Matthias Baur - cp's OEIS frontend

User: Matthias Baur

Matthias Baur has authored 2 sequences.

A327836 Least k > 0 such that n^k == 1 (mod (n+1)^(n+1)).

1, 18, 64, 1250, 3888, 235298, 2097152, 86093442, 250000000, 51874849202, 743008370688, 46596170244962, 396857386627072, 58385852050781250, 1152921504606846976, 97322383751333736962, 273238944967337066496, 208254700595822483065682, 5242880000000000000000000, 556436858893903097274392802

Offset: 1

Matthias Baur, Sep 27 2019

nonn

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.

			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.

a:= n-> (t-> numtheory[order](n, t^t))(n+1):
seq(a(n), n=1..20);  # Alois P. Heinz, Sep 27 2019

a[n_] := MultiplicativeOrder[n, (n+1)^(n+1)];
Table[a[n], {n, 1, 19}] (* Jean-François Alcover, Feb 02 2025, after PARI code *)

a(n) = znorder(Mod(n, (n+1)^(n+1))); \\ Daniel Suteu, Sep 27 2019

More terms from Daniel Suteu, Sep 27 2019

A322985 Numbers k such that 123456789*10^k+1 is prime.

Original entry on oeis.org

1, 5, 17, 23, 25, 28, 91, 187, 287, 398, 899, 1364, 2921, 5125, 5890, 8780, 14881, 35689, 46669, 71861, 111710

Offset: 1

Matthias Baur, Jan 01 2019

a(22) > 1.3*10^5. All numbers up to this bound were sieved using newpgen and sr1sieve. Remaining numbers were checked for primality using Jean Penné's LLR application (BLS (N-1/N+1) test).

			1 is a term because 1234567891 is prime.
2 is not a term because 12345678901 is composite (it is divisible by 857).

Cf. A248349, A248350, A321806.

Select[Range@ 1400, PrimeQ[123456789*10^# + 1] &] (* Michael De Vlieger, Jan 04 2019 *)

from sympy.ntheory.primetest import isprime
for n in range(1,1000):
    if isprime(123456789*10**n+1):
        print(n, end=', ') # Stefano Spezia, Jan 05 2019

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User: Matthias Baur

A327836 Least k > 0 such that n^k == 1 (mod (n+1)^(n+1)).

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