A270547 Denominators of r-Egyptian fraction expansion for sqrt(1/3), where r(k) = 1/(2k-1).

2, 5, 19, 909, 866969, 1391730212321, 3778000141026848185401614, 40721846314736913560423097666261109869051562216449, 2752670766792898979726597410050364093079267724156167495561617776950485371872363605942870860465128310

Offset

1,1

Comments

Suppose that r is a sequence of rational numbers r(k) <= 1 for k >= 1, and that x is an irrational number in (0,1). Let f(0) = x, n(k) = floor(r(k)/f(k-1)), and f(k) = f(k-1) - r(k)/n(k). Then x = r(1)/n(1) + r(2)/n(2) + r(3)/n(3) + ... , the r-Egyptian fraction for x.

See A269993 for a guide to related sequences.

Links

Clark Kimberling, Table of n, a(n) for n = 1..11

Eric Weisstein's World of Mathematics, Egyptian Fraction

Index entries for sequences related to Egyptian fractions

Example

sqrt(1/3) = 1/(1*2) + 1/(3*5) + 1/(5*19) + 1/(7*909) + ...

Mathematica

r[k_] := 1/(2k-1); f[x_, 0] = x; z = 10;
n[x_, k_] := n[x, k] = Ceiling[r[k]/f[x, k - 1]]
f[x_, k_] := f[x, k] = f[x, k - 1] - r[k]/n[x, k]
x = Sqrt(1/3); Table[n[x, k], {k, 1, z}]

Prog

(PARI) r(k) = 1/(2*k-1);
f(k, x) = if (k==0, x, f(k-1, x) - r(k)/a(k, x); );
a(k, x=sqrt(1/3)) = ceil(r(k)/f(k-1, x)); \\ Michel Marcus, Apr 03 2016

Crossrefs

Cf. A269993, A005408, A020760.

Sequence in context: A119563 A270398 A269997 * A177875 A187602 A260140

Adjacent sequences: A270544 A270545 A270546 * A270548 A270549 A270550

Keyword

nonn,frac,easy

Author

Clark Kimberling, Apr 02 2016

Status

approved