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A089499
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a(0)=0; a(1)=1; a(2n)=4*Sum_{k=0...n-1}a(2n+1); a(2n+1)=a(2n)+a(2n-1).
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1
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0, 1, 4, 5, 24, 29, 140, 169, 816, 985, 4756, 5741, 27720, 33461, 161564, 195025, 941664, 1136689, 5488420, 6625109, 31988856, 38613965, 186444716, 225058681, 1086679440, 1311738121, 6333631924, 7645370045, 36915112104, 44560482149
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| 1, 4, 5, 24, 29, 140,...= numerators in convergents to (sqrt(8) - 2) = continued fraction [1, 4, 1, 4, 1, 4,...]; where sqrt(8) - 2 = .828427124... = the inradius of a right triangle with hypotenuse 6, legs sqrt(32) and 2. Denominators of convergents to [1, 4, 1, 4, 1, 4,...] = A041011 starting (1, 5, 6, 29, 35,...). - Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 22 2007
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FORMULA
| For n>0, a(n)=A001333(n)+A084068(n-1)*(-1)^n; e.g. 29=41-12. a(n)*a(n+1)=A046729(n); cf. A001333. a(2n+1)=A001653(n); a(2n)=A005319(n).
a(1) = 1, a(2n) = 4*a(2n-1) + a(2n-2); a(2n-1) = a(2n-2) + a(2n-3). Given the 2 X 2 matrix X = [1, 4; 1, 5], [a(2n-1), a(2n)] = top row of X^n. The sequence starting (1, 4, 5, 24, 29,...) = numerators in continued fraction [1, 4, 1, 4, 1, 4,...] = (sqrt(8) - 2) = .828427124... E.g. X^3 = [29, 140; 35, 169], where 29/35, 140/169 are convergents to (sqrt(8)-2). - Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 22 2007
a(1) = 1, a(2n) = 4*a(2n-1) + a(2n-2); a(2n-1) = a(2n-2) + a(2n-3). - Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 23 2007
a(n)=A0000129(n)*A000034(n+1). a(n)=6*a(n-2)-a(n-4). G.f.: -x*(-1-4*x+x^2)/((x^2-2*x-1)*(x^2+2*x-1)). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 08 2009]
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MATHEMATICA
| Numerator[NestList[(4/(4+#))&, 0, 60]] [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Apr 13 2010]
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CROSSREFS
| Cf. A041011.
Sequence in context: A039583 A042123 A041531 * A042601 A164054 A047168
Adjacent sequences: A089496 A089497 A089498 * A089500 A089501 A089502
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KEYWORD
| nonn
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AUTHOR
| Charlie Marion (charliem(AT)bestweb.net), Nov 11 2003
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EXTENSIONS
| Corrected by T. D. Noe (noe(AT)sspectra.com), Nov 08 2006
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