A033318 -id:A033318 - cp's OEIS frontend

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

Eric W. Weisstein

nonn

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

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

Cf. A000037, A033313, A033318.

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 *)

A067872 Least m > 0 for which m*n^2 + 1 is a square.

Original entry on oeis.org

3, 2, 7, 3, 23, 8, 47, 15, 79, 24, 119, 2, 167, 48, 3, 63, 287, 80, 359, 6, 88, 120, 527, 28, 623, 168, 727, 12, 839, 44, 959, 255, 216, 288, 8, 20, 1367, 360, 19, 77, 1679, 22, 1847, 30, 208, 528, 2207, 7, 2399, 624, 128, 42, 2807, 728, 696, 3, 160, 840, 3479, 11

Offset: 1

Lekraj Beedassy, Feb 25 2002

Least m > 0 for which x^2 - m*y^2 = 1 has a solution with y = n.
For n > 1, a(n) <= n^2-2. - Chai Wah Wu, Jan 26 2016
Considering the greatest prime factor (gpf) of terms, two curves are evident.  One has gpf ~ n^2 (see a(9949)), the other has gpf ~ n^2/7 (see a(9923)). See the gpf plot link. - Bill McEachen, Apr 13 2025

			a(4)=3, based on 3*4^2 + 1 = 7^2.

T. D. Noe and Chai Wah Wu, Table of n, a(n) for n = 1..10000 n = 1..500 from T. D. Noe
Bill McEachen, Plot of gpf(sequence terms)

Cf. A033318, A068310.

Cf. A010052.

a067872 n = (until ((== 1) . a010052 . (+ 1)) (+ nn) nn) `div` nn
            where nn = n ^ 2
-- Reinhard Zumkeller, Jun 28 2013

a[n_] := For[m=1, True, m++, If[IntegerQ[Sqrt[m*n^2+1]], Return[m]]]; Table[a[n], {n, 100}]
lm[n_]:=Module[{m=1},While[!IntegerQ[Sqrt[m n^2+1]],m++];m]; Array[lm,60] (* Harvey P. Dale, Feb 24 2013 *)

def A067872(n):
    y, x, n2 = n*(n+2), 2*n+3, n**2
    m, r = divmod(y,n2)
    while r:
        y += x
        x += 2
        m, r = divmod(y,n2)
    return m # Chai Wah Wu, Jan 25 2016

For n a power of an odd prime, a(n) = n^2 - 2. For n twice a power of an odd prime, a(n) = (n/2)^2 - 1. - T. D. Noe, Sep 13 2007

Edited by Dean Hickerson, Mar 19 2002

A033319 Incrementally largest values of minimal y satisfying Pell equation x^2-Dy^2=1.

Original entry on oeis.org

0, 2, 4, 6, 180, 1820, 3588, 9100, 226153980, 15140424455100, 183567298683461940, 9562401173878027020, 42094239791738433660, 1238789998647218582160, 189073995951839020880499780706260

Offset: 1

Eric W. Weisstein

nonn

Records in A033317 (or A002349).

Ray Chandler, Table of n, a(n) for n = 1..63
N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references)
Eric Weisstein's World of Mathematics, Pell Equation.

Cf. A000037, A033317, A033318, A002349, A002350, A033316.

PellSolve[(m_Integer)?Positive] := Module[{cf, n, s}, cf = ContinuedFraction[Sqrt[m]]; n = Length[Last[cf]]; If[n == 0, Return[{}]]; If[OddQ[n], n = 2 n]; s = FromContinuedFraction[ ContinuedFraction[ Sqrt[m], n]]; {Numerator[s], Denominator[s]}];
yy = DeleteCases[PellSolve /@ Range[10^5], {}][[All, 2]];
Reap[Module[{y, record = 0}, Sow[0]; For[i = 1, i <= Length@yy, i++, y = yy[[i]]; If[y > record, record = y; Sow[y]]]]][[2, 1]] (* Jean-François Alcover, Nov 21 2020, after N. J. A. Sloane in A002350 *)

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

A067872 Least m > 0 for which m*n^2 + 1 is a square.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

Python

Formula

Extensions

A033319 Incrementally largest values of minimal y satisfying Pell equation x^2-Dy^2=1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica