

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.


4



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
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OFFSET

1,2


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).


REFERENCES

Michael J. Jacobson, Jr. and Hugh C. Williams, Solving the Pell Equation, Springer, 2009, pages 117.


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.182192.


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)  (pA091338(p)) for p is an odd prime.  A.H.M. Smeets, Aug 02 2018


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 *)


PROG

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


CROSSREFS

Cf. A091338, A298210, A298212.
Sequence in context: A278910 A235669 A118088 * A088212 A085208 A257302
Adjacent sequences: A298208 A298209 A298210 * A298212 A298213 A298214


KEYWORD

nonn


AUTHOR

A.H.M. Smeets, Jan 15 2018


STATUS

approved



