Bernoulli numbers B_k with denominators of the form 2^n - 2

[Tomasz Ordowski]

Hello! 

Conjecture: 

Denominator(B_k) = 2^n - 2 if and only if 

n = 2^m + 1 for 0 <= m <= 5 and k is a power of 2. 

Cf. https://oeis.org/A219196  > 1.  

See https://oeis.org/A074777 

Best, 

Tom Ordo 

[Tomasz Ordowski]

PS. Let us define (using nice symmetry): 

Primes p such that 2^p - 2 is squarefree

and for every prime factor q of 2^p - 2

we have q - 1 | p - 1 or p - 1 | q - 1.  

Cf. A099596 - OEIS (here p odd). 

[Geoffrey Caveney]

The "only if...k is a power of 2" condition is false: counterexamples include the denominators of B_14, B_26, B_34, B_38, B_62, etc. = 6, the denominators of B_68, B_76, B_124, etc. = 30, and the denominators of B_304, B_496, B_608, etc. = 510. One might rather conjecture that, for each given denominator value 2^n - 2, the values of k that are powers of 2 are all smaller than the values of k that are not powers of 2. 

Based on the comments in A219196 and A074777, it seems that your conjecture in general, with the above condition removed, may be equivalent to the conjecture that there do not exist any Fermat primes > 65537. The fact that the values 2, 6, 30, 510, 131070, and 8589934590 are each the product of 2 and the first m Fermat primes would also explain the squarefree and q - 1 | p - 1 or p - 1 | q - 1 conditions that you mention.

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[Tomasz Ordowski]

Yes, Geoffrey, thank you for clarifying my conjecture, 

but the PS issue needs better justification as to whether 

such symmetric odd primes p are exactly A099596. 

Primes p such that 2^p - 2 is squarefree

and for every prime factor q of 2^p - 2

we have q - 1 | p - 1 or p - 1 | q - 1. 

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[Tomasz Ordowski]

PS. I found an exception; 2^137-2 is not square-free. 

However, my definition is interesting in itself and 

could be the name of a new OEIS sequence.

Primes p such that 2^p - 2 is squarefree

and for every prime factor q of 2^p - 2

we have q - 1 | p - 1 or p - 1 | q - 1. 

[Tomasz Ordowski]

Hello Everyone! 

It is known that the denominator of the Bernoulli number B_{2n} is of the form 2^m - 2 if and only if m = 2^k + 1 for k = 1, 2, 3, 4, 5. 

Indeed, by the von Staudt-Clausen theorem, 
Den(B_{2n}) = Product_{p-1 | 2n} p. 
Hence Den(B_{2n}) = 2 * Product_{F_i -1 | 2n} F_i, where the product runs over the Fermat primes F_i. Since F_5 = 2^(2^5) + 1 is  composite, the only possible Fermat-prime factors are 3, 5, 17, 257, and 65537, whose product is 
2 * 3 * 5 * 17 * 257 * 65537 = 2^33 - 2 = 2^(2^5+1) - 2. 

Now define a(n), for n > 0, as the largest integer k such that 
2^(2^k+1) - 2 divides Den(B_{2n}). 

It follows immediately that a(n) <= 5 for every n. 

Moreover, the condition 
2^(2^k+1) - 2 divides Den(B_{2n}) 
is equivalent to 
2^(k+1) divides 2n, 
that is, 
2^k divides n. 
Therefore, 
a(n) = min(v_2(n), 5), 
where v_2(n) denotes the exponent of 2 in n. 

Consequently, the sequence is periodic with the least period 
P = 2^15 = 32768, 
since min(v_2(n + 2^15), 5) = min(v_2(n), 5) for every n, while no smaller power of two has this property. 

Has this sequence, or the simple description a(n) = min(v_2(n), 5), appeared before in the literature or in OEIS? 

Are there any  interesting generalizations that come to mind? 

Best, 
Tom Ordo 

[M F Hasler]

On Fri, Jul 24, 2026, 05:50 Tomasz Ordowski <tomaszo...@gmail.com> wrote:

Hello Everyone! 

It is known that the denominator of the Bernoulli number B_{2n} is of the form 2^m - 2 if and only if m = 2^k + 1 for k = 1, 2, 3, 4, 5. 

As Geoffrey wrote, the "if" part is not true.  "A if B" means B => A.

But you want A => B.

You must say 

" if the denominator of B(2n) is of the form 2^m – 2, then m = ... ".

Or, "the denominator... can be ... ONLY IF ..."

Because, consider for example 

m = 3 = 2¹ + 1.

Now tell me, is the denominator of B(2n) of the form 2^m – 2 or not ?

You can't tell !

Because your "necessary and sufficient condition" actually doesn't tell anything about *the* denominator of B(2n).

It tells something about the number 2^m–2, 

if that number is a denominator of (some) B(2n).

Maybe it's easier to see better why the first is absurd, by considering a simpler example of the same form:

" the least prime divisor of n is 2m+1 if and only if 2m+1 is prime."

That's exactly the form of your statement.

But you mean "only if m = ..., then it can be ..."

- Maximilian

[Tomasz Ordowski]

Yes, Maximilian, formally you are right, but this inaccuracy does not affect the rest of my post, which I recommend reading.

If the denominator of the Bernoulli number B_{2n} is of the form 2^m - 2, then m = 2^k + 1 for k = 1, 2, 3, 4, 5. 

Indeed, by the von Staudt-Clausen theorem, 
Den(B_{2n}) = Product_{p-1 | 2n} p. 
Hence Den(B_{2n}) = 2 * Product_{F_i -1 | 2n} F_i, where the product runs over the Fermat primes F_i. Since F_5 = 2^(2^5) + 1 is  composite, the only possible Fermat-prime factors are 3, 5, 17, 257, and 65537, whose product is 
2 * 3 * 5 * 17 * 257 * 65537 = 2^33 - 2 = 2^(2^5+1) - 2. 

Now define a(n), for n > 0, as the largest integer k such that 
2^(2^k+1) - 2 divides Den(B_{2n}). 

It follows immediately that a(n) <= 5 for every n. 

Moreover, the condition 
2^(2^k+1) - 2 divides Den(B_{2n}) 
is equivalent to 
2^(k+1) divides 2n, 
that is, 
2^k divides n. 
Therefore, 
a(n) = min(v_2(n), 5), 
where v_2(n) denotes the exponent of 2 in n. 

Consequently, the sequence is periodic with the least period 
P = 2^15 = 32768, 
since min(v_2(n + 2^15), 5) = min(v_2(n), 5) for every n, while no smaller power of two has this property. 

Has this sequence, or the simple description a(n) = min(v_2(n), 5), appeared before in the literature or in OEIS? 

Are there any interesting generalizations that come to mind? 

Best, 
Tom Ordo 

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[Tomasz Ordowski]

PS. Primes p such that 2^p - 2 is squarefree and 

if prime q | 2^p - 2, then q - 1 | p - 1 or p - 1 | q - 1.

Cf. A334797 - OEIS (add square-free requirement).