0,3

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

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]

Cf. A041011.

Sequence in context: A039583 A042123 A041531 this_sequence A042601 A164054 A047168

Adjacent sequences: A089496 A089497 A089498 this_sequence A089500 A089501 A089502

nonn

Charlie Marion (charliem(AT)bestweb.net), Nov 11 2003

Corrected by T. D. Noe (noe(AT)sspectra.com), Nov 08 2006