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

Author

N. J. A. Sloane, Walter Gilbert

Keywords

• nonn

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

Examples

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

Links

• Kevin Ryde, Table of n, a(n) for n = 1..6000
• Kevin Ryde, PARI/GP Code and Notes

Crossrefs

Cf. A000045 (Fibonacci).

Programs

Mathematica

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

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