A217954 T(n,k) = number of n-element 0..3 arrays with each element the minimum of k adjacent elements of a random 0..3 array of n+k-1 elements.
4, 4, 16, 4, 16, 64, 4, 16, 50, 256, 4, 16, 50, 144, 1024, 4, 16, 50, 130, 422, 4096, 4, 16, 50, 130, 310, 1268, 16384, 4, 16, 50, 130, 296, 736, 3823, 65536, 4, 16, 50, 130, 296, 624, 1821, 11472, 262144, 4, 16, 50, 130, 296, 610, 1289, 4673, 34350, 1048576, 4, 16
Offset: 1
Examples
Some solutions for n=4 k=4 ..0....1....0....1....1....0....1....2....0....1....1....1....0....0....3....0 ..0....1....1....1....3....3....2....2....2....1....2....2....1....2....3....2 ..1....3....3....2....3....2....3....3....2....1....1....3....1....2....2....3 ..1....1....3....3....0....0....0....1....0....0....1....3....0....1....2....0
Links
- R. H. Hardin, Table of n, a(n) for n = 1..1275
Formula
Empirical for column k:
k=2: a(n) = 4*a(n-1) -6*a(n-2) +10*a(n-3) -5*a(n-4) +6*a(n-5) -a(n-6) +a(n-7)
k=3: a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) +5*a(n-4) -4*a(n-5) +6*a(n-6) +4*a(n-7) +2*a(n-9) +a(n-10)
k=4: a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4) +6*a(n-5) -4*a(n-6) +6*a(n-7) +4*a(n-8) +5*a(n-9) +a(n-10) +3*a(n-11) +2*a(n-12) +a(n-13)
k=5: a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4) +6*a(n-6) -4*a(n-7) +6*a(n-8) +4*a(n-9) +5*a(n-10) +6*a(n-11) +2*a(n-12) +4*a(n-13) +3*a(n-14) +2*a(n-15) +a(n-16)
k=6: a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4) +6*a(n-7) -4*a(n-8) +6*a(n-9) +4*a(n-10) +5*a(n-11) +6*a(n-12) +7*a(n-13) +3*a(n-14) +5*a(n-15) +4*a(n-16) +3*a(n-17) +2*a(n-18) +a(n-19)
k=7: a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4) +6*a(n-8) -4*a(n-9) +6*a(n-10) +4*a(n-11) +5*a(n-12) +6*a(n-13) +7*a(n-14) +8*a(n-15) +4*a(n-16) +6*a(n-17) +5*a(n-18) +4*a(n-19) +3*a(n-20) +2*a(n-21) +a(n-22)
Diagonal: a(n) = (1/720)*n^6 + (1/48)*n^5 + (23/144)*n^4 + (9/16)*n^3 + (241/180)*n^2 + (11/12)*n + 1
Comments