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Search: id:A086902
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| A086902 |
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a(n) = 7a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 7, a(n) = [(7+sqrt(53))/2]^n + [(7-sqrt(53))/2]^n. |
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2
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| 2, 7, 51, 364, 2599, 18557, 132498, 946043, 6754799, 48229636, 344362251, 2458765393, 17555720002, 125348805407, 894997357851, 6390330310364, 45627309530399, 325781497023157, 2326097788692498, 16608466017870643
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OFFSET
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0,1
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COMMENT
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a(n+1)/a(n) converges to (7+sqrt(53))/2 = 7.14005... a(0)/a(1)=2/7; a(1)/a(2)=7/51; a(2)/a(3)=51/364; a(3)/a(4)=364/2599; ... etc. Lim a(n)/a(n+1) as n approaches infinity = 0.1400549... = 2/(7+sqrt(53)) = (sqrt(53)-7)/2.
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LINKS
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Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for sequences related to Chebyshev polynomials.
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FORMULA
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a(n) = 7a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 7, a(n) = [(7+sqrt(53))/2]^n + [(7-sqrt(53))/2]^n.
E.g.f. : 2exp(7x/2)cosh(sqrt(53)x/2); a(n)=2^(1-n)sum{k=0..floor(n/2), C(n, 2k)53^k7^(n-2k)}. a(n)=2T(n, 7i/2)(-i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry (pbarry(AT)wit.ie), Nov 15 2003
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EXAMPLE
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a(4) = 2599 = 7a(3) + a(2) = 7*364 + 51 = [(7+sqrt(53))/2]^4 + [(7-sqrt(53))/2]^4 =
2598.999615 + 0.000385 = 2599
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CROSSREFS
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Cf. A058316.
Sequence in context: A045598 A139008 A058721 this_sequence A138737 A046662 A118191
Adjacent sequences: A086899 A086900 A086901 this_sequence A086903 A086904 A086905
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KEYWORD
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easy,nonn
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AUTHOR
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Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Sep 18 2003
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EXTENSIONS
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More terms from Ray Chandler (rayjchandler(AT)sbcglobal.net), Feb 14 2004
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