A229012 T(n,k) = number of arrays of median of three adjacent elements of some length n+2 0..k array, with no adjacent equal elements in the latter.
2, 3, 2, 4, 7, 2, 5, 14, 15, 2, 6, 23, 46, 31, 2, 7, 34, 101, 130, 57, 2, 8, 47, 186, 359, 332, 105, 2, 9, 62, 307, 794, 1145, 830, 193, 2, 10, 79, 470, 1527, 3002, 3527, 2054, 353, 2, 11, 98, 681, 2666, 6635, 10860, 10735, 5108, 653, 2, 12, 119, 946, 4335, 13040, 27379
Offset: 1
Examples
Some solutions for n=4 k=4 ..1....1....0....3....4....1....2....3....2....3....3....1....1....0....2....2 ..4....1....1....1....3....0....2....1....0....0....2....3....4....4....0....1 ..0....1....3....3....3....2....0....1....4....2....2....4....1....1....3....3 ..3....2....3....1....0....0....1....3....3....1....0....3....4....4....2....3
Links
- R. H. Hardin, Table of n, a(n) for n = 1..925
Crossrefs
Row 2 is A008865(n+1).
Formula
Empirical for column k:
k=1: a(n) = a(n-1)
k=2: [order 13]
k=3: [order 27]
k=4: [order 46]
k=5: [order 69]
k=6: [order 95] for n>97
Empirical for row n:
n=1: a(n) = 1*n + 1
n=2: a(n) = 1*n^2 + 2*n - 1
n=3: a(n) = 1*n^3 + 3*n^2 - 3*n + 1
n=4: a(n) = (2/3)*n^4 + (10/3)*n^3 - (5/3)*n^2 + (2/3)*n - 1
n=5: [polynomial of degree 5]
n=6: [polynomial of degree 6]
n=7: [polynomial of degree 7]
Comments