A298212 Smallest n such that A060645(a(n)) = 0 (mod n), i.e., x=A023039(a(n)) and y=A060645(a(n)) is the fundamental solution of the Pell equation x^2 - 5*(n*y)^2 = 1.

1, 1, 2, 1, 5, 2, 4, 2, 2, 5, 5, 2, 7, 4, 10, 4, 3, 2, 3, 5, 4, 5, 4, 2, 25, 7, 6, 4, 7, 10, 5, 8, 10, 3, 20, 2, 19, 3, 14, 10, 10, 4, 22, 5, 10, 4, 8, 4, 28, 25, 6, 7, 9, 6, 5, 4, 6, 7, 29, 10, 5, 5, 4, 16, 35, 10, 34, 3, 4, 20, 35, 2, 37, 19, 50

Offset: 1

Author

A.H.M. Smeets, Jan 15 2018

Keywords

• nonn

Comments

The fundamental solution of the Pell equation x^2 - 5*(n*y)^2 = 1, is the smallest solution of x^2 - 5*y^2 = 1 satisfying y = 0 (mod n).

References

• Michael J. Jacobson, Jr. and Hugh C. Williams, Solving the Pell Equation, Springer, 2009, pages 1-17.

Links

• A.H.M. Smeets, Table of n, a(n) for n = 1..20000
• H. W. Lenstra Jr., Solving the Pell Equation, Notices of the AMS, Vol.49, No.2, Feb. 2002, pp. 182-192.

Crossrefs

Cf. A298210, A298211.

Programs

Mathematica

b[n_] := b[n] = Switch[n, 0, 0, 1, 4, _, 18 b[n - 1] - b[n - 2]];
a[n_] := For[k = 1, True, k++, If[Mod[b[k], n] == 0, Return[k]]];
a /@ Range[100] (* Jean-François Alcover, Nov 16 2019 *)

Python

xf, yf = 9, 4
x, n = 2*xf, 0
while n < 20000:
    n = n+1
    y1, y0, i = 0, yf, 1
    while y0%n != 0:
        y1, y0, i = y0, x*y0-y1, i+1
    print(n, i)

Formula

a(n) <= n.

a(A000351(n)) = A000351(n).

A023039(a(n)) = A002350(5*n^2).

A060645(a(n)) = A002349(5*n^2).

if n | m then a(n) | a(m).

a(5^m) = 5^m for m>=0.

In general: if p is prime and p = 1 (mod 4) then: a(n) = n iff n = p^m, for m>=0.