A373649 -id:A373649 - cp's OEIS frontend

A373648 Integers k such that there are i groups of order k+i up to isomorphism, for i=1,2.

2, 4, 12, 32, 36, 60, 72, 84, 132, 140, 144, 156, 176, 192, 212, 216, 276, 312, 344, 392, 396, 420, 444, 456, 480, 500, 536, 540, 552, 560, 564, 612, 660, 672, 696, 704, 716, 732, 744, 756, 792, 816, 864, 876, 884, 912, 932, 956, 972, 996, 1040, 1092, 1140, 1152, 1172, 1200

Offset: 1

Robin Jones, Jun 12 2024

nonn

All the terms are even. - Robin Jones, Apr 18 2025

			2 is a term since there is 1 group of order 3 up to isomorphism, 2 of order 4.

Robin Jones, Table of n, a(n) for n = 1..500

Equals A296023 - 1.
Cf. A373649 (i=1,2,3), A373650 (i=1,2,3,4), A381335 (i=1,2,3,4,5).
Subsequence of A003277 - 1.

A373650 Integers k such that there are i groups of order k+i up to isomorphism, for i=1,2,3,4.

Original entry on oeis.org

72, 20664, 66600, 84744, 89784, 141240, 175032, 232680, 271272, 288072, 378984, 428472, 620472, 697320, 740520, 789672, 792360, 1016472, 1063272, 1207704, 1250472, 1304472, 1338600, 1570584, 1617672, 1628472, 1844472, 2150712, 2186472, 2283672, 2399112, 2427672

Offset: 1

Robin Jones, Jun 12 2024

nonn

Comment from Robin Jones, May 05 2025: (Start)
Each term is a multiple of 24. No terms are multiples of 48. That is, each term is congruent to 24 mod 48. Equivalently, 8 divides a(n), 3 divides a(n), but 16 does not divide a(n), for all n.
Each term is congruent to 0, 2 or 4 modulo 5. Terms can't be congruent to 5 modulo 7.
Conjectures (verified to 783 terms):
Terms are never 3 mod 7.
Terms are never 5 or 9 mod 11.
Terms are never 9 or 11 mod 13.
Terms are never 13 or 15 mod 17.
Terms are never 17 mod 19.
Terms are never 19 or 21 mod 23.
In fact it looks like it could be true that for any prime p > 3, terms are never congruent to p-2 mod p.
If n = p-2 mod p, then n+2 = 2pm for some m, since n is even. Then if m > 1, one can show that this will always have 4 or more groups. Thus m=1 and n=2p-2. One must show that n can never be of the form 2p-2 if this conjecture is true.
It looks like the terms are also unevenly distributed modulo 5, 7, 11, and so on, within the classes that terms can belong to. For example, modulo 5, it seems like terms are most commonly congruent to 2 modulo 5, and it is relatively rare for terms to be congruent to 4 modulo 5 (see plots in Links).
(End)

			72 is in this sequence as there is 1 group of order 73 up to isomorphism, 2 of order 74, 3 of order 75, 4 of order 76.

David Radcliffe, Table of n, a(n) for n = 1..6405 (terms 1..1418 and 1420..5512 from Robin Jones).
Robin Jones, Graph of n, a(n) mod 5 for n = 1..783
Robin Jones, Graph of n, a(n) mod 7 for n = 1..783
Robin Jones, Graph of n, a(n) mod 11 for n = 1..783

Cf. A373648 (i=1,2), A373649 (i=1,2,3), A381335 (i=1,2,3,4,5).

for x in [1 .. 100000] do //get the terms up to 100000
    if NumberOfSmallGroups(x+1) eq 1 then
        if NumberOfSmallGroups(x+2) eq 2 then
            if NumberOfSmallGroups(x+3) eq 3 then
                if NumberOfSmallGroups(x+4) eq 4 then
                    x;
                end if;
            end if;
        end if;
    end if;
end for; // Robin Jones, Apr 18 2025

A381335 Integers k such that there are i groups of order k+i up to isomorphism, for i=1,2,3,4,5.

