A249707 T(n,k)=Number of length n+3 0..k arrays with every four consecutive terms having the maximum of some two terms equal to the minimum of the remaining two terms.
10, 39, 14, 100, 69, 20, 205, 208, 125, 28, 366, 485, 440, 221, 38, 595, 966, 1153, 896, 377, 52, 904, 1729, 2524, 2601, 1724, 659, 72, 1305, 2864, 4893, 6172, 5425, 3440, 1177, 100, 1810, 4473, 8688, 12789, 13666, 11925, 7056, 2119, 138, 2431, 6670, 14433
Offset: 1
Examples
Some solutions for n=6 k=4 ..3....3....2....4....1....4....3....4....3....1....0....3....3....2....3....2 ..1....3....4....1....4....2....3....1....4....1....2....0....3....1....3....1 ..0....3....0....1....1....0....2....1....3....2....2....0....3....1....4....2 ..1....2....2....1....1....2....4....1....0....1....3....0....4....0....0....4 ..1....4....2....4....0....4....3....2....3....1....2....0....2....1....3....2 ..1....3....2....1....3....2....3....1....3....1....1....0....3....3....3....1 ..0....3....0....1....1....1....3....1....3....3....2....0....3....1....4....2 ..4....3....4....1....1....2....2....1....1....1....3....0....4....1....3....4 ..1....0....2....0....0....2....3....4....4....0....2....1....1....0....2....2
Links
- R. H. Hardin, Table of n, a(n) for n = 1..8549
Formula
Empirical for column k:
k=1: a(n) = a(n-1) +a(n-4)
k=2: [order 10]
k=3: [order 17]
k=4: [order 24]
k=5: [order 30]
k=6: [order 37]
k=7: [order 43]
Empirical for row n:
n=1: a(n) = 2*n^3 + 4*n^2 + 3*n + 1
n=2: a(n) = (1/2)*n^4 + 4*n^3 + (11/2)*n^2 + 3*n + 1
n=3: a(n) = (1/15)*n^5 + 2*n^4 + 7*n^3 + 7*n^2 + (44/15)*n + 1
n=4: a(n) = (7/15)*n^5 + 5*n^4 + 11*n^3 + 8*n^2 + (38/15)*n + 1
n=5: a(n) = (5/3)*n^5 + 10*n^4 + 16*n^3 + 8*n^2 + (4/3)*n + 1
n=6: a(n) = (1/5)*n^6 + (73/15)*n^5 + 18*n^4 + 22*n^3 + (34/5)*n^2 - (13/15)*n + 1
n=7: a(n) = (1/70)*n^7 + (19/15)*n^6 + (178/15)*n^5 + 30*n^4 + (851/30)*n^3 + (56/15)*n^2 - (446/105)*n + 1
Comments