Poppins numbers and A229088

[Allan Wechsler]

The multiperfect numbers are those n such that n divides sigma(n) (the sum of divisors of n). That is, when the fraction sigma(n)/n is reduced to lowest terms, the denominator being 1 means n is multiperfect.

The denominator of sigma(n)/n is at oeis.org/A017666 .

Some numbers just fail to be multiperfect, in the sense that when you reduce sigma(n)/n to lowest terms, almost everything cancels. An example is:

L = 951870502263811526107776159523473265965115414366224136036102963200

= 2^32 3^13 5^2 7^6 11^2 17 19 23 29 31 41 89 163 263 307 547^2 613 1093 2141 4733 599479

In this case, sigma(L)/L = 25/4. I have been jokingly thinking of numbers like this as "Poppins numbers", remembering that Mary Poppins thought of herself a "practically perfect in every way". I don't have a strong definition of these -- that is, I have no clear criterion for how small the denominator has to be before we take notice of the number as a near miss in the search for multi-perfection.

OEIS, though, thinks that if the denominator is 6 or less, that's interesting.

Numbers for which this denominator (let's call it D) is 1 are of course the multiperfect numbers themselves, oeis.org/A007691 .

Numbers for which D = 2 are striking enough that they actually have a name, "hemiperfect" numbers. They are at oeis.org/A159907 .

D = 3 doesn't have a name, but I propose calling them "tritoperfect" on analogy with "hemiperfect". See oeis.org/A245775 .

D = 4, 5, and 6 are at oeis.org/A229088 , oeis.org/A067237 , and oeis.org/A262356, respectively. We could call them tetartoperfect, pemptoperfect, and hektoperfect numbers (from the Greek words for "third", "fourth", and "fifth").

BUT: The identity of A229088 with the tetartoperfect numbers is only conjectural. This sequence was added to OEIS with a different original interpretation. The title interpretation is that these are the numbers k, such that sigma(k) and antisigma(k) (the sum of the nondivisors of k less than k) have the same residue modulo k. 

Now the punchline: my clumsy calculations suggest that the number L given above is a counterexample to the conjecture. I am pretty sure that this number is tetartoperfect, and I am also pretty sure that it doesn't belong in A229088, because the residue mod L of its antisigma is exactly twice the sigma residue, not the same number.

Can somebody confirm this? I'm feeling all thumbs and I half suspect I have made a stupid arithmetic error.

Also: if anybody thinks the D = 7 "hebdomoperfect" numbers are interesting, the first few are 7, 14, 42, 56, 168, 280, 588, 840, 2520.

Everybody who searches for multiperfect numbers stumbles on Poppinses like L from time to time, and it would be nice to have a place to record them.

-- Allan

[M F Hasler]

On Tue, May 26, 2026 at 4:45 PM Allan Wechsler <...> wrote:

(...) Some numbers just fail to be multiperfect, in the sense that when you reduce sigma(n)/n to lowest terms, almost everything cancels.(...)

You might also be interested in 

A174292 = Spoof-perfect numbers: Freestyle perfect numbers (A058007) which are not perfect numbers (A000396).

also discussed on  OEIS wiki: Spoof perfect numbers.

Descartes found one odd Spoof Perfect Number, N = 3² ⋅  7² ⋅  11² ⋅  13² ⋅  22021,

which would be perfect, i.e.,  σ(N)  =  σ(3²) ⋅ σ(7²) ⋅ σ(11²) ⋅ σ(13²) ⋅ σ(22021)  =?=  2 N,

if 22021 was prime <=>  σ(22021)  =?=  1 + 22021  (but actually  22021  = 19²  ⋅  61).

To me, the most interesting challenge is to find other *odd* freestyle perfect numbers,

question first asked by John Leech according to B1, pp. 44-45, UPiNT2, R.K.Guy

[but obviously already Descartes implicitly asked that question...]

and also subject of Carlos Rivera's primepuzzles.net/puzzles/puzz_111.htm.

