A252177 T(n,k)=Number of length n+2 0..k arrays with the sum of the maximum minus the minimum of adjacent triples multiplied by some arrangement of +-1 equal to zero.
2, 3, 12, 4, 49, 12, 5, 132, 83, 40, 6, 285, 264, 369, 56, 7, 536, 687, 1872, 957, 144, 8, 917, 1428, 6361, 6820, 3217, 240, 9, 1464, 2729, 17092, 30315, 31420, 9295, 544, 10, 2217, 4680, 39109, 100894, 179225, 123826, 28977, 992, 11, 3220, 7661, 79672
Offset: 1
Examples
Some solutions for n=5 k=4 ..0....0....2....1....1....1....1....3....3....2....2....2....1....3....0....4 ..3....2....1....2....0....1....0....4....4....3....2....2....3....0....2....2 ..3....3....4....3....2....2....1....1....4....3....3....4....2....2....3....3 ..3....0....0....3....4....2....0....3....4....2....1....3....2....4....4....4 ..0....1....3....0....1....3....1....1....1....4....0....2....2....3....2....0 ..0....2....1....3....2....3....3....2....1....0....2....3....4....0....2....4 ..3....2....2....1....3....1....1....0....2....1....4....2....1....0....1....0
Links
- R. H. Hardin, Table of n, a(n) for n = 1..181
Crossrefs
Column 1 is A251421
Formula
Empirical for column k:
k=1: a(n) = 2*a(n-1) +2*a(n-2) -4*a(n-3)
k=2: [order 14]
k=3: [order 39] for n>40
Empirical for row n:
n=1: a(n) = n + 1
n=2: a(n) = (1/6)*n^4 + (7/3)*n^3 + (29/6)*n^2 + (11/3)*n + 1
n=3: a(n) = 3*a(n-1) -8*a(n-3) +6*a(n-4) +6*a(n-5) -8*a(n-6) +3*a(n-8) -a(n-9); also a polynomial of degree 5 plus a quasipolynomial of degree 2 with period 2
n=4: [linear recurrence of order 23; also a polynomial of degree 6 plus a quasipolynomial of degree 3 with period 12]
Comments