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A072256
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a(n) = 10*a(n-1) - a(n-2); a(0) = a(1) = 1.
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18
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1, 1, 9, 89, 881, 8721, 86329, 854569, 8459361, 83739041, 828931049, 8205571449, 81226783441, 804062262961, 7959395846169, 78789896198729, 779939566141121, 7720605765212481, 76426118085983689, 756540575094624409
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| Any k in the sequence is followed by 5*k + 2sqrt{2(3*k^2 - 1)}. Gives solutions for x in 3*x^2 - 2*y^2 = 1. Corresponding y is given by A054320(n).
a(n) = L(n-1,10), where L is defined as in A108299; see also A054320 for L(n,-10). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Jun 01 2005
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5,6,7,8,9} which do not end in 0. - Tanya Khovanova (tanyakh(AT)yahoo.com), Jan 10 2007
a(n) = A138288(n-1) for n > 0. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Mar 12 2008
For n>= 2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(8)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. [From John M. Campbell, Jul 08 2011]
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REFERENCES
| S. J. Cyvin and I. Gutman, Kekule structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 283).
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LINKS
| Index entries for sequences related to linear recurrences with constant coefficients
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
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FORMULA
| a(n) = (3-sqrt(6))/6 * (5+2*sqrt(6))^n + (3+sqrt(6))/6 * (5-2*sqrt(6))^n.
a(n)={2*A031138(n) + 1}/3 = sqrt(2*A054320(n)^2 + 1)/3), n>=1.
a(n) = U(n-1, 5)-U(n-2, 5) = T(2*n-1, sqrt(3))/sqrt(3) with Chebyshev's U- and T- polynomials and U(-1, x) := 0, U(-2, x) := -1, T(-1, x) := x.
G.f.: (1-9*x)/(1-10*x+x^2).
For all members x of the sequence, 6*x^2 - 2 is a square. Lim. n -> Inf. a(n)/a(n-1) = 5 + 2*sqrt(6). - Gregory V. Richardson (omomom(AT)hotmail.com), Oct 10 2002
Let q(n, x)=sum(i=0, n, x^(n-i)*binomial(2*n-i, i)); then q(n, 8)=a(n+1) - Benoit Cloitre (benoit7848c(AT)orange.fr), Nov 10 2002
a(n)a(n+3) = 80 + a(n+1)a(n+2). - R. Stephan, May 29 2004
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MATHEMATICA
| a[n_] := a[n] = 10a[n - 1] - a[n - 2]; a[0] = a[1] = 1; Table[ a[n], {n, 0, 20}]
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PROG
| (Other) sage: [lucas_number1(n, 10, 1)-lucas_number1(n-1, 10, 1) for n in xrange(0, 20)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Nov 10 2009]
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CROSSREFS
| Cf. A054320, A031138.
Row 10 of array A094954.
First differences of A004189.
A072256(n)=sqrt(A046172(n)) [From Weisenhorn Paul (paulweisenhorn(AT)online.de), May 15 2009]
Sequence in context: A198967 A015584 * A138288 A059482 A109002 A142991
Adjacent sequences: A072253 A072254 A072255 * A072257 A072258 A072259
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KEYWORD
| nonn
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AUTHOR
| Lekraj Beedassy (blekraj(AT)yahoo.com), Jul 08 2002
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EXTENSIONS
| Edited by Robert G. Wilson v (rgwv(AT)rgwv.com), Jul 17 2002
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