn

%O 1,1

%A _R. H. Hardin_, Apr 20 2011