cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A298210 Smallest n such that A001542(a(n)) == 0 (mod n), i.e., x=A001541(a(n)) and y=A001542(a(n)) is the fundamental solution of the Pell equation x^2 - 2*(n*y)^2 = 1.

Original entry on oeis.org

1, 1, 2, 2, 3, 2, 3, 4, 6, 3, 6, 2, 7, 3, 6, 8, 4, 6, 10, 6, 6, 6, 11, 4, 15, 7, 18, 6, 5, 6, 15, 16, 6, 4, 3, 6, 19, 10, 14, 12, 5, 6, 22, 6, 6, 11, 23, 8, 21, 15, 4, 14, 27, 18, 6, 12, 10, 5, 10, 6, 31, 15, 6, 32, 21, 6, 34, 4, 22, 3, 35, 12, 18, 19, 30
Offset: 1

Views

Author

A.H.M. Smeets, Jan 15 2018

Keywords

Comments

The fundamental solution of the Pell equation x^2 - 2*(n*y)^2 = 1, is the smallest solution of x^2 - 2*y^2 = 1 satisfying y == 0 (mod n).
If n is prime (i.e., n in A000040) then a(n) divides (n - Legendre symbol (n/2)); the Legendre symbol (n/2), or more general Kronecker symbol (n/2) is A091337(n). - A.H.M. Smeets, Jan 23 2018
From A.H.M. Smeets, Jan 23 2018: (Start)
Stronger, but conjectured:
If n is prime (i.e., in A000040) and n in {2,3,5,7,11,13,19,23} (mod 24) then (n - Legendre symbol (n/2)) / a(n) == 2 (mod 4).
If n is a safe prime (i.e., in A005385) and n in {7,23} (mod 24) then (n - Legendre symbol (n/2)) / a(n) = 2, i.e., a(n) is a Sophie Germain prime (A005384).
If n is prime (i.e., in A000040) and n in {1,17} (mod 24) then (n - Legendre symbol (n/2)) / a(n) == 0 (mod 4). (End)

References

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

Links

Crossrefs

Cf. A298211, A298212.

Programs

  • Mathematica
    b[n_] := b[n] = Switch[n, 0, 0, 1, 2, _, 6 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 = 3, 2
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) <= A000010(n) < n. - A.H.M. Smeets, Jan 23 2018
A001541(a(n)) = A002350(2*n^2).
A001542(a(n)) = A002349(2*n^2).
if n | m then a(n) | a(m).
a(2^(m+1)) = 2^m for m>=0.

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.

Original entry on oeis.org

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

Views

Author

A.H.M. Smeets, Jan 15 2018

Keywords

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

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)
Showing 1-2 of 2 results.

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