OFFSET
1,1
COMMENTS
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.
From Kevin P. Thompson, May 27 2022: (Start)
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)
LINKS
FactorDB, Status of 20301625^56-20301624^56
Kevin P. Thompson, Factorizations to support known terms for n = 1..95
EXAMPLE
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.
MATHEMATICA
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Thomas Ordowski, Nov 13 2018
EXTENSIONS
a(12)-a(23) from Amiram Eldar, Nov 13 2018
a(24)-a(55) from Kevin P. Thompson, May 27 2022
STATUS
approved