About 13 years ago I have added some considerations in A174292, which actually concern the definition of the freestyle perfect numbers A058007 -- namely, what are reasonable conditions for mistakenly considering a composite factor prime: It seems quite natural to exclude the possibility that a very small prime be a divisor; in particular:

- the "spoof prime" factor should (IMHO quite obviously) not be even;

- none of the smaller prime factors of the number should divide the "spoof prime"(composite) factor:

  If you know that 3 divides N, you would naturally factor out all factors of 3.

- maybe the "spoof prime" factors should necessarily be the largest factor(s)

- no two factors should have a gcd > 1 (because testing primality may be considered nontrivial, but the Euclidean algorithm is straightforwardly applied to a pair of even huge numbers, to reveal very quickly a gcd.

Descartes' 198585576189 satisfies all of these criteria, but  none of them is required in the definition of A058007; yet no other odd term is known!

- Maximilian

[Allan Wechsler]

Pontus von Brömssen was kind enough to check my work, and he concluded that my number L did indeed meet the defining criterion for A229088; I repeated my arithmetic more carefully and I now concur with him that L is not a counterexample to the conjecture, so the conjecture still stands. I'm going to spell out the conjecture here, in case anybody wants to take a whack at proving it.

Let sigma(m) be the sum of all the positive integer divisors of m, as usual.

Let antisigma(m) be the sum of all positive integer nondivisors of m not exceeding m. In other words, antisigma(m) + sigma(m) = m(m+1)/2.

Call m tetartoperfect when sigma(m)/m, reduced to lowest terms, has a denominator of 4.

Let a%b denote the remainder when dividing a by b. Call m a Krizek number (after the author of A229088) if antisigma(m)  % m = sigma(m) % m.

Conjecture (Jaroslav Krizek): All tetartoperfect numbers are Krizek numbers, and all Krizek numbers except 1 are tetartoperfect.

Max -- the spoof-perfect numbers are interesting but my particular toolkit is bad for making progress on them.

-- Allan 

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[Martin Fuller]

The conjecture is true iff there are no odd Krizek numbers.

Lemma:

A. Krizek numbers are tetartoperfect iff they are even.

B. All tetartoperfect numbers are multiples of 4 and are Krizek numbers.

Proof of A: Let k be a Krizek number.

  sigma(k) = k(k+1)/2 - sigma(k) (mod k).

  2*sigma(k) = k(k+1)/2 (mod k).

  2*sigma(k) = ak + k(k+1)/2.

  4*sigma(k) = (2a + k + 1)k.

  sigma(k)/k = (2a + k + 1)/4.

This has the required form iff k is even.

  

Proof of B: Let k be a tetartoperfect number.

  sigma(k)/k = (2a+1)/4.

  4*sigma(k) = (2a+1)k.

LHS is a multiple of 4, and (2a+1) is odd, therefore k must be a multiple of 4.

  4*sigma(k) = (2a-k)k + (k+1)k.

  2*sigma(k) = (a-k/2)k + k(k+1)/2.

  2*sigma(k) = k(k+1)/2 (mod k).

  sigma(k) = k(k+1)/2 - sigma(k) (mod k).

Martin Fuller

[Allan Wechsler]

Martin,

Excellent! I tried a couple of times and got mired, and I'm grateful that you pushed it through. The insight that this depends on whether there are any odd Krizek numbers seems to be the key.

All,

1. Doesn't Martin Fuller's proof of the conditional version of Krizek's conjecture belong in the comments to A229088?

2. I occasionally find fairly large Poppins numbers in the course of my search for multiperfect numbers. But I have no guarantees that they are contiguous with the ones listed in the various relevant sequences. Would it be worthwhile to upload auxiliary files for these extras (known to be members of the sequences, but exact position not known)?

-- Allan

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[Max Alekseyev]

It's easy to see that the odd Krizek numbers are exactly the odd multiperfect numbers. And no odd multiperfect numbers > 1 are currently known.

Regards,

Max

On Wed, Jul 29, 2026 at 11:16 AM 'Martin Fuller' via SeqFan <seq...@googlegroups.com> wrote:

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