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A094287
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Number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 7 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 1, s(n) = 1.
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0
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1, 1, 2, 4, 9, 21, 51, 127, 323, 835, 2188, 5798, 15510, 41822, 113531, 309937, 850118, 2340918, 6466953, 17913087, 49726649, 138287113, 385126811, 1073832695, 2996974774, 8370739326, 23394528640, 65415732100, 182989086965, 512046072481, 1433197869570
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OFFSET
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0,3
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COMMENTS
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In general, a(n) = (2/m)*Sum_{k=1..m} sin(Pi*k/m)^2(1+2*cos(Pi*k/m))^n counts the (s(0), s(1), ..., s(n)) such that 0 < s(i) < m and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 1, s(n) = 1. Here, m=7.
a(n) is the number of Motzkin n-paths of height <= 5. - Alois P. Heinz, Nov 24 2023
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LINKS
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FORMULA
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a(n) = (2/7)*Sum_{k=1..6} sin(Pi*k/7)^2(1+2*cos(Pi*k/7))^n.
Conjecture: a(n)= +6*a(n-1) -10*a(n-2) +9*a(n-4) -2*a(n-5) -a(n-6) with g.f. -x*(-1+4*x-2*x^2-5*x^3+2*x^4+x^5) / ( (x^3+3*x^2-4*x+1)*(x^3-x^2-2*x+1) ). - R. J. Mathar, Dec 20 2011
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MATHEMATICA
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f[n_] := FullSimplify[ TrigToExp[(2/7)*Sum[ Sin[Pi*k/7]^2(1 + 2Cos[Pi*k/7])^n, {k, 1, 6}]]]; Table[ f[n], {n, 28}] (* Robert G. Wilson v, Jun 18 2004 *)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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