A242239 T(n,k)=Number of length n+k+1 0..k arrays with every value 0..k appearing at least once in every consecutive k+2 elements, and new values 0..k introduced in order.
3, 6, 5, 10, 12, 8, 15, 22, 22, 13, 21, 35, 43, 40, 21, 28, 51, 71, 82, 74, 34, 36, 70, 106, 139, 157, 136, 55, 45, 92, 148, 211, 271, 304, 250, 89, 55, 117, 197, 298, 416, 531, 586, 460, 144, 66, 145, 253, 400, 592, 821, 1047, 1129, 846, 233, 78, 176, 316, 517, 799
Offset: 1
Examples
Some solutions for n=5 k=4 ..0....0....0....0....0....0....0....0....0....0....0....0....0....0....0....0 ..1....1....1....1....1....1....1....1....1....1....1....1....1....0....1....1 ..2....2....0....2....2....1....0....0....2....0....2....2....0....1....2....0 ..1....3....2....3....3....2....2....2....3....2....3....3....2....2....3....2 ..3....4....3....4....4....3....3....3....0....3....0....0....3....3....4....3 ..4....0....4....1....0....4....4....4....4....4....4....4....4....4....1....4 ..0....2....2....0....1....0....2....0....2....1....2....1....0....2....0....1 ..2....1....1....0....2....1....1....1....1....0....1....2....1....0....2....0 ..2....3....0....2....3....0....0....1....0....0....3....4....2....1....3....4 ..1....0....0....3....0....2....2....2....3....2....2....3....3....1....0....2
Links
- R. H. Hardin, Table of n, a(n) for n = 1..731
Crossrefs
Formula
Empirical for column k:
k=1: a(n) = a(n-1) +a(n-2)
k=2: a(n) = a(n-1) +a(n-2) +a(n-3)
k=3: a(n) = a(n-1) +a(n-2) +a(n-3) +a(n-4)
k=4: a(n) = a(n-1) +a(n-2) +a(n-3) +a(n-4) +a(n-5)
k=5: a(n) = a(n-1) +a(n-2) +a(n-3) +a(n-4) +a(n-5) +a(n-6)
k=6: a(n) = a(n-1) +a(n-2) +a(n-3) +a(n-4) +a(n-5) +a(n-6) +a(n-7)
k=7: a(n) = a(n-1) +a(n-2) +a(n-3) +a(n-4) +a(n-5) +a(n-6) +a(n-7) +a(n-8)
k=8: a(n) = a(n-1) +a(n-2) +a(n-3) +a(n-4) +a(n-5) +a(n-6) +a(n-7) +a(n-8) +a(n-9)
k=9: a(n) = a(n-1) +a(n-2) +a(n-3) +a(n-4) +a(n-5) +a(n-6) +a(n-7) +a(n-8) +a(n-9) +a(n-10)
Empirical for row n:
n=1: a(n) = (1/2)*n^2 + (3/2)*n + 1
n=2: a(n) = (3/2)*n^2 + (5/2)*n + 1
n=3: a(n) = (7/2)*n^2 + (7/2)*n + 1
n=4: a(n) = (15/2)*n^2 + (9/2)*n + 1
n=5: a(n) = (31/2)*n^2 + (11/2)*n + 1 for n>1
n=6: a(n) = (63/2)*n^2 + (13/2)*n + 1 for n>2
n=7: a(n) = (127/2)*n^2 + (15/2)*n + 1 for n>3
n=8: a(n) = (255/2)*n^2 + (17/2)*n + 1 for n>4
n=9: a(n) = (511/2)*n^2 + (19/2)*n + 1 for n>5
n=10: a(n) = (1023/2)*n^2 + (21/2)*n + 1 for n>6
n=11: a(n) = (2047/2)*n^2 + (23/2)*n + 1 for n>7
n=12: a(n) = (4095/2)*n^2 + (25/2)*n + 1 for n>8
n=13: a(n) = (8191/2)*n^2 + (27/2)*n + 1 for n>9
n=14: a(n) = (16383/2)*n^2 + (29/2)*n + 1 for n>10
n=15: a(n) = (32767/2)*n^2 + (31/2)*n + 1 for n>11
Empirical large-k generalization, for k>n-4: T(n,k) = ((2^n-1)/2)*k^2 + ((2*n+1)/2)*k + 1
Empirical recurrence generalization, for column k: a(n) = sum {i in 1..k+1} a(n-i)
Comments