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A038725
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a(n)=6a(n-1)-a(n-2), n >= 2, a(0)=1, a(1)=2.
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5
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1, 2, 11, 64, 373, 2174, 12671, 73852, 430441, 2508794, 14622323, 85225144, 496728541, 2895146102, 16874148071, 98349742324, 573224305873, 3340996092914, 19472752251611, 113495517416752, 661500352248901
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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REFERENCES
| I. Adler, Three Diophantine equations - Part II, Fib. Quart.,7 (1969), pps. 181-193.
A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pps. 122-125, 194-196.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7(1969), pps. 231-242.
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LINKS
| Index entries for sequences related to linear recurrences with constant coefficients
Tanya Khovanova, Recursive Sequences
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FORMULA
| a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3); a(n) = ((4-sqrt(2))/8)*(3+2*sqrt(2))^(n-1)+((4+sqrt(2))/8)*(3-2*sqrt(2))^(n-1). - Antonio A. Olivares (olivares14031(AT)yahoo.com), Mar 29 2008
Sequence satisfies -7 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 6*u*v. - Michael Somos Sep 28 2008
G.f.: (1 - 4*x) / (1 - 6*x + x^2). a(n) = (7 + a(n-1)^2) / a(n-2). - Michael Somos Sep 28 2008
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MAPLE
| a[0]:=1: a[1]:=2: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 26 2006
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MATHEMATICA
| Union[Flatten[NestList[{#[[2]], #[[3]], 6#[[3]]-#[[2]]}&, {1, 2, 11}, 25]]] (* From Harvey P. Dale, Mar 4 2011 *)
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PROG
| (PARI) {a(n) = real((3 + 2*quadgen(8))^n * (1 - quadgen(8) / 4))} /* Michael Somos Sep 28 2008 */
(PARI) {a(n) = polchebyshev(n, 1, 3) - polchebyshev(n-1, 2, 3)} /* Michael Somos Sep 28 2008 */
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CROSSREFS
| Cf. A001653 and A001541.
A038723(n) = a(-n).
Sequence in context: A080049 A126745 A179120 * A161947 A001565 A199412
Adjacent sequences: A038722 A038723 A038724 * A038726 A038727 A038728
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KEYWORD
| easy,nonn
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AUTHOR
| Barry E. Williams, May 02 2000
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EXTENSIONS
| More terms from James A. Sellers (sellersj(AT)math.psu.edu), May 03 2000
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