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A200724
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Expansion of 1/(1-35*x+x^2).
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3
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1, 35, 1224, 42805, 1496951, 52350480, 1830769849, 64024594235, 2239030028376, 78302026398925, 2738331893933999, 95763314261291040, 3348977667251252401, 117118455039532542995, 4095796948716387752424, 143235774750034038791845, 5009156319302474969962151
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OFFSET
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0,2
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COMMENTS
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A Diophantine property of these numbers: (a(n+1)-a(n-1))^2 - 1221*a(n)^2 = 4. (See also comment in A200441.)
a(n) equals the number of 01-avoiding words of length n on alphabet {0,1,...,34}. - Milan Janjic, Jan 26 2015
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LINKS
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Bruno Berselli, Table of n, a(n) for n = 0..500
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (35,-1).
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FORMULA
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G.f.: 1/(1-35*x+x^2).
a(n) = 35*a(n-1)-a(n-2) with a(0)=1, a(1)=35.
a(n) = -a(-n-2) = (t^(n+1)-1/t^(n+1))/(t-1/t) where t=(35+sqrt(1221))/2.
a(n) = sum((-1)^k*binomial(n-k, k)*35^(n-2k), k=0..floor(n/2)).
a(n) = Sum_{k, 0<=k<=n} A101950(n,k)*34^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = 1/33*(33 + sqrt(1221)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = 1/70*(33 + sqrt(1221)). - Peter Bala, Dec 23 2012
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MATHEMATICA
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LinearRecurrence[{35, -1}, {1, 35}, 17]
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PROG
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(PARI) Vec(1/(1-35*x+x^2)+O(x^17))
(Magma) Z<x>:=PolynomialRing(Integers()); N<r>:=NumberField(x^2-1221); S:=[(((35+r)/2)^n-1/((35+r)/2)^n)/r: n in [1..17]]; [Integers()!S[j]: j in [1..#S]];
(Maxima) makelist(sum((-1)^k*binomial(n-k, k)*35^(n-2*k), k, 0, floor(n/2)), n, 0, 16);
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CROSSREFS
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Cf. A029547, A144128.
Sequence in context: A170754 A218737 A126158 * A207492 A207889 A208112
Adjacent sequences: A200721 A200722 A200723 * A200725 A200726 A200727
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KEYWORD
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nonn,easy
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AUTHOR
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Bruno Berselli, Nov 21 2011
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STATUS
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approved
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