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%I #47 Apr 26 2022 03:03:12
%S 2,24,2,240,2,504,2,480,2,264,2,65520,2,24,2,16320,2,28728,2,13200,2,
%T 552,2,131040,2,24,2,6960,2,171864,2,32640,2,24,2,138181680,2,24,2,
%U 1082400,2,151704,2,5520,2,1128,2,4455360,2,264,2,12720,2,86184,2,13920
%N Largest number m such that a^n == 1 (mod m) whenever a is coprime to m.
%C a(m) divides the Jordan function J_m(n) for all n except when n is a prime dividing a(m) or m=2, n=4; it is the largest number dividing all but finitely many values of J_m(n). For m > 0, a(m) also divides Sum_{k=1}^n J_m(k) for n >= the largest exceptional value. - _Franklin T. Adams-Watters_, Dec 10 2005
%C The numbers m with this property are the divisors of a(n) that are not divisors of a(r) for r<n.
%D R. C. Vaughan and T. D. Wooley, Waring's problem: a survey, pp. 285-324 of Surveys in Number Theory (Urbana, May 21, 2000), ed. M. A. Bennett et al., Peters, 2003. (The function K(n), see p. 303.)
%H Antti Karttunen, <a href="/A079612/b079612.txt">Table of n, a(n) for n = 1..16384</a>
%H Shigeki Akiyama and Hajime Kaneko, <a href="https://arxiv.org/abs/2204.11267">Curious congruences on cyclotomic polynomials</a>, arXiv:2204.11267 [math.NT], 2022. See Proposition 1 p. 5-6.
%H P. J. Cameron and D. A. Preece, <a href="http://www.maths.qmul.ac.uk/~pjc/csgnotes/lambda.pdf">Notes on primitive lambda-roots</a>, 2009. See lambda*() in theorem 5.2 (b) p. 8.
%H Joris van der Hoeven and Grégoire Lecerf, <a href="https://hal.archives-ouvertes.fr/hal-02382117">Sparse polynomial interpolation. Exploring fast heuristic algorithms over finite fields</a>, Simon Fraser University (BC Canada) / Institut Polytechnique de Paris (France, 2019) hal-02382117.
%H R. C. Vaughan and T. D. Wooley, <a href="https://www.people.maths.bris.ac.uk/~matdw/2002%20wps.pdf">Waring's problem: a survey</a>, The function K(n), see p. 19.
%F a(n) = 2 for n odd; for n even, a(n) = product of 2^(t+2) (where 2^t exactly divides n) and p^(t+1) (where p runs through all odd primes such that p-1 divides n and p^t exactly divides n).
%F From _Antti Karttunen_, Dec 19 2018: (Start)
%F a(n) = A185633(n)*(2-A000035(n)).
%F It also seems that for n > 1, a(n) = 2*A075180(n-1). (End)
%F We have 2*A075180(2n-1) = A006863(n) by definition, and A006863(n) = a(2n) by the comments in A006863. Hence a(n) = 2*A075180(n-1) for all even n. For all odd n > 1, we have a(n) = 2, which is also equal to 2*A075180(n-1). So the formula above is true. - _Jianing Song_, Apr 05 2021
%o (PARI) a(n) = {if (n%2, 2, res = 1; forprime(p=2, n+1, if (!(n % (p-1)), t = valuation(n, p); if (p==2, if (t, res *= p^(t+2)), res *= p^(t+1)););); res;);} \\ _Michel Marcus_, May 12 2018
%Y Cf. A006863 (bisection except for initial term); A059379 (Jordan function).
%Y Cf. A075180, A115000, A115001, A115002, A115003.
%Y Cf. A143407, A143408, A185633, A322315.
%K nonn
%O 1,1
%A _N. J. A. Sloane_, Jan 29 2003
%E Edited by _Franklin T. Adams-Watters_, Dec 10 2005
%E Definition corrected by _T. D. Noe_, Aug 13 2008
%E Rather arbitrary term a(0) removed by _Max Alekseyev_, May 27 2010
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