A033314 - OEIS

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

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OFFSET

2,1

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.

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;