Primitive function a(n) for denominators of B_{2n}

[Tomasz Ordowski]

Hello All Thinkers! 

Let a(n) = 2n / lambda(Den(B_{2n})), for n > 0. 

1,1,1,2,1,1,7,1,1,1,1,2,13,1,1,2,17,1,19,1,1,...*  

Equivalently, a(n) = n / m, for the smallest m  

such that Den(B_{2m}) = Den(B_{2n}).

Numbers 2m such that a(m) = 1 

are A248614 > 1. 

Primes A248614(n)+1 are all primes. 

Composites A248614(n)+1 are A317210.  

Their asymptotic natural density is also zero.  

How to characterize the set of values a(n)? 

(*) Find more terms to see their diversity. 

Best,

Tom Ordo  

[Tomasz Ordowski]

PS. Instead of this unfortunate title, I suggest the new name:
Index redundancy function a(n) for denominators of B_{2n}. 

[DONG HAOXUAN]

Hi Tomasz,

I spent some time looking at this problem, and it seems to have a rather clean arithmetic reformulation.

Define

H(n) = lcm { d : d divides n and 2d+1 is prime }.

Then von Staudt-Clausen gives exactly

a(n) = n / H(n).

A useful feature is that H is idempotent:

H(H(n)) = H(n).

For a fixed positive integer r, define the canonical sequence

m_0 = 1,
m_{k+1} = H(r m_k).

Then r occurs as a value of a(n) if and only if this sequence eventually stabilizes.

Indeed, if it stabilizes at m, then

H(m) = H(rm) = m,

so

a(rm) = r.

More importantly, if M is any other witness satisfying

H(M) = H(rM) = M,

then m_k divides M for every k. Thus the canonical sequence is not merely a search heuristic: if it grows forever, no witness exists.

This also gives several general results.

1. Every prime power occurs as a value:

q^t is in the range of a(n)

for every prime q and every t >= 1.

In fact, every prime-power value occurs infinitely many times.

2. The range is divisor-closed.
3. H(n) is exactly the greatest divisor m of n satisfying H(m) = m. Thus a(n) can be viewed as the quotient left after removing the largest "stable" divisor of n.

I do not yet have a complete characterization of the range.

The first value for which the canonical closure behaves very differently is 30. All values 1,...,29 occur. For r = 30, the closure has continued to grow through every stage I have checked.

However, I do not think it is currently justified to claim that 30 is absent. If the closure for 30 were infinite, this would in particular imply infinitely many primes of the form

1 + 2^(a+1) 3^b 5^c,

so the remaining question runs into a genuine shifted-prime distribution problem.

Thus at present I would describe the situation as follows: the original Bernoulli-denominator question reduces to an idempotent arithmetic closure problem, for which a substantial part of the structure can be determined, while the complete range still seems open.

I am continuing to look at the full range question.

Best,
Jason

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[Tomasz Ordowski]

Nice!

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