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A247536 Number of length 4+3 0..n arrays with some disjoint pairs in every consecutive four terms having the same sum.

%I A247536 #6 Jul 23 2025 11:36:01
%S A247536 8,81,364,1007,2164,3997,6584,10219,14852,20847,28108,37095,47564,
%T A247536 60087,74428,91101,109760,131243,154956,181677,211024,243709,279136,
%U A247536 318445,360676,406933,456648,510683,568172,630613,696744,767859,843244,923955
%N A247536 Number of length 4+3 0..n arrays with some disjoint pairs in every consecutive four terms having the same sum.
%C A247536 Row 4 of A247533
%H A247536 R. H. Hardin, <a href="/A247536/b247536.txt">Table of n, a(n) for n = 1..349</a>
%F A247536 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 A247536 Also as a cubic plus a linear quasipolynomial with period 420, first 12 listed:
%F A247536 Empirical for n mod 420 = 0: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (4243/210)*n + 1
%F A247536 Empirical for n mod 420 = 1: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (2003/210)*n + (622/45)
%F A247536 Empirical for n mod 420 = 2: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (3263/210)*n + (3998/315)
%F A247536 Empirical for n mod 420 = 3: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (2003/210)*n + (148/5)
%F A247536 Empirical for n mod 420 = 4: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (4243/210)*n - (3821/315)
%F A247536 Empirical for n mod 420 = 5: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (341/70)*n + (3580/63)
%F A247536 Empirical for n mod 420 = 6: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (4243/210)*n - (404/35)
%F A247536 Empirical for n mod 420 = 7: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (2003/210)*n + (784/45)
%F A247536 Empirical for n mod 420 = 8: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (3263/210)*n + (1187/45)
%F A247536 Empirical for n mod 420 = 9: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (2003/210)*n + (886/35)
%F A247536 Empirical for n mod 420 = 10: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (4243/210)*n - (184/9)
%F A247536 Empirical for n mod 420 = 11: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (341/70)*n + (20294/315)
%e A247536 Some solutions for n=6
%e A247536 ..2....1....3....6....3....3....2....0....4....5....5....0....2....4....5....1
%e A247536 ..0....0....5....3....5....6....4....2....3....6....1....2....0....0....4....5
%e A247536 ..4....3....3....0....2....2....3....4....2....3....4....1....2....3....5....0
%e A247536 ..2....2....5....3....6....5....5....2....5....2....2....3....4....1....6....6
%e A247536 ..2....1....3....0....3....3....4....0....4....5....5....2....2....2....5....1
%e A247536 ..0....4....5....3....5....4....6....2....1....0....1....0....4....0....4....5
%e A247536 ..0....5....3....0....2....2....3....4....0....3....4....1....2....3....3....2
%K A247536 nonn
%O A247536 1,1
%A A247536 _R. H. Hardin_, Sep 18 2014