Le mar. 5 janv. 2021 à 20:30, Allan Wechsler <acwacw@gmail.com> a écrit :
While we are here -- I am as big a fan as anybody of not excluding the zero case whenever possible. But this case just seems weird to me. We have two sequences, A027623 and A037234, which differ only in the zero entry, which is 1 for the former and 0 for the latter. Clearly there must be angels-dancing-on-pins arguments lurking here, but I can't think of any. There aren't any rings with no elements, are there?
I totally agree, AFAIK (and as the description of that sequence explicitly states) a ring is an abelian group (R,+) with a second law * etc., and a group must have a neutral element (which would then be the 0 of the ring). (See also A000001(0) = 0 : there are 0 groups with 0 elements.) I have no idea why one would let A027623(0) = 1. There are clearly 0 rings with 0 elements, since a ring has at least one element, the zero. If there was a ring with 0 elements, then forgetting about the multiplication we would have an abelian group with 0 elements, and thus A000001(0) >= 1. Anyway, the sequence's definition treats a(0)=1 as a special case: It does not claim that there is a ring with 0 elements. ("Number of rings..." is the definition of the terms a(n) with n >= 1, only.) The only suspicion of an explanation I can imagine is that this strange a(0)=1 was (erroneously) introduced because of the phrase "does not need to have a 1" (like, "we don't need a 1, so we can have a ring with only 0" - but of course, not 0 elements!). As a side note, f(0) = 1 is also in contradiction with the "mult" keyword saying that this is a multiplicative function, i.e., f(x y) = f(x) f(y) when gcd(x,y)=1. Considering y=0, this implies f(x)=1 for all x, if f(0)=1. (Usually multiplicative functions aren't even defined for index/argument 0, but if they are, either the case xy=0 must be excluded in the above equation, or f(0) must be equal to 0.) - Maximilian