A298211 Smallest n such that A001353(a(n)) == 0 (mod n), i.e., x=A001075(a(n)) and y=A001353(a(n)) is the fundamental solution of the Pell equation x^2 - 3*(n*y)^2 = 1.

1, 2, 3, 2, 3, 6, 4, 4, 9, 6, 5, 6, 6, 4, 3, 8, 9, 18, 5, 6, 12, 10, 11, 12, 15, 6, 27, 4, 15, 6, 16, 16, 15, 18, 12, 18, 18, 10, 6, 12, 7, 12, 11, 10, 9, 22, 23, 24, 28, 30, 9, 6, 9, 54, 15, 4, 15, 30, 29, 6, 30, 16, 36, 32, 6, 30, 17, 18, 33, 12, 7, 36, 18, 18, 15

Offset: 1

Author

A.H.M. Smeets, Jan 15 2018

Keywords

• nonn

Comments

The fundamental solution of the Pell equation x^2 - 3*(n*y)^2 = 1 is the smallest solution of x^2 - 3*y^2 = 1 satisfying y == 0 (mod n).

For primes p > 2, 2^p-1 is a Mersenne prime if and only if a(2^p-1) = 2^(p-1). For example, a(7) = 4, a(31) = 16, a(127) = 64, but a(2047) = 495 < 1024. - Jianing Song, Jun 02 2022

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, p.182-192.

Crossrefs

Cf. A091338, A298210, A298212.

Programs

Mathematica

With[{s = Array[ChebyshevU[-1 + #, 2] &, 75]}, Table[FirstPosition[s, k_ /; Divisible[k, n]][[1]], {n, Length@ s}]] (* Michael De Vlieger, Jan 15 2018, after Eric W. Weisstein at A001353 *)

Python

xf, yf = 2, 1
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(A038754(n)) = A038754(n).

A001075(a(n)) = A002350(3*n^2).

A001353(a(n)) = A002349(3*n^2).

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

a(3^m) = 3^m and a(2*3^m) = 2*3^m for m>=0.

In general: if p is prime and p == 3 (mod 4) then: a(n) = n iff n = p^m or n = 2*p^m, for m>=0.

a(k*A005385(n)) = a(k)*A005384(n) for n>2 and k > 0 (conjectured).

a(p) | (p-A091338(p)) for p is an odd prime. - A.H.M. Smeets, Aug 02 2018

From Jianing Song, Jun 02 2022: (Start)

a(p) | (p-A091338(p))/2 for p is an odd prime > 3.

a(p^e) = a(p)*p^(e-r) for e >= r, where r is the largest number such that a(p^r) = a(p). r can be greater than 1, for p = 2, 103, 2297860813 (Cf. A238490).

If gcd(m,n) = 1, then a(m*n) = lcm(a(m),a(n)). (End)