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A298211
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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.
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4
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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
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1,2
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COMMENTS
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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
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REFERENCES
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Michael J. Jacobson, Jr. and Hugh C. Williams, Solving the Pell Equation, Springer, 2009, pages 1-17.
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LINKS
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FORMULA
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a(n) <= n.
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(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)
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MATHEMATICA
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PROG
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(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)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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