login
Number of nondecreasing arrangements of n+2 numbers in 0..4 with the last equal to 4 and each after the second equal to the sum of one or two of the preceding four.
1

%I #8 May 02 2018 09:22:40

%S 7,12,20,32,49,70,94,120,148,178,210,244,280,318,358,400,444,490,538,

%T 588,640,694,750,808,868,930,994,1060,1128,1198,1270,1344,1420,1498,

%U 1578,1660,1744,1830,1918,2008,2100,2194,2290,2388,2488,2590,2694,2800,2908

%N Number of nondecreasing arrangements of n+2 numbers in 0..4 with the last equal to 4 and each after the second equal to the sum of one or two of the preceding four.

%C Column 4 of A189326.

%H R. H. Hardin, <a href="/A189321/b189321.txt">Table of n, a(n) for n = 1..200</a>

%F Empirical: a(n) = n^2 + 11*n - 32 for n>5.

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

%e Some solutions for n=3:

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

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

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

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

%e ..4....4....4....4....4....4....4....4....4....4....4....4....4....4....4....4

%Y Cf. A189326.

%K nonn

%O 1,1

%A _R. H. Hardin_, Apr 20 2011