A263693 T(n,k)=Number of length n arrays of permutations of 0..n-1 with each element moved by -k to k places and every three consecutive elements having its maximum within 3 of its minimum.
1, 1, 2, 1, 2, 3, 1, 2, 6, 5, 1, 2, 6, 14, 7, 1, 2, 6, 24, 14, 11, 1, 2, 6, 24, 18, 16, 16, 1, 2, 6, 24, 36, 18, 22, 25, 1, 2, 6, 24, 36, 20, 24, 36, 37, 1, 2, 6, 24, 36, 36, 24, 40, 56, 57, 1, 2, 6, 24, 36, 36, 27, 40, 64, 85, 85, 1, 2, 6, 24, 36, 36, 48, 40, 64, 100, 125, 130, 1, 2, 6, 24, 36
Offset: 1
Examples
Some solutions for n=7 k=4 ..1....0....0....3....1....0....0....0....0....0....0....0....1....0....1....2 ..0....2....1....0....0....1....1....1....1....2....1....2....0....1....0....0 ..2....1....2....1....3....2....2....2....2....1....2....1....2....2....3....1 ..3....3....3....2....2....4....3....4....3....3....4....4....3....4....2....3 ..4....4....4....4....4....3....5....5....4....4....3....3....5....5....5....4 ..6....5....6....5....5....6....4....6....5....6....5....6....6....3....4....6 ..5....6....5....6....6....5....6....3....6....5....6....5....4....6....6....5
Links
- R. H. Hardin, Table of n, a(n) for n = 1..653
Crossrefs
Column 1 is A130137(n-1).
Formula
Empirical for diagonal: a(n) = a(n-1) +a(n-2) -a(n-3) +a(n-4) for n>12
Empirical for column k:
k=1: a(n) = a(n-1) +a(n-2) -a(n-3) +a(n-4)
k=2: a(n) = a(n-1) +a(n-2) -a(n-3) +a(n-4) for n>12
k=3: a(n) = a(n-1) +a(n-2) -a(n-3) +a(n-4) for n>12
k=4: a(n) = a(n-1) +a(n-2) -a(n-3) +a(n-4) for n>12
k=5: a(n) = a(n-1) +a(n-2) -a(n-3) +a(n-4) for n>12
k=6: a(n) = a(n-1) +a(n-2) -a(n-3) +a(n-4) for n>12
k=7: a(n) = a(n-1) +a(n-2) -a(n-3) +a(n-4) for n>13
Comments