A247997 Number of length 2+5 0..n arrays with no disjoint triples in any consecutive six terms having the same sum

32, 702, 5316, 27800, 104620, 329742, 884032, 2131356, 4664480, 9508130, 18168932, 33008212, 57264516, 95672090, 154419880, 242095992, 369529512, 551174206, 804749300, 1153181480, 1623975972, 2251830342, 3077638456, 4151941100

Offset

1,1

Comments

Row 2 of A247995

Links

R. H. Hardin, Table of n, a(n) for n = 1..45

Formula

Empirical: a(n) = 2*a(n-1) +2*a(n-2) -4*a(n-3) -3*a(n-4) +7*a(n-6) +4*a(n-7) -5*a(n-8) -4*a(n-9) -5*a(n-10) +4*a(n-11) +7*a(n-12) -3*a(n-14) -4*a(n-15) +2*a(n-16) +2*a(n-17) -a(n-18)

Empirical for n mod 12 = 0: a(n) = n^7 - 4*n^6 + (1523/32)*n^5 - (5325/32)*n^4 + (4180/9)*n^3 - (12625/24)*n^2 + (641/6)*n

Empirical for n mod 12 = 1: a(n) = n^7 - 4*n^6 + (1523/32)*n^5 - (5325/32)*n^4 + (64225/144)*n^3 - (20525/48)*n^2 - (24079/96)*n + (111235/288)

Empirical for n mod 12 = 2: a(n) = n^7 - 4*n^6 + (1523/32)*n^5 - (5325/32)*n^4 + (4180/9)*n^3 - (12625/24)*n^2 + (107/2)*n + (2260/9)

Empirical for n mod 12 = 3: a(n) = n^7 - 4*n^6 + (1523/32)*n^5 - (5325/32)*n^4 + (64225/144)*n^3 - (20525/48)*n^2 - (24079/96)*n + (16555/32)

Empirical for n mod 12 = 4: a(n) = n^7 - 4*n^6 + (1523/32)*n^5 - (5325/32)*n^4 + (4180/9)*n^3 - (12625/24)*n^2 + (641/6)*n - (640/9)

Empirical for n mod 12 = 5: a(n) = n^7 - 4*n^6 + (1523/32)*n^5 - (5325/32)*n^4 + (64225/144)*n^3 - (20525/48)*n^2 - (9733/32)*n + (209795/288)

Empirical for n mod 12 = 6: a(n) = n^7 - 4*n^6 + (1523/32)*n^5 - (5325/32)*n^4 + (4180/9)*n^3 - (12625/24)*n^2 + (641/6)*n - 20

Empirical for n mod 12 = 7: a(n) = n^7 - 4*n^6 + (1523/32)*n^5 - (5325/32)*n^4 + (64225/144)*n^3 - (20525/48)*n^2 - (24079/96)*n + (128515/288)

Empirical for n mod 12 = 8: a(n) = n^7 - 4*n^6 + (1523/32)*n^5 - (5325/32)*n^4 + (4180/9)*n^3 - (12625/24)*n^2 + (107/2)*n + (2440/9)

Empirical for n mod 12 = 9: a(n) = n^7 - 4*n^6 + (1523/32)*n^5 - (5325/32)*n^4 + (64225/144)*n^3 - (20525/48)*n^2 - (24079/96)*n + (14635/32)

Empirical for n mod 12 = 10: a(n) = n^7 - 4*n^6 + (1523/32)*n^5 - (5325/32)*n^4 + (4180/9)*n^3 - (12625/24)*n^2 + (641/6)*n - (820/9)

Empirical for n mod 12 = 11: a(n) = n^7 - 4*n^6 + (1523/32)*n^5 - (5325/32)*n^4 + (64225/144)*n^3 - (20525/48)*n^2 - (9733/32)*n + (227075/288)

Example

Some solutions for n=5
..4....0....4....5....2....5....5....2....3....2....1....3....1....3....3....2
..1....5....5....4....0....1....3....2....2....0....1....4....2....1....5....3
..1....4....5....2....1....4....0....4....1....0....4....3....0....5....5....2
..5....0....0....1....5....2....0....3....3....1....5....5....3....3....2....3
..2....3....5....5....1....0....3....1....1....1....0....3....2....4....4....0
..4....5....3....0....2....3....4....1....3....1....0....5....5....1....2....1
..2....0....0....3....4....5....3....2....3....2....4....1....0....1....3....0

Crossrefs

Sequence in context: A025031 A025008 A020984 * A199708 A264093 A062261

Adjacent sequences: A247994 A247995 A247996 * A247998 A247999 A248000

Keyword

nonn

Author

R. H. Hardin, Sep 28 2014

Status

approved