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%I #46 May 11 2023 18:20:16
%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).
%C 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
%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
%F From _Jianing Song_, Jun 02 2022: (Start)
%F a(p) | (p-A091338(p))/2 for p is an odd prime > 3.
%F 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).
%F If gcd(m,n) = 1, then a(m*n) = lcm(a(m),a(n)). (End)
%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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