A298429 Numbers n such that there are precisely 12 groups of orders n and n+1.

[Brendan]

This sequence says  30135, 76312, 130890, 173445, 356610

However, the Maple code (after adding the obvious missing left paren) gives
 30135, 128270, 162924, 173445, 243224, 325325, 477015, ...

Can someone please investigate this with GAP, Magma, or something else?

Brendan.

[Robert Israel]

The Maple documentation for NumGroups says "In the current implementation, the smallest value of n for which the number of groups of order n cannot be computed is n = 2048."  So it looks like we might not be able to rely on Maple to compute values >= 2048 in this sequence.  

.

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[Neil Sloane]

As you all know, this is sequence A000001, and the b-file only goes up to n=2047,

so that is probably as far as the numbers are known.

Best regards

Neil 

Neil J. A. Sloane, Chairman, OEIS Foundation.

Also Visiting Scientist, Math. Dept., Rutgers University, 

Email: njas...@gmail.com

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[Allan Wechsler]

I'm a little less pessimistic than some about this particular sequence.

It's true that A1 is notoriously difficult for numbers with a lot of small prime divisors with high multiplicity. The problem is, at least partly, the astronomically high number of possibilities. A1(2^n) is challenging because it's gigantic.

But 12 isn't gigantic. We know in advance that A1(2048) is way bigger than 12, so we don't have to devote any attention to it. For several small values of k, we have complete characterizations of numbers n such that A1(n) = k. I don't think we have that for k = 12, but just checking that A1(n) = 12 isn't a gargantuan project like calculating A1 for big highly-divisible n.

The data for oeis.org/A249555, the sequence of all n with A1(n) = 12, goes way farther than 2048. Entries in this sequence seem to display only a few prime signatures, and a search confined to "numbers like that" seems like a promising approach.

A249555 doesn't say much about what's known, but I will guess that a lot is known, which might make extending A298429 less daunting.

--Allan

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[Misha Lavrov]

Despite the skepticism about Maple, I think it is actually the sequence that is wrong here.

Consider 76312. Can it be that A1(76312) = A1(76313) = 12? Well, 76313 appears in A350245 (Numbers p^2*q, p > q odd primes such that q divides p+1): it is 67^2*17. According to A350245, in such a case there are only three groups of order 76313: they are C_76313, (C_67 x C_67) x C_17, and a semidirect product (C_67 x C_67) : C_17.

Or am I missing something?

Best,

Misha

P.S. It's possible that Maple cannot correctly compute A1(2048) because there are so many groups of order 2048, but can correctly count the number of groups of order n for most n much higher than 2048.

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[Christian Sievers]

Maple's NumGroups returns FAIL if the number is not known.

If one of the two numbers that are checked is unknown, this is still fine if the other is known and different from 12.

The maple code should be extended to check this.

If both values are unknown, or one is and the other is 12, the number can be output as a problematic case that may be handled by other methods.

For example, the number of groups of order 2048 is unknown, but for 2047 and 2049 it is 1.

If either were unknown, we could still argue that A1(2048)>12.

BTW, an easy way to do this is to note that there are already 56 = # of partitions of 11 abelian groups of order 2048=2^11 (cf. A688).

My experiments with GAP didn't go far. Maybe I'm missing an extra package to get more results.

The first value I couldn't decide was 2024, with the nuber of groups for both 2024 and 2045 unknown.

I can confirm the numbers the sequence and the maple computation have in common, i.e. 30135 and 173445.

For the other terms of the sequence as given, the number of groups of order n+1 is 3 or 2.

The other results of the maple code are possible (I'd guess they are true).

(But we really need to check if there are unsure values between them.)

For experiments with GAP, note that its NumberSmallGroups function raises an error if called with an order for which

the number is not known. A function that returns fail in that case can be defined like this:

NumberGroups := function ( n )
    if NumberSmallGroupsAvailable( n ) then
        return NumberSmallGroups( n );
    else
        return fail;
    fi;
end;

(And right now I'm finding that the sglppow package adds a little, but only some prime powers, not enough for this task.)

All the best

Christian

[Brendan]

The NumGroups function in Maple first looks in a table of small orders, then it applies some collection of rules based on the prime factorization, and finally it gives up and returns FAIL.

As someone said, for many orders it is clear that the count is greater than 12 even if the precise count is unclear. I modified the code to eliminate orders divisible by the fourth power of a prime.  Here are the first cases where n and n+1 both give 12 or FAIL.

The items are n, [factors of n], [factors of n+1], [NumGroups(n), NumGroups(n+1)].

30135, [[3, 1], [5, 1], [7, 2], [41, 1]], [[2, 3], [3767, 1]], [12, 12]
51219, [[3, 3], [7, 1], [271, 1]], [[2, 2], [5, 1], [13, 1], [197, 1]], [FAIL, FAIL]
51975, [[3, 3], [5, 2], [7, 1], [11, 1]], [[2, 3], [73, 1], [89, 1]], [FAIL, FAIL]
52184, [[2, 3], [11, 1], [593, 1]], [[3, 1], [5, 1], [7, 2], [71, 1]], [FAIL, FAIL]
52839, [[3, 3], [19, 1], [103, 1]], [[2, 3], [5, 1], [1321, 1]], [FAIL, FAIL]
52983, [[3, 2], [7, 1], [29, 2]], [[2, 3], [37, 1], [179, 1]], [FAIL, FAIL]
53000, [[2, 3], [5, 3], [53, 1]], [[3, 3], [13, 1], [151, 1]], [FAIL, FAIL]
53108, [[2, 2], [11, 1], [17, 1], [71, 1]], [[3, 3], [7, 1], [281, 1]], [FAIL, FAIL]
53624, [[2, 3], [6703, 1]], [[3, 1], [5, 3], [11, 1], [13, 1]], [12, FAIL]
54404, [[2, 2], [7, 1], [29, 1], [67, 1]], [[3, 3], [5, 1], [13, 1], [31, 1]], [FAIL, FAIL]
54872, [[2, 3], [19, 3]], [[3, 2], [7, 1], [13, 1], [67, 1]], [FAIL, FAIL]

Then there is the question of bugs in Maple.  I'm pessimistic about getting a reliable sequence.

What to do?

Brendan.

[Christian Sievers]

Oh, I didn't expect the first unsure value so early.

But we can still examine further.

There are (according to GAP) 13 groups of order 3^3*7, the direct product of each of these with the

cyclic group of order 271 should give 13 nonisomorphic groups of order 51219, so that number is out.

It seems that all the cases here can be eliminated by one of {n, n+1} having a divisor with more than 12 groups of that order.

So I'm suggesting using a function like this (GAP):

Has12Groups := function ( n )
    local x;
    x := NumberGroups( n );
    if x <> fail then
        return x = 12;
    fi;

   if ForAny( List( DivisorsInt( n ), NumberGroups ), r -> r <> fail and r > 12 ) then

        return false;

   else

        return fail;

   fi;

end;

Alas, with GAP's data, the first number I can't handle is n=3992, where

[Has12Groups(n),Has12Groups(n+1)] is [true,fail].

But the same idea with maple should be more successful.

Bye

Christian

[Brendan]

Is it a theorem that "m divides n" implies that the number of groups of order n is at least equal to the number of order m?

The case when m and n are coprime is clear, but the general case exceeds my weak group theory.

Brendan.

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[Christian Sievers]

It seemed obvious to me, but then I wasn't completely sure and wrote "should be".

But now I am sure. As suggested, we can form the direct product with the cyclic group of order n/m to get groups of order n.

And they will be non-isomorphic, even if it is not as obvious as I expected.

For any finite group G and any groups H, K, we have that H x G ~ K x G implies H ~ K (where ~ means isomorphic),

a result from Hirshon with an elementary proof when all groups are finite, see https://math.stackexchange.com/a/1260307.

I also wondered if there is any chance that the current sequence is what was meant,

and the name just doesn't give the intended meaning?

Christian

[Brendan]

Maple claims that there are 627 groups of order 19208=2^3.7^4, but 1146 groups of order 9604=2^2.7^4.

I'm even more suspicious about this example:

3 groups of order 50925=3.5^2.7.97 but 6 groups of order 525=50925/97.

Is this plausible?  Are the 6 direct products H x Z[97] for |H|=525 not non-isomorphic?

Btw, I'm using Maple 2024.

Brendan.

[Allan Wechsler]

I am not a great group theorist, but I confess I have always believed intuitively that for positive k, A1(n) <= A1(kn). My handwaving pseudo-proof is, to each group G of order n, associate G' = G x C(k) of order kn. Now I have to assert that if G and H are not isomorphic, then neither are G' and H'. This feels extremely plausible to me, but if anybody knows a counterexample, please disabuse me!

-- Allan.

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[Ruud H.G. van Tol]

On 2026-02-27 18:59, Neil Sloane wrote:
> As you all know, this is sequence A000001, and the b-file only goes
up to n=2047,
> so that is probably as far as the numbers are known.

> [...]

In the linked https://oeis.org/A000001/a000001_4.txt
there are about 50k values, marked with a flag where unknown.

-- Ruud

[Hugo Pfoertner]

Isn't there a similar problem with A298430 and A298431? These sequences also rely on the assumption that Maple is able to correctly count the number of groups with orders > 50000.

[David Radcliffe]

This argument works because every finite group can be expressed uniquely as a direct product of indecomposable subgroups, up to isomorphism and reordering of factors. This is a special case of the Krull-Schmidt theorem.

