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Search: id:A100237
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| A100237 |
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Secondary diagonal of triangle A100235 divided by row number: a(n) = A100235(n+1,n)/(n+1) for n>=0. |
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3
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| 1, 4, 21, 109, 566, 2939, 15261, 79244, 411481, 2136649, 11094726, 57610279, 299146121, 1553340884, 8065850541, 41882593589, 217478818486, 1129276686019, 5863862248581, 30448587928924, 158106801893201, 820982597394929
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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G.f. equals the ratio of the g.f.s of any two adjacent diagonals of triangle A100235.
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LINKS
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Index entries for sequences related to linear recurrences with constant coefficients
Tanya Khovanova, Recursive Sequences
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FORMULA
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a(n) = 5*a(n-1) + a(n-2) for n>1, with a(0)=1, a(1)=4. G.f.: A(x)=(1-x)/(1-5*x-x^2).
Numerators in continued fraction [1, 4, 5, 5, 5,...]. Continued fraction [1, 4, 5, 5, 5,...] = .807417596433..., the inradius of a right triangle with legs 2 and 5. n-th convergent (n>0) to [1, 4, 5, 5, 5,...] = A100237(n)/A052918(n), the first few being: 1/1, 4/5, 21/26, 109/135,... - Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 21 2007
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MAPLE
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a[0]:=1: a[1]:=4: for n from 2 to 26 do a[n]:=5*a[n-1]+a[n-2] od: seq(a[n], n=0..21); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 26 2006
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PROGRAM
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(PARI) a(n)=polcoeff((1-x)/(1-5*x-x^2)+x*O(x^n), n)
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CROSSREFS
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Cf. A100234, A100235, A100236.
First differences of A052918.
Cf. A052918.
Sequence in context: A014986 A015531 A083425 this_sequence A117381 A010908 A136786
Adjacent sequences: A100234 A100235 A100236 this_sequence A100238 A100239 A100240
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KEYWORD
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nonn
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Nov 30 2004
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