A079612 Largest number m such that a^n == 1 (mod m) whenever a is coprime to m.

2, 24, 2, 240, 2, 504, 2, 480, 2, 264, 2, 65520, 2, 24, 2, 16320, 2, 28728, 2, 13200, 2, 552, 2, 131040, 2, 24, 2, 6960, 2, 171864, 2, 32640, 2, 24, 2, 138181680, 2, 24, 2, 1082400, 2, 151704, 2, 5520, 2, 1128, 2, 4455360, 2, 264, 2, 12720, 2, 86184, 2, 13920

Offset: 1

Author

N. J. A. Sloane, Jan 29 2003

Keywords

• nonn

Comments

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

The numbers m with this property are the divisors of a(n) that are not divisors of a(r) for r

References

• 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.)

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16384
• Shigeki Akiyama and Hajime Kaneko, Curious congruences on cyclotomic polynomials, arXiv:2204.11267 [math.NT], 2022. See Proposition 1 p. 5-6.
• P. J. Cameron and D. A. Preece, Notes on primitive lambda-roots, 2009. See lambda*() in theorem 5.2 (b) p. 8.
• Joris van der Hoeven and Grégoire Lecerf, Sparse polynomial interpolation. Exploring fast heuristic algorithms over finite fields, Simon Fraser University (BC Canada) / Institut Polytechnique de Paris (France, 2019) hal-02382117.
• R. C. Vaughan and T. D. Wooley, Waring's problem: a survey, The function K(n), see p. 19.

Crossrefs

Cf. A006863 (bisection except for initial term); A059379 (Jordan function).

Cf. A075180, A115000, A115001, A115002, A115003.

Cf. A143407, A143408, A185633, A322315.

Programs

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

Formula

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).

From Antti Karttunen, Dec 19 2018: (Start)

a(n) = A185633(n)*(2-A000035(n)).

It also seems that for n > 1, a(n) = 2*A075180(n-1). (End)

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

Extensions

Edited by Franklin T. Adams-Watters, Dec 10 2005

Definition corrected by T. D. Noe, Aug 13 2008

Rather arbitrary term a(0) removed by Max Alekseyev, May 27 2010

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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