Hello!
The first primes p satisfying
lambda(Den(B_{p-1})/p) = p - 1
are
2, 13, 31, 37, 61, 67, 113, 127, ...
Recall that, for n > 1, the divisibility
n | Den(B_{n-1})
holds exactly when n is either a prime or a Carmichael number. Indeed, by the von Staudt–Clausen theorem this is equivalent, for composite n, to Korselt's criterion.
There is a related, more general condition
lambda(Den(B_{n-1})) = n - 1.
This condition holds for every prime n and for many composite n, but according to Pomerance, the set of such n has asymptotic density zero.
For a prime p, the factor p can therefore be removed from the Bernoulli denominator, leading to the first condition. Similarly below...
Question: Are there any Carmichael numbers k satisfying
lambda(Den(B_{k-1})/k) = k - 1 ?
More generally, does this condition have any interesting connection with the structure of Carmichael numbers?
Best,
Tom Ordo
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PS. Bernoulli-primitive primes:
13, 31, 37, 61, 67, 113, 127, 139, 157, 181, 199, 211, 241, 277, 281, 307, 331, 337, 349, 397, 401, 409, 421, 433, 461, 463, 499, 521, 523, 541, 547, 571, ...
Theorem. For an odd prime p, let D = Den(B_{p−1})/p. Then lambda(D) = p − 1 if and only if D is not the denominator of any Bernoulli number B_k with k < p − 1.
Proof. By von Staudt–Clausen, D is squarefree and q | D iff q − 1 | p − 1. Hence the least k for which D is a Bernoulli denominator is lambda(D).
The same holds for Carmichael numbers n: by Korselt's criterion, n is squarefree and q − 1 | n − 1 for every q | n. Thus, with D = Den(B_{n−1})/n, lambda(D) = n − 1 iff D is not the denominator of any B_k with k < n − 1.
I suggest calling these Bernoulli-primitive primes and Bernoulli-primitive Carmichael numbers.
What proportion of primes and Carmichael numbers are Bernoulli-primitive in this sense?
It seems that, unlike the Bernoulli-primitive primes, almost all Carmichael numbers (except 1105, 63973, ...) are primitive in this sense (more exceptions needed).
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