A033314 Least D in the Pellian x^2 - D*y^2 = 1 for which x has least solution n.

3, 2, 15, 6, 35, 12, 7, 5, 11, 30, 143, 42, 195, 14, 255, 18, 323, 10, 399, 110, 483, 33, 23, 39, 27, 182, 87, 210, 899, 60, 1023, 17, 1155, 34, 1295, 38, 1443, 95, 1599, 105, 1763, 462, 215, 506, 235, 138, 47, 96, 51, 26, 2703, 78, 2915, 21, 3135, 203, 3363

Offset: 2

Author

Eric W. Weisstein

Keywords

• nonn

Comments

The i-th solution pair V(i) = [x(i), y(i)] to the Pellian x^2 - D*y^2 = 1 for a given least solution x(1) = n may be generated through the recurrence V(i+2) = 2*n*V(i+1) - V(i) taking V(0) = [1, 0] and V(1) = [n, sqrt((n^2-1)/a(n))]. V(i) stands for the numerator and denominator of the 2i-th convergent of the continued fraction expansion of sqrt(D).

Thus setting n = 3, for instance, we have D = a(3) = 2 and V(1) = [3, 2] so that along with V(0) = [1, 0] recurrence V(i+2) = 6*V(i+1) - V(i) generates [A001333(2k), A000129(2k)]. Similarly, setting n = 9 generates [A023039, A060645], respectively the numerator and denominator of the 2i-th convergent of sqrt(a(9)), i.e., sqrt(5). - Lekraj Beedassy, Feb 26 2002

Links

• Ray Chandler, Table of n, a(n) for n = 2..1001
• Eric Weisstein's World of Mathematics, Pell Equation.

Crossrefs

Cf. A000037, A033313, A033318.

Programs

Mathematica

squarefreepart[n_] :=
  Times @@ Power @@@ ({#[[1]], Mod[#[[2]], 2]} & /@ FactorInteger[n]);
pellminx[d_] := Module[{q, p, z}, {q, p} = ContinuedFraction[Sqrt[d]];
  If[OddQ[p // Length], p = Join[p, p]];
  z = FromContinuedFraction[Join[{q}, Drop[p, -1]]]; Numerator[z]]
NMAX = 60; a = {};
For[n = 2, n <= NMAX, n++, s = squarefreepart[n^2 - 1];
sd = s Divisors[Sqrt[(n^2 - 1)/s]]^2;
t = Sort[Transpose[{sd, pellminx[#] & /@ sd}]];
AppendTo[a, Select[t, #[[2]] == n &, 1][[1, 1]]]
]; a (* Herbert Kociemba, Jun 05 2022 *)