A190041 T(n,k)=Number of nondecreasing arrangements of n+2 numbers in 0..k with the last equal to k and each after the second equal to the sum of one or two of the preceding three.
2, 4, 2, 5, 6, 2, 7, 7, 8, 2, 8, 12, 10, 10, 2, 10, 12, 18, 14, 12, 2, 11, 18, 16, 27, 18, 14, 2, 13, 17, 30, 23, 39, 22, 16, 2, 14, 24, 22, 47, 33, 53, 26, 18, 2, 16, 22, 40, 31, 72, 45, 69, 30, 20, 2, 17, 30, 31, 65, 43, 107, 57, 87, 34, 22, 2, 19, 27, 49, 49, 105, 60, 151, 69, 107
Offset: 1
Examples
Some solutions for n=5 k=3 ..2....3....1....1....0....1....0....0....0....0....1....1....1....1....1....1 ..3....3....1....1....1....2....1....1....1....3....1....3....1....1....1....2 ..3....3....2....1....1....3....1....1....1....3....1....3....1....1....2....2 ..3....3....3....1....1....3....2....1....2....3....1....3....1....2....2....3 ..3....3....3....2....1....3....3....2....2....3....1....3....2....3....3....3 ..3....3....3....3....2....3....3....3....3....3....2....3....2....3....3....3 ..3....3....3....3....3....3....3....3....3....3....3....3....3....3....3....3
Links
- R. H. Hardin, Table of n, a(n) for n = 1..4434
Formula
Empirical: T(n,1) = 2
Empirical: T(n,2) = 2*n + 2
Empirical: T(n,3) = 4*n - 2 for n>2
Empirical: T(n,4) = n^2 + 3*n - 1 for n>3
Empirical: T(n,5) = 12*n - 27 for n>4
Empirical: T(n,6) = 4*n^2 - 8*n + 11 for n>5
Empirical: T(n,7) = 20*n - 60 for n>5
Empirical: T(n,8) = (1/3)*n^3 + 2*n^2 + (50/3)*n - 83 for n>6
Comments