A079936 Greedy frac multiples of sqrt(5): a(1)=1, sum(n>0,frac(a(n)*x))=1 at x=sqrt(5).

1, 2, 5, 13, 17, 34, 305, 610, 1597, 4181, 5473, 10946, 98209, 196418, 514229, 1346269, 1762289, 3524578, 31622993, 63245986, 165580141, 433494437, 567451585, 1134903170, 10182505537, 20365011074, 53316291173, 139583862445

Offset: 1

Author

Benoit Cloitre and Paul D. Hanna, Jan 21 2003

Keywords

• nonn
• easy

Comments

The n-th greedy frac multiple of x is the smallest integer that does not cause sum(k=1..n,frac(a(k)*x)) to exceed unity; an infinite number of terms appear as the denominators of the convergents to the continued fraction of x.

Examples

			a(4) = 13 since frac(1x) + frac(2x) + frac(5x) + frac(13x) < 1, while frac(1x) + frac(2x) + frac(5x) + frac(k*x) > 1 for all k>5 and k<13.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,322,0,0,0,0,0,-1).

Crossrefs

Cf. A001076 (denominators of convergents to sqrt(5)), A079934, A079935, A079937.

Formula

For n>=0, a(6n+1)=A001076(4n+1); a(6n+2)=2a(6n+1); a(6n+3)=A001076(4n+1)+A001076(4n+2); a(6n+4)=A001076(4n+3)-A001076(4n+2); a(6n+5)=A001076(4n+3); a(6n+6)=2a(6n+5). Asymptotics: a(6n) -> 2*sqrt(5)*(tau)^(12n-3); a(6n+2)/a(6n+1) -> (tau)^2; a(6n+3)/a(6n+2) -> (tau)^2; a(6n+4)/a(6n+3) -> (tau)^2/2; a(6n+6)/a(6n+5) -> (tau)^6/2; where tau = (1+sqrt(5))/2.

G.f.: -x*(x -1)*(2*x^10 +3*x^9 +8*x^8 +21*x^7 +55*x^6 +72*x^5 +38*x^4 +21*x^3 +8*x^2 +3*x +1) / (x^12 -322*x^6 +1). - Colin Barker, Jun 16 2013