OFFSET
1,3
COMMENTS
When n is a square, the trivial solution x=1, y=0 is taken; otherwise we take the least x that satisfies either the +1 or -1 equation. - T. D. Noe, May 19 2007
Apparently the generating function of the sequence of the denominators of continued fraction convergents to sqrt(n) is always rational and of form p(x)/[1 - C*x^m + (-1)^m * x^(2m)], or equivalently, the denominators satisfy the linear recurrence b(n+2m) = C*b(n+m) - (-1)^m * b(n). If so, then it seems that a(n) is half the value of C for each nonsquare n, or 1. See A003285 for the conjecture regarding m. The same conjectures apply to the sequences of the numerators of continued fraction convergents to sqrt(n). - Ralf Stephan, Dec 12 2013
The conjecture is true, cf. link. - Jan Ritsema van Eck, Mar 08 2021
REFERENCES
A. Cayley, Report of a committee appointed for the purpose of carrying on the tables connected with the Pellian equation ..., Collected Mathematical Papers. Vols. 1-13, Cambridge Univ. Press, London, 1889-1897, Vol. 13, pp. 430-443.
C. F. Degen, Canon Pellianus. Hafniae, Copenhagen, 1817.
D. H. Lehmer, Guide to Tables in the Theory of Numbers. Bulletin No. 105, National Research Council, Washington, DC, 1941, p. 55.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
T. D. Noe, Table of n, a(n) for n = 1..10000
A. Cayley, Report of a committee appointed for the purpose of carrying on the tables connected with the Pellian equation ..., Collected Mathematical Papers. Vols. 1-13, Cambridge Univ. Press, London, 1889-1897, Vol. 13, pp. 430-443. (Annotated scanned copy)
Jan Ritsema van Eck, Proof of conjecture in A006702, Mar 08 2021
MATHEMATICA
r[x_, n_] := Reduce[y > 0 && (x^2 - n*y^2 == -1 || x^2 - n*y^2 == 1 ), y, Integers];
a[n_ /; IntegerQ[ Sqrt[n]]] = 1;
a[n_] := a[n] = (k = 1; While[ r[k, n] === False, k++]; k);
Table[ Print[ a[n] ]; a[n], {n, 1, 67}] (* Jean-François Alcover, Jan 30 2012 *)
nmax = 500;
nconv = 200; (* The number of convergents 'nconv' should be increased if the linear recurrence is not found for some terms. *)
a[n_] := a[n] = Module[{lr}, If[IntegerQ[Sqrt[n]], 1, lr = FindLinearRecurrence[Numerator[Convergents[Sqrt[n], nconv]]]; SelectFirst[lr, #>1&]/2]];
Table[Print[n, " ", a[n] ]; a[n], {n, 1, nmax}] (* Jean-François Alcover, Feb 22 2021 *)
CROSSREFS
KEYWORD
nonn,nice,easy
AUTHOR
EXTENSIONS
Corrected and extended by T. D. Noe, May 19 2007
STATUS
approved