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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.

%I A373650 #73 Aug 01 2025 11:09:41
%S A373650 72,20664,66600,84744,89784,141240,175032,232680,271272,288072,378984,
%T A373650 428472,620472,697320,740520,789672,792360,1016472,1063272,1207704,
%U A373650 1250472,1304472,1338600,1570584,1617672,1628472,1844472,2150712,2186472,2283672,2399112,2427672
%N A373650 Integers k such that there are i groups of order k+i up to isomorphism, for i=1,2,3,4.
%C A373650 Comment from _Robin Jones_, May 05 2025: (Start)
%C A373650 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.
%C A373650 Each term is congruent to 0, 2 or 4 modulo 5. Terms can't be congruent to 5 modulo 7.
%C A373650 Conjectures (verified to 783 terms):
%C A373650 Terms are never 3 mod 7.
%C A373650 Terms are never 5 or 9 mod 11.
%C A373650 Terms are never 9 or 11 mod 13.
%C A373650 Terms are never 13 or 15 mod 17.
%C A373650 Terms are never 17 mod 19.
%C A373650 Terms are never 19 or 21 mod 23.
%C A373650 In fact it looks like it could be true that for any prime p > 3, terms are never congruent to p-2 mod p.
%C A373650 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.
%C A373650 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).
%C A373650 (End)
%H A373650 David Radcliffe, <a href="/A373650/b373650.txt">Table of n, a(n) for n = 1..6405</a> (terms 1..1418 and 1420..5512 from Robin Jones).
%H A373650 Robin Jones, <a href="/A373650/a373650.png">Graph of n, a(n) mod 5 for n = 1..783</a>
%H A373650 Robin Jones, <a href="/A373650/a373650_1.png">Graph of n, a(n) mod 7 for n = 1..783</a>
%H A373650 Robin Jones, <a href="/A373650/a373650_2.png">Graph of n, a(n) mod 11 for n = 1..783</a>
%e A373650 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.
%o A373650 (Magma) for x in [1 .. 100000] do //get the terms up to 100000
%o A373650     if NumberOfSmallGroups(x+1) eq 1 then
%o A373650         if NumberOfSmallGroups(x+2) eq 2 then
%o A373650             if NumberOfSmallGroups(x+3) eq 3 then
%o A373650                 if NumberOfSmallGroups(x+4) eq 4 then
%o A373650                     x;
%o A373650                 end if;
%o A373650             end if;
%o A373650         end if;
%o A373650     end if;
%o A373650 end for; // _Robin Jones_, Apr 18 2025
%Y A373650 Cf. A373648 (i=1,2), A373649 (i=1,2,3), A381335 (i=1,2,3,4,5).
%K A373650 nonn
%O A373650 1,1
%A A373650 _Robin Jones_, Jun 12 2024