Talk:Prime signatures

Prime signature of 1

MathWorld says it is by convention {1}. I would have preferred { } (empty multiset, or empty bag). What is the argument for {1}? — Daniel Forgues 07:15, 22 December 2010 (UTC)

Is it only so that A118914 (which is the concatenation of the prime signatures for $\scriptstyle n\,\geq \,1\,$) can include the prime signature of 1 in a visible way. — Daniel Forgues 07:30, 22 December 2010 (UTC)

• DF>> What is the argument for {1}? <<DF. The true answer is that 1 has no prime signature. (And that it is completely irrelevant what MathWorld says.) Your problem arises from the fact that you made the second step before the first. Give a precise definition what a 'prime factorization' is, base the notion of 'prime signature' on this and your problem will be vanished. Peter Luschny 08:11, 22 December 2010 (UTC)

I actually did put the definition of the prime factorization first on the page, that's why I considered that the prime signature of 1 should be { }, not {1} as MathWorld says and A118914 also implies for 1 as the first term of the concatenation of prime signatures. — Daniel Forgues 08:41, 22 December 2010 (UTC)

I will remove the first term 1 from A118914 (I don't see why the prime signature of 1 should by convention be {1}, it is so obviously { } to me) because we have the concatenation of

{{ }, {1}, {1}, {2}, {1}, {1, 1}, ...}

giving

{1, 1, 2, 1, 1, ...} — Daniel Forgues 05:33, 23 December 2010 (UTC)

Also, in A025487, A036035 and A046523, 1 is considered having a different prime signature than 2, implying { } for 1 and {1} for 2). — Daniel Forgues 05:37, 23 December 2010 (UTC)

• On the page about Omega you write: "Obviously Omega(1) = 0, since 1 is the product of no primes." So you say that 1 has no prime factorization. I agree.

Next on the page 'Prime signature' you start with the first sentence: "A positive integer n, with prime factorization..." As 1 has no prime factorisation you exclude 1 from your considerations and thus you do not assign a prime signature to 1. This is consequential and fine.

However, some sentences later you write: "The prime signature of 1 is the partition of 0". How that? And then all this mess with MathWorld starts.

In your table this is reflected: n = 1 | | {}^[2] | The second column is empty and the third column should be equally be empty, instead it is filled with the empty set. If you delete the empty set from the third column the reference to MathWorld will also vanish and you can also delete this unholy mess about MathWorld alltogether from the page. It is neither the task of the OEIS Wiki to correct or to criticize MathWorld nor does the OEIS Wiki rely on MathWorld. Peter Luschny 01:30, 24 December 2010 (UTC)

• Saying that 1 is the product of no primes (product of 0 primes) is equivalent to saying that the prime factorization of 1 is the empty product of primes. $\scriptstyle \omega (n)\,$ is the count of distinct prime factors of n. So $\scriptstyle \omega (1)\,$ is the count of distinct prime factors of 1, i.e. the empty sum, which is defined as the additive identity, 0. The ancient Greeks did not consider 0 to be a number, so for them $\scriptstyle \omega (1)\,$ would be undefined. We don't consider $\scriptstyle \omega (1)\,$ to be undefined (since $\scriptstyle \omega (1)\,$ is 0). Same way, we don't say that 1 has no prime factorization, we say that the prime factorization of 1 is the empty product of primes.

The concepts of empty sum and empty product, bona fide mathematical concepts, are key here. Without the concept of empty product, 1 would have no prime factorization.

Also, it is obvious that the prime signature of n is a partition of Omega(n), and this works too for Omega(1) = 0, since 0 has exactly one partition, the empty partition of positive integers, whose sum of parts is the empty sum, defined as the additive identity, 0.

The prime signature of n is the multiset (bag) of positive exponents of distinct prime factors of n, so the prime signature of 1 is the empty multiset (empty bag) since 1 is the empty product of primes, the empty product being the multiplicative identity, 1.

The concepts of empty set and empty multiset (empty bag), bona fide mathematical concepts, are also key here. Without the concept of empty multiset, 1 would have no prime signature.

The Greeks used to say that 5 - 5 was undefined, since they did not consider 0 as a number. A similar situation happens without the concepts of empty sum, empty product, empty set, empty multiset (empty bag), ...

The mess (which I am longing to clean up) comes from:

• MathWorld: wrongly defines the prime signature of 1 as {1} (I was considering contacting Eric Weisstein to correct this... you say I shouldn't...)
• A118914 Concatenation of {{1}, {1}, {1}, {2}, ...} giving {1, 1, 1, 2, ...}, whose first term corresponds wrongly to {1} as the a prime signature for 1. (I was considering deleting the first term to get it right... )
• A025487, A036035, A046523 which rightly consider the prime signature of 1 to be different from 2 (implying { } for 1 and {1} for 2)

I will remove all references to MathWorld's page for prime signature.

— Daniel Forgues 06:12, 24 December 2010 (UTC)

I finally edited A118914... — Daniel Forgues 07:27, 24 December 2010 (UTC)

Prime signature convention: sorted or reverse sorted?

... the prime signature of $\scriptstyle n\,$, which by convention is sorted in order (sometimes the alternative convention, sorted in reverse order seems to be used.) Confusion ahead... — Daniel Forgues, 22 December 2010

I might have misread. An argument in favor of using the reverse order sort would have been that the prime signature would be in the same order as the exponents of the smallest number of a given prime signature, e.g.

420 = 2^2 * 3^1 * 5^1 * 7^1 is smallest number with prime signature {1, 1, 1, 2}, which I think is where I got confused (and where it would be convenient to have {2, 1, 1, 1} as prime signature with reverse order convention.) — Daniel Forgues 05:33, 23 December 2010 (UTC)

Since the prime signature is a partition of $\scriptstyle \Omega (n)\,$, the order doesn't change the signature, so we are free to choose the sort order which is more convenient (no confusion.) — Daniel Forgues 08:52, 23 December 2010 (UTC)