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Number of length 4+3 0..n arrays with some disjoint pairs in every consecutive four terms having the same sum
1

%I #4 Sep 18 2014 14:13:28

%S 8,81,364,1007,2164,3997,6584,10219,14852,20847,28108,37095,47564,

%T 60087,74428,91101,109760,131243,154956,181677,211024,243709,279136,

%U 318445,360676,406933,456648,510683,568172,630613,696744,767859,843244,923955

%N Number of length 4+3 0..n arrays with some disjoint pairs in every consecutive four terms having the same sum

%C Row 4 of A247533

%H R. H. Hardin, <a href="/A247536/b247536.txt">Table of n, a(n) for n = 1..349</a>

%F Empirical: a(n) = -a(n-1) -a(n-2) +a(n-4) +2*a(n-5) +3*a(n-6) +3*a(n-7) +2*a(n-8) -2*a(n-10) -4*a(n-11) -4*a(n-12) -4*a(n-13) -2*a(n-14) +2*a(n-16) +3*a(n-17) +3*a(n-18) +2*a(n-19) +a(n-20) -a(n-22) -a(n-23) -a(n-24)

%F Also as a cubic plus a linear quasipolynomial with period 420, first 12 listed:

%F Empirical for n mod 420 = 0: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (4243/210)*n + 1

%F Empirical for n mod 420 = 1: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (2003/210)*n + (622/45)

%F Empirical for n mod 420 = 2: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (3263/210)*n + (3998/315)

%F Empirical for n mod 420 = 3: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (2003/210)*n + (148/5)

%F Empirical for n mod 420 = 4: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (4243/210)*n - (3821/315)

%F Empirical for n mod 420 = 5: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (341/70)*n + (3580/63)

%F Empirical for n mod 420 = 6: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (4243/210)*n - (404/35)

%F Empirical for n mod 420 = 7: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (2003/210)*n + (784/45)

%F Empirical for n mod 420 = 8: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (3263/210)*n + (1187/45)

%F Empirical for n mod 420 = 9: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (2003/210)*n + (886/35)

%F Empirical for n mod 420 = 10: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (4243/210)*n - (184/9)

%F Empirical for n mod 420 = 11: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (341/70)*n + (20294/315)

%e Some solutions for n=6

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

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

%e ..4....3....3....0....2....2....3....4....2....3....4....1....2....3....5....0

%e ..2....2....5....3....6....5....5....2....5....2....2....3....4....1....6....6

%e ..2....1....3....0....3....3....4....0....4....5....5....2....2....2....5....1

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

%e ..0....5....3....0....2....2....3....4....0....3....4....1....2....3....3....2

%K nonn

%O 1,1

%A _R. H. Hardin_, Sep 18 2014