A166062 a(n) = denominator(Bernoulli(prime(n) - 1)).

2, 6, 30, 42, 66, 2730, 510, 798, 138, 870, 14322, 1919190, 13530, 1806, 282, 1590, 354, 56786730, 64722, 4686, 140100870, 3318, 498, 61410, 4501770, 33330, 4326, 642, 209191710, 1671270, 4357878, 8646, 4110, 274386, 4470, 2162622, 1794590070, 130074

Offset: 1

Author

Paul Curtz, Oct 05 2009

Keywords

• nonn

Comments

Divisibility through terms of A008578 is a consequence of the Staudt-Clausen theorem.

(Vaguely similar divisibility properties are considered in A165248 and A165943.)

The first 250 entries are all different. Is this true in general?

Would sorting the entries yield the full A090801?

a(n) > 1 is the largest number k such that x*y^p == y*x^p (mod k) for all integers x and y, where p = prime(n). Example: x*y^19 == y*x^19 (mod 798). - Michel Lagneau, Apr 19 2012

Comment from Herbert Kociemba, May 29 2020: (Start)

For each n there is exactly one member of the sequence whose factorization has prime(n) as its largest prime factor, namely a(n). From this we conclude:

1. All elements of the sequence are different.

2. Not all denominators of Bernoulli numbers appear in this sequence. For example the denominator of B(20), 330=2*3*5*11 never appears because the unique sequence element with largest prime divisor 11=prime(5) is a(5)=2*3*11. (End)

Links

• Peter Luschny, Table of n, a(n) for n = 1..1000
• Eric Weisstein's World of Mathematics, von Staudt-Clausen Theorem.

Crossrefs

Cf. A071772, A110936.

Programs

Maple

seq(denom(bernoulli(ithprime(n)-1)), n=1..38); # Peter Luschny, Jul 14 2019

Mathematica

Table[Denominator[BernoulliB[n - 1]], {n, Prime[Range[38]]}] (* Harvey P. Dale, Apr 22 2012 *)
Table[GCD @@ Table[(n^k - n), {n, 2, 13}], {k, Prime[Range[100]]}] (* Increase n to 80 and k to 1000 for first thousand terms. - Herbert Kociemba, May 05 2020 *)
a[i_] := Times @@ Select[Prime[Range[i]], Mod[Prime[i] - 1, # - 1] == 0&]; Table[a[i], {i, 1, 100}](* Herbert Kociemba, May 06 2020 *)

PARI

a(n)=denominator(bernfrac(prime(n)-1)) \\ Charles R Greathouse IV, Apr 30 2012

Formula

a(n) = A027642(A008578(n) - 1).

Extensions

Edited by Peter Luschny, Jul 14 2019