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 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

%I

%S 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,

%T 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,

%U 4,15,30,29,6,30,16,36,32,6,30,17,18,33,12,7,36,18,18,15

%N 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.

%C 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).

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

%H A.H.M. Smeets, <a href="/A298211/b298211.txt">Table of n, a(n) for n = 1..20000</a>

%H H. W. Lenstra Jr., <a href="http://www.ams.org/notices/200202/fea-lenstra.pdf">Solving the Pell Equation</a>, Notices of the AMS, Vol.49, No.2, Feb. 2002, p.182-192.

%F a(n) <= n.

%F a(A038754(n)) = A038754(n).

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

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

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

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

%F 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.

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

%F a(p) | (p-A091338(p)) for p is an odd prime. - _A.H.M. Smeets_, Aug 02 2018

%t 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 *)

%o Python:

%o xf, yf = 2, 1

%o x, n = 2*xf, 0

%o while n < 20000:

%o ....n = n+1

%o ....y1, y0, i = 0, yf, 1

%o ....while y0%n != 0:

%o ........y1, y0, i = y0, x*y0-y1, i+1

%o ....print(n, i)

%Y Cf. A091338, A298210, A298212.

%K nonn

%O 1,2

%A _A.H.M. Smeets_, Jan 15 2018

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