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A321576
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a(n) is the smallest b > 1 such that b^n - (b-1)^n has all divisors d == 1 (mod n).
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1
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2, 2, 2, 3, 2, 4, 2, 45, 3, 6, 2, 301, 2, 15, 10, 121, 2, 64, 2, 2101, 7, 12, 2, 1900081, 6, 27, 18, 225, 2, 9241, 2, 31825, 12, 52, 31, 537850405, 2, 96, 26, 13568281, 2, 232, 2, 35421, 486, 24, 2, 4164776161, 7, 2101, 68, 10765, 2, 145180, 1925
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OFFSET
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1,1
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COMMENTS
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For n > 1, a(n) is the least b > 1 such that b^n - (b-1)^n has all prime divisors p == 1 (mod n).
If n is prime, then a(n) = 2. Conjecture: If n is composite, then a(n) > 2.
Sequence continues for n = 56..95 (unconfirmed terms marked with a '?'): 20301625?, 171, 30, 2, ?, 2, 156, 18298, 405825?, 442, 361285?, 2, 8365, 553, 392106?, 2, ?, 2, 75, 4975?, 31351?, 1914, 247339?, 2, ?, 1513?, 42, 2, ?, 391, 87, 406?, ?, 2, ?, 39, ?, 63, 142, 145
a(60) > 1.3831*10^10.
a(72) > 1.34*10^8.
a(80) > 10^8.
a(84) > 2.29*10^8.
a(88) > 10^7.
a(90) > 10^8.
a(92) > 10^6. (End)
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LINKS
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EXAMPLE
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a(6) = 4 since b^n - (b-1)^n = 4^6 - 3^6 = 3367 has divisors 1, 7, 13, 37, 91, 259, 481, and 3367, each of which is congruent to 1 (mod 6), and b = 4 is the smallest such number satisfying this requirement.
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MATHEMATICA
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primes[n_]:=First@# & /@ FactorInteger[n]; bQ[m_, n_]:=AllTrue[primes[m] -1, Divisible[#, n]&] ; a[n_]:=Module[{b=2}, While[!bQ[b^n - (b-1)^n, n], b++]; b]; Array[a, 100] (* Amiram Eldar, Nov 13 2018 *)
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PROG
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(PARI) A321576(n)=if(n<4||isprime(n), 2, for(b=2, oo, Set(factor(b^n-(b-1)^n)[, 1]%n)==[1]&&return(b))) \\ M. F. Hasler, Nov 18 2018
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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