

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.


5



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
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OFFSET

1,3


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
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 117.


LINKS



FORMULA

if n  m then a(n)  a(m).
a(2^(m+1)) = 2^m for m>=0.


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]]];


PROG

(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*y0y1, i+1
print(n, i)


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



