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A079612
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Largest number m such that a^n == 1 (mod m) whenever a is coprime to m.
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11
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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
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OFFSET
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1,1
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COMMENTS
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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<n.
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REFERENCES
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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.)
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LINKS
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FORMULA
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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).
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
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PROG
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(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
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CROSSREFS
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Cf. A006863 (bisection except for initial term); A059379 (Jordan function).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Definition corrected by T. D. Noe, Aug 13 2008
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STATUS
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approved
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