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 A249234 Number of length 1+5 0..n arrays with no six consecutive terms having two times the sum of any two elements equal to the sum of the remaining four. 1
 42, 486, 2772, 10620, 32070, 81402, 183696, 376752, 718530, 1289430, 2201892, 3603396, 5688582, 8702250, 12954360, 18823392, 26773506, 37358622, 51242100, 69200820, 92147742, 121136346, 157385352, 202283880, 257418690, 324579342 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 LINKS R. H. Hardin, Table of n, a(n) for n = 1..92 FORMULA Empirical: a(n) = 4*a(n-1) - 4*a(n-2) - 3*a(n-3) + 6*a(n-4) - 6*a(n-7) + 3*a(n-8) + 4*a(n-9) - 4*a(n-10) + a(n-11). Empirical for n mod 6 = 0: a(n) = n^6 + (1/4)*n^5 + (115/4)*n^4 - (80/3)*n^3 + 35*n^2 + 7*n Empirical for n mod 6 = 1: a(n) = n^6 + (1/4)*n^5 + (115/4)*n^4 - (80/3)*n^3 + 35*n^2 + (163/4)*n - (445/12) Empirical for n mod 6 = 2: a(n) = n^6 + (1/4)*n^5 + (115/4)*n^4 - (80/3)*n^3 + 35*n^2 + 7*n + (40/3) Empirical for n mod 6 = 3: a(n) = n^6 + (1/4)*n^5 + (115/4)*n^4 - (80/3)*n^3 + 35*n^2 + (163/4)*n - (255/4) Empirical for n mod 6 = 4: a(n) = n^6 + (1/4)*n^5 + (115/4)*n^4 - (80/3)*n^3 + 35*n^2 + 7*n + (80/3) Empirical for n mod 6 = 5: a(n) = n^6 + (1/4)*n^5 + (115/4)*n^4 - (80/3)*n^3 + 35*n^2 + (163/4)*n - (605/12). Empirical g.f.: 6*x*(7 + 53*x + 166*x^2 + 267*x^3 + 314*x^4 + 167*x^5 + 266*x^6 + 53*x^7 + 147*x^8) / ((1 - x)^7*(1 + x)^2*(1 + x + x^2)). - Colin Barker, Nov 09 2018 EXAMPLE Some solutions for n=7: 6 4 6 2 2 0 6 2 4 0 6 4 6 2 1 4 0 2 2 4 5 6 2 5 7 2 6 4 6 1 4 0 1 6 6 2 4 7 4 0 1 2 6 7 1 5 2 0 4 5 5 7 1 0 2 0 2 6 6 0 4 1 0 5 1 7 2 2 0 4 6 6 3 6 4 6 2 6 7 3 5 2 5 6 2 7 2 3 3 3 6 7 7 2 0 2 CROSSREFS Row 1 of A249233. Sequence in context: A204566 A348833 A249233 * A250324 A086944 A263289 Adjacent sequences: A249231 A249232 A249233 * A249235 A249236 A249237 KEYWORD nonn AUTHOR R. H. Hardin, Oct 23 2014 STATUS approved

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Last modified May 28 12:54 EDT 2024. Contains 372913 sequences. (Running on oeis4.)