Original entry on oeis.org

2814120, 22411272, 29436120, 46906920, 58734120, 59558520, 71510520, 106822200, 109673064, 117873720, 200250120, 213805272

Offset: 1

Robin Jones, Apr 19 2025

a(13) > 220000000 if it exists.
Each term is a multiple of 24. No terms are multiples of 48.
Each term is congruent to 0, 2 or 4 modulo 5. Terms can't be congruent to 5 modulo 7. I think they also can't be congruent to 3 modulo 7, but I haven't proven that yet.

			2814120 is in this sequence as there is 1 group of order 2814121 up to isomorphism, 2 of order 2814122, 3 of order 2814123, 4 of order 2814124, 5 of order 2814125.

Cf. A373648 (i=1,2), A373649 (i=1,2,3), A373650 (i=1,2,3,4).

A296024 Numbers n such that there is precisely 1 group of order n, 2 of order n + 1 and 3 of order n + 2.

Original entry on oeis.org

73, 865, 2065, 2173, 3973, 7933, 10333, 12633, 15121, 16537, 17473, 19237, 20317, 20337, 20665, 23773, 23881, 24421, 25093, 28921, 31477, 33133, 35137, 36877, 38173, 41017, 41773, 42061, 46021

Offset: 1

Muniru A Asiru, Dec 03 2017

nonn

Equivalently, lower member of consecutive terms of A296023.

Being a subsequence of A003277, all the terms are odd.

			73 is in the sequence because 73 is a cyclic number, A000001(74) = 2 and A000001(75) = 3.
865 is in the sequence because 865 is a cyclic number, A000001(866) = 2 and A000001(867) = 3.
20317 is in the sequence because 20317 is a cyclic number, A000001(20318) = 2 and A000001(20319) = 3.

H. U. Besche, B. Eick and E. A. O'Brien, A Millennium Project: Constructing Small Groups, Internat. J. Algebra and Computation, 12 (2002), 623-644.
Gordon Royle, Numbers of Small Groups
Index entries for sequences related to groups

Cf. A000001, A003277. Subsequence of A296023.

Equals A373649 + 1.

with(GroupTheory): with(numtheory):
for n from 1 to 10^5 do if [NumGroups(n), NumGroups(n+1), NumGroups(n+2)]=[1, 2, 3]  then print(n); fi; od;

Sequence is { n | A000001(n) = 1, A000001(n+1) = 2, A000001(n+2) = 3 }.

A383350 a(n) is the smallest integer k such that there are k+i groups of order a(n)+i, for i=1,...,n.

Original entry on oeis.org

0, 2, 72, 72, 2814120, 29436120

Offset: 1

Robin Jones, Apr 24 2025

The sequence is finite. For any multiple of 32, there are more than 32 groups of that order. Thus, the sequence 1,2,...,32 can't appear in A000001, and this sequence is at most 31 terms long.
The sequence is either 6 or 7 terms long. This can be shown by first showing every entry of A373650 is congruent to 24 mod 48. It then follows that if n is such that A000001(n+i) = i for i=1,2,3,4, then n+8 is a multiple of 16. But then A000001(n+8) >= 14, so we can't have A000001(n+i) = i for i=1,2,3,4,8.
From a(2) onwards, each entry is a multiple of 24, but not a multiple of 48.
a(7) > 223000000 if it exists.
Each entry is congruent to 0, 2 or 4 modulo 5.

			a(1) = 0 since there is 1 group of order 1.
a(2) = 2 since there is 1 group of order 3, 2 groups of order 4.

Cf. A373648, A373649, A373650, A381335.

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A373648 Integers k such that there are i groups of order k+i up to isomorphism, for i=1,2.

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A373650 Integers k such that there are i groups of order k+i up to isomorphism, for i=1,2,3,4.

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Magma

A381335 Integers k such that there are i groups of order k+i up to isomorphism, for i=1,2,3,4,5.

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A296024 Numbers n such that there is precisely 1 group of order n, 2 of order n + 1 and 3 of order n + 2.

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A383350 a(n) is the smallest integer k such that there are k+i groups of order a(n)+i, for i=1,...,n.

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