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A249086 Number of length 1+5 0..n arrays with every six consecutive terms having two times the sum of some two elements equal to the sum of the remaining four. 1
22, 243, 1324, 5005, 14586, 36247, 78448, 154689, 281470, 482131, 784092, 1223413, 1840954, 2688375, 3822856, 5314177, 7238718, 9687259, 12757900, 16565301, 21232162, 26899543, 33717624, 41856745, 51497086, 62841147, 76101988, 91516789 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
LINKS
FORMULA
Empirical: a(n) = 3*a(n-1) - a(n-2) - 4*a(n-3) + 2*a(n-4) + 2*a(n-5) + 2*a(n-6) - 4*a(n-7) - a(n-8) + 3*a(n-9) - a(n-10).
Empirical for n mod 6 = 0: a(n) = (23/4)*n^5 - (55/4)*n^4 + (140/3)*n^3 - 20*n^2 - n + 1
Empirical for n mod 6 = 1: a(n) = (23/4)*n^5 - (55/4)*n^4 + (140/3)*n^3 - 20*n^2 - (139/4)*n + (457/12)
Empirical for n mod 6 = 2: a(n) = (23/4)*n^5 - (55/4)*n^4 + (140/3)*n^3 - 20*n^2 - n - (37/3)
Empirical for n mod 6 = 3: a(n) = (23/4)*n^5 - (55/4)*n^4 + (140/3)*n^3 - 20*n^2 - (139/4)*n + (259/4)
Empirical for n mod 6 = 4: a(n) = (23/4)*n^5 - (55/4)*n^4 + (140/3)*n^3 - 20*n^2 - n - (77/3)
Empirical for n mod 6 = 5: a(n) = (23/4)*n^5 - (55/4)*n^4 + (140/3)*n^3 - 20*n^2 - (139/4)*n + (617/12).
Empirical g.f.: x*(22 + 177*x + 617*x^2 + 1364*x^3 + 1823*x^4 + 2260*x^5 + 1135*x^6 + 880*x^7 + 3*x^8 - x^9) / ((1 - x)^6*(1 + x)^2*(1 + x + x^2)). - Colin Barker, Nov 09 2018
EXAMPLE
Some solutions for n=6:
4 3 1 3 4 4 6 3 6 1 1 1 1 3 4 6
6 2 2 6 1 3 1 1 0 2 5 4 6 1 0 2
5 1 0 6 6 6 4 6 6 0 6 0 5 0 4 0
0 5 5 5 1 4 5 2 3 5 0 1 3 2 3 4
5 6 1 5 3 2 0 0 5 2 1 3 4 5 6 2
1 1 6 2 3 2 5 3 4 2 5 0 5 4 1 4
CROSSREFS
Row 1 of A249085.
Sequence in context: A126903 A035706 A249085 * A300624 A249087 A348707
KEYWORD
nonn
AUTHOR
R. H. Hardin, Oct 20 2014
STATUS
approved

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Last modified June 17 02:48 EDT 2024. Contains 373432 sequences. (Running on oeis4.)