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A327836
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Least k > 0 such that n^k == 1 (mod (n+1)^(n+1)).
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0
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1, 18, 64, 1250, 3888, 235298, 2097152, 86093442, 250000000, 51874849202, 743008370688, 46596170244962, 396857386627072, 58385852050781250, 1152921504606846976, 97322383751333736962, 273238944967337066496, 208254700595822483065682, 5242880000000000000000000
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OFFSET
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1,2
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COMMENTS
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Alternative description: For each n, a(n) gives the first k such that n^k-1 has (n+1)^(n+1) as a factor.
As n^(m*k)-1 = (n^k)^m-1 is divisible by n^k-1 for all m >= 1, all integer multiples k = m*a(n), m >= 1, also give n^k == 1 (mod (n+1)^(n+1)).
Conjecture: a(n) <= 2*(n+1)^n.
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LINKS
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EXAMPLE
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For n=2: 2^18-1 has the factor 27=3^3.
For n=3: 3^64-1 has the factor 256=2^8=4^4.
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MAPLE
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a:= n-> (t-> numtheory[order](n, t^t))(n+1):
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PROG
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(PARI) a(n) = znorder(Mod(n, (n+1)^(n+1))); \\ Daniel Suteu, Sep 27 2019
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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