Express G, H, and C(k) as products of indecomposable subgroups:

   G = G1 × ⋯  × Gr, 

   H = H1 × ⋯  × Hs, 

   C(k) = C1 × ⋯ × Ct.

Then we have

    G' = (G1 × ⋯ × Gr)   ×   (C1 × ⋯ × Ct)  and

    H' = (H1 × ⋯ × Hs)   ×   (C1 × ⋯ × Ct)

If G' and H' are isomorphic, then their decompositions are equivalent. This means that the multisets of isomorphism types

    {|G1|, ..., |Gr|, |C1|, ..., |Ct|}   and

    {|H1|, ..., |Hs|, |C1|, ..., |Ct|}

are equal.

This implies that {|G1|, ..., |Gr|} and {|H1|, ..., |Hs|} are equal as multisets, hence G and H are isomorphic.

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[Christian Sievers]

So, there is also a GAP package Cubefree which provides the command NumberCFGroups,

and then there is a package gnu at https://github.com/olexandr-konovalov/gnu/

which combines all this and more, and is also the source of the data mentioned by Ruud.

I had to comment out the IsCubeFree function in lib/gnu.g and add some bogus values for

stuff like ArchiveURL (in PackageInfo.g) that LoadPackage insisted on, but then I could use

a variant of Has12Groups that calls Gnu(n) and takes into account that Gnu returns false instead

of fail when a value is unknown.

That allowed for the first time to find the term 30135 without any previous possible term.

However, after that came several possible terms, the first one being 61624.

It also allowed to confirm all but one of the terms computed by maple.

And better: the package GrpConst has a function ConstructAllGroups that I could use to

reject all the extra possible terms I had found (15 of them, the last one 88478), and it also

confirmed the last still unconfirmed term from the maple result.

I may later try to speed up my code and automate the extra ConstructAllGroups checking.

So far, my calculations confirm that all the terms of the maple result are part of the sequence,

and show that there are no other terms up to 88478.

(And looking into the history of the sequence makes it very clear that this is indeed 

the sequence that was intended.)

Bye

Christian

[Brendan]

I have reported the Maple bug to MapleSoft.

In the past I have found them very responsive and they usually fix bugs by the next release.

Brendan.

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[Christian Sievers]

According to my computations with GAP, the sequence should start with

30135, 128270, 134385 (!)

A298430 (same with 13 groups of order n and n+1) should start with

82323, 95876, 130724, 144675

Currently, the sequence entry only has 82323.

The second term given there, 390446, is also part of the sequence.

I can confirm the first nine terms of A298431 (same with 14 groups).

This includes their position, i.e. there are no missing terms between them.

There are more sequences like this, I don't know why only these have been put in question.

For now, I have only computed these.

All the best

Christian

[Allan Wechsler]

I note that in A298429, one thing that could be done to speed up the search is to look for odd entries of A249555, because entries in A298429 must be either these values or one less. I was slightly surprised to find that the odd subsequence of A249555 (8085, 11325, 17457, 22365, 22725, 27885, 29095, 30135, 33165, 37191, 41745, 45075, ...) is not present in OEIS.

-- Allan

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[Christian Sievers]

I found two more terms for
A298429 (k with 12 groups of order k and 12 groups of order k+1).

The terms below 500000 are
30135, 128270, 134385, 162924, 173445, 243224, 316503, 325325, 359864, 477015.

I'm certain enough of these terms that I would normally go ahead and
edit the sequence.

This strangely seems to be the only sequence of this kind that not only
misses some terms, but includes wrong ones.
And confusingly, a pink box comment suggests yet another set of terms.

A298430 (13 groups) has more missing terms and starts with
82323, 95876, 130724, 144675, 222452, 390446, 395955.

The other terms given there are also in the sequence, but I see no reason
to believe there are no more missing terms around them.

A298431 (14 groups) has missing terms and starts with
4328, 22311, 29864, 57896, 75368, 99368, 120807, 130664, 131943, 152295,
157287, 164072, 180327, 184232, 186344, 212456, 235431, 236583, 239912,
257895, 268712, 276392, 304551, 331112, 338792, 348776, 381927, 387495,
415592, 424808, 438632, 456615, 457767, 466472, 483303

(and I'm still checking the last 11000 numbers below 500000).

Some of the other sequences only give terms below 50000 and have
the right number of terms, so they are probably fine. I will check
them later.
I have also computed more terms for the ones with only a few terms.

But I found missing terms in A298427 (9 groups) and A298428 (10 groups).

Interesstingly, there is no sequence for k with 7 groups of order k and k+1.
Indeed, there is no such k up to 500000.
I wonder if there is a nice reason for that, and if there is any such k.

All the best

Christian