A010342 Numbers k such that all terms in the periodic part of the continued fraction for sqrt(k) except the final term are 1.

3, 7, 8, 13, 15, 24, 32, 35, 48, 58, 63, 74, 75, 80, 99, 120, 135, 136, 143, 168, 185, 195, 215, 224, 255, 288, 312, 323, 346, 360, 399, 425, 427, 440, 483, 528, 557, 560, 575, 624, 675, 711, 728, 783, 818, 819, 840, 880, 899, 960

Offset

1,1

Comments

Theorem: If (b-1)/(2q-1) = F(m)/F(m+1) then sqrt(q^2+b) = [q;1,1,...,1,1,2q,...], where F(m) are the Fibonacci numbers and the period contains m ones. - Thomas Ordowski, Jun 09 2012

Terms are all and only k = ((d*F(m+1) + 1)/2)^2 + d*F(m) + 1 for d>=1 odd, and m>=1 with m == 0 or 1 (mod 3) (so F(m+1) odd), and consequently lim_{n->oo} n/sqrt(a(n)) = A360957 - 1 = 1.696383... - Kevin Ryde, Mar 07 2023

Links

Kevin Ryde, Table of n, a(n) for n = 1..6000

Kevin Ryde, PARI/GP Code and Notes

Formula

sqrt(k) = [q;1,1,...,1,1,2q,...] = sqrt(q^2+b), where (2q-1)/(b-1) = F(m+1)/F(m) for m=1,3,4,6,7,9,10,12,13,... The period contains m ones. F(m) is the m-th Fibonacci number. Note that this formula does not generate all terms of this sequence. - Thomas Ordowski, Jun 08 2012

sqrt(k) = [q;1,1,...,1,1,2q,...] with m ones in its repeating continued fraction expansion precisely when q=floor(sqrt(k)) and k=q^2+2q*F(m)/F(m+1)+F(m-1)/F(m+1). Such k are integral precisely when 2q-1 is divisible by F(m+1). - Gary Walsh, Jan 06 2023

Example

(2q-1)/(b-1) = 1/1 so b=2q. Let q=1, b=2; k = q^2 + b = 3.

Mathematica

fQ[n_] := Union@ Most@ Last@ ContinuedFraction@ Sqrt[1/n] == {1}; Select[ Range@ 1000, fQ] (* Robert G. Wilson v, Jun 07 2012 *)

Crossrefs

Cf. A000045 (Fibonacci).

Sequence in context: A050014 A132017 A244592 * A108873 A002312 A252496

Adjacent sequences: A010339 A010340 A010341 * A010343 A010344 A010345

Keyword

nonn

Author

N. J. A. Sloane, Walter Gilbert

Status

approved