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Search: id:A094789
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| A094789 |
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Number of (s(0), s(1), ..., s(2n+1)) such that 0 < s(i) < 7 and |s(i) - s(i-1)| = 1 for i = 1,2,....,2n+1, s(0) = 1, s(2n+1) = 4. |
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+0
6
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| 1, 4, 14, 47, 155, 507, 1652, 5373, 17460, 56714, 184183, 598091, 1942071, 6305992, 20475625, 66484244, 215873462, 700937471, 2275930827, 7389902771, 23994866364, 77910846021, 252974934692, 821404463698, 2667083556359
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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In general a(n)= 2/m*Sum_{r=1..m-1} Sin(r*j*Pi/m)Sin(r*k*Pi/m)(2Cos(r*Pi/m))^(2n+1)) counts (s(0), s(1), ..., s(2n+1)) such that 0 < s(i) < m and |s(i) - s(i-1)| = 1 for i = 1,2,....,2n+1, s(0) = j, s(2n+1) = k.
With interpolated zeros (0,0,0,1,0,4,0,14,...) counts walks of length n between the start and fourth nodes on P_6. - Paul Barry (pbarry(AT)wit.ie), Jan 26 2005
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FORMULA
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a(n)=2/7*Sum_{k=1..6} Sin(Pi*k/7)Sin(4Pi*k/7)(2Cos(Pi*k/7))^(2n+1); a(n) = 5a(n-1)-6a(n-2)+a(n-3); G.f.: x(-1+x)/(-1+5x-6x^2+x^3)
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MATHEMATICA
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f[n_] := FullSimplify[ TrigToExp[(2/7)Sum[ Sin[Pi*k/7]Sin[4Pi*k/7](2Cos[Pi*k/7])^(2n + 1), {k, 1, 6}]]]; Table[ f[n], {n, 25}] (from Robert G. Wilson v Jun 18 2004)
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CROSSREFS
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Cf. A094790, A080937, A005021.
Sequence in context: A121299 A046718 A104487 this_sequence A082574 A137284 A121095
Adjacent sequences: A094786 A094787 A094788 this_sequence A094790 A094791 A094792
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KEYWORD
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nonn
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AUTHOR
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Herbert Kociemba (kociemba(AT)t-online.de), Jun 11 2004
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