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A248492 Number of length 3+5 0..n arrays with some three disjoint pairs in each consecutive six terms having the same sum 1
22, 183, 988, 3301, 8370, 17923, 33520, 55665, 87310, 129991, 185340, 256861, 346258, 456195, 590920, 751873, 942030, 1167247, 1428940, 1730397, 2078794, 2475475, 2924280, 3433177, 4003774, 4640463, 5351908, 6140413, 7010538, 7972267 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

Row 3 of A248489

LINKS

R. H. Hardin, Table of n, a(n) for n = 1..210

FORMULA

Empirical: a(n) = -a(n-2) +a(n-3) +2*a(n-5) +a(n-6) +2*a(n-7) +a(n-8) -a(n-10) -3*a(n-11) -2*a(n-12) -4*a(n-13) -a(n-14) -2*a(n-15) +2*a(n-16) +a(n-17) +4*a(n-18) +2*a(n-19) +3*a(n-20) +a(n-21) -a(n-23) -2*a(n-24) -a(n-25) -2*a(n-26) -a(n-28) +a(n-29) +a(n-31)

Also a quartic polynomial plus a linear quasipolynomial with period 840, the first 12 being:

Empirical for n mod 840 = 0: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (1769/35)*n + 1

Empirical for n mod 840 = 1: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (1664/35)*n + (133093/630)

Empirical for n mod 840 = 2: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (10907/105)*n + (175159/315)

Empirical for n mod 840 = 3: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (2084/35)*n + (7001/14)

Empirical for n mod 840 = 4: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (1769/35)*n + (73067/315)

Empirical for n mod 840 = 5: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (10592/105)*n + (2311/18)

Empirical for n mod 840 = 6: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (1769/35)*n - (4969/35)

Empirical for n mod 840 = 7: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (2084/35)*n + (30739/90)

Empirical for n mod 840 = 8: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (10907/105)*n + (27191/63)

Empirical for n mod 840 = 9: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (1664/35)*n + (10141/70)

Empirical for n mod 840 = 10: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (1769/35)*n + (20035/63)

Empirical for n mod 840 = 11: a(n) = (15/2)*n^4 + (701/9)*n^3 - (23846/105)*n^2 - (11852/105)*n + (367733/630)

EXAMPLE

Some solutions for n=6

..5....6....4....4....4....2....4....6....4....0....4....5....3....3....6....3

..2....5....3....0....0....2....6....5....2....2....3....4....3....5....1....4

..2....4....6....0....1....3....5....4....4....1....2....4....4....4....2....5

..1....5....2....2....2....4....4....1....0....0....3....5....5....2....5....3

..5....6....5....4....3....4....3....2....6....1....2....3....1....1....4....4

..6....4....4....2....2....3....2....3....2....2....1....3....2....3....3....2

..5....3....1....4....1....5....1....0....4....3....4....2....3....3....0....3

..2....2....0....0....3....2....0....5....2....2....3....4....6....5....1....4

CROSSREFS

Sequence in context: A248490 A191012 A248491 * A248493 A248494 A248495

Adjacent sequences:  A248489 A248490 A248491 * A248493 A248494 A248495

KEYWORD

nonn

AUTHOR

R. H. Hardin, Oct 07 2014

STATUS

approved

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Last modified October 23 08:40 EDT 2021. Contains 348211 sequences. (Running on oeis4.)