A189326 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 four.
2, 4, 2, 5, 6, 2, 7, 7, 8, 2, 8, 12, 11, 10, 2, 10, 12, 20, 16, 12, 2, 11, 18, 21, 32, 22, 14, 2, 13, 17, 36, 33, 49, 28, 16, 2, 14, 24, 31, 64, 54, 70, 34, 18, 2, 16, 22, 49, 51, 110, 84, 94, 40, 20, 2, 17, 30, 42, 95, 91, 179, 119, 120, 46, 22, 2, 19, 27, 63, 76, 179, 157, 275, 157
Offset: 1
Examples
Some solutions for n=5 k=3 ..1....0....1....0....1....1....0....1....0....0....3....0....1....1....1....1 ..2....1....1....1....2....1....1....3....1....1....3....1....1....1....1....2 ..2....1....1....1....2....1....1....3....1....1....3....1....2....2....1....3 ..3....1....1....1....2....1....2....3....2....2....3....1....2....2....2....3 ..3....2....1....1....3....2....2....3....2....3....3....2....3....2....3....3 ..3....2....2....2....3....3....2....3....3....3....3....3....3....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..3692
Crossrefs
Row 1 is A001651(n+1)
Formula
Empirical: T(n,1) = 2
Empirical: T(n,2) = 2*n + 2
Empirical: T(n,3) = 6*n - 8 for n>3
Empirical: T(n,4) = n^2 + 11*n - 32 for n>5
Empirical: T(n,5) = 38*n - 147 for n>6
Empirical: T(n,6) = 6*n^2 + 34*n - 264 for n>8
Empirical: T(n,7) = 140*n - 751 for n>8
Empirical: T(n,8) = (1/3)*n^3 + 10*n^2 + (587/3)*n - 1558 for n>10
Comments