A107901 Length of continued fraction for prime(n)/n.

1, 2, 3, 3, 2, 2, 3, 4, 4, 3, 4, 2, 3, 2, 3, 3, 3, 5, 4, 5, 3, 5, 6, 5, 4, 5, 5, 6, 4, 5, 3, 4, 5, 3, 4, 3, 3, 4, 5, 3, 6, 4, 5, 6, 6, 3, 3, 7, 8, 8, 5, 5, 6, 6, 4, 6, 7, 4, 7, 5, 7, 7, 5, 4, 6, 4, 5, 5, 3, 3, 4, 3, 3, 4, 4, 3, 3, 3, 3, 4, 6, 4, 4, 4, 3, 7, 5, 5, 6, 4, 6, 5, 5, 4, 4, 5, 3, 4, 6, 7, 5, 5, 5, 4, 6

Offset

1,2

Links

David A. Corneth, Table of n, a(n) for n = 1..10000

Eric Weisstein's World of Mathematics, Continued Fraction.

Formula

a(n) = A107898(n) - 1.

Example

From David A. Corneth, Jan 09 2024: (Start)
a(10) = 3 as prime(10)/10 = 29/10 whose continued fraction is [2; 1, 9] having length 3
a(18) = 5 as prime(18)/18 = 61/18 whose continued fraction is [3; 2, 1, 1, 3] having length 5. (End)

Mathematica

A107901[n_]=Length[ContinuedFraction[Prime[n]/n]

Prog

(PARI) a(n) = #contfrac(prime(n)/n) \\ David A. Corneth, Jan 09 2024
(PARI) first(n) = my(res = primes(n)); for(i = 1, n, res[i] = #contfrac(res[i]/i)); res \\ David A. Corneth, Jan 09 2024

Crossrefs

Cf. A107898, A107899, A107900.

Sequence in context: A030620 A348241 A110764 * A334236 A030423 A130631

Adjacent sequences: A107898 A107899 A107900 * A107902 A107903 A107904

Keyword

nonn

Author

Zak Seidov, May 26 2005

Status

approved