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A321575
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Nexus primary pretenders: a(n) is the smallest composite k such that n^k - (n-1)^k == 1 (mod k).
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1
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9, 4, 341, 4, 6, 4, 9, 4, 14, 4, 6, 4, 9, 4, 21, 4, 6, 4, 9, 4, 15, 4, 6, 4, 9, 4, 10, 4, 6, 4, 9, 4, 62, 4, 6, 4, 9, 4, 49, 4, 6, 4, 9, 4, 33, 4, 6, 4, 9, 4, 14, 4, 6, 4, 9, 4, 10, 4, 6, 4, 9, 4, 65, 4, 6, 4, 9, 4, 49, 4, 6, 4, 9, 4, 111, 4, 6, 4, 9, 4, 15
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OFFSET
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0,1
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COMMENTS
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The sequence is bounded, namely a(n) <= 561 (the smallest Carmichael number), since if n^k == n (mod k) and (n-1)^k == n-1 (mod k), then n^k - (n-1)^k == 1 (mod k).
Problem: find all distinct terms of the sequence. Is this sequence periodic like the primary pretenders?
Note that a(n) > 9 if and only if n == 2 (mod 6). We have a(6m+2) = 341, 14, 21, 15, 10, 62, 49, 33, 14, 10, 65, 49, 111, 15, 10, ... for m >= 0. Found a(n) = 561 for the smallest n = 6*70+2 = 422.
Since a(n) depends only on the residues of n mod k for composites k <= 561, it must be periodic with period at most the lcm of those composites.
Up to n=2*10^6, the last term to appear for the first time is 478 = a(184748).
Conjecture: the only terms of the sequence that are not squarefree are 4, 9 and 49. (End)
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LINKS
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FORMULA
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a(n) = 4 iff n == 1,3,5 (mod 6), thus n is odd.
a(n) = 6 iff n == 4 (mod 6).
a(n) = 9 iff n == 0 (mod 6).
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MAPLE
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Comps:= remove(isprime, [$4..561]):
f:= proc(n) local k;
for k in Comps do if n&^k - (n-1)&^k - 1 mod k = 0 then return k fi od
end proc:
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MATHEMATICA
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a[n_]:=Module[{k=4}, While[PrimeQ[k] || Mod[n^k-(n-1)^k, k]!=1, k++]; k]; Array[a, 100, 0] (* Amiram Eldar, Nov 13 2018 *)
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PROG
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(PARI) a(n)=forcomposite(k=4, , Mod(n, k)^k-Mod(n-1, k)^k==1&&return(k)) \\ M. F. Hasler, Nov 13 2018
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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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