A polynomial operator P is a differential operator or of differential type if deg(P p(x)) = deg(p(x)) - 1 (Combinatorial Theory, Martin Aigner, p. 100, Springer-Verlag, 1979). (Examples: Dx(xn) = n·xn-1, for CONTINUOUS derivative Dx; D(s(xn)) = n·s(xn-1), where Dis the FINITE or DISCRETE derivative and s(x) is the STIRLING FUNCTION OF THE 2ND KIND, for FUNCTAND x.)I'll show that Euler's f-function is a differential operator on my Partorial-function. And challenge Madsters to explore the consequences.
To DEFINE the Partorial-function, I first DEFINE my Primorial-function.
DEF. #n is THE NTH PRIMORIAL IF, AND ONLY IF, #n IS THE PRODUCT OF THE FIRST n PRIMES, n = 1,2,3,..., with #0 = 1, BY DEFINITION. Thus, #1 = 2; #2 = 2·3 = 6; #3 = 2·3·5 = 30; etc. Note that #(n+1) = #n·pn, where pn is THE nTH PRIME.
(In Mathematica,#[n_] := Prime[n] * #[n-1]; #[1] := 2. Thus, #[30] = 31610054640417607788145206291543662493274686990.)
DEF. ##n is THE NTH PARTORIAL IF, AND ONLY IF, ##n IS THE PRODUCT OF THE FIRST n PRIMORIALS, n = 1,2,3,... Thus, ##1 = 2; ##2 = 22·3 = 12; ##3 = 23·32·5 = 360; etc. Note that ##(n+1) = ##n·(#n), where #n is THE nTH PRIMORIAL.
(In Mathematica, ##[n_] := #[n] * ##[n - 1]; #[1] := 2. Thus, #[10] = 37481813439427687898244906452608585200000000. I chose the label "partorial" to INDICATE that THIS FUNCTION CANONICALLY PARTITIONS THE FACTORIAL FUNCTION. Thus, THE DENSITY FUNCTION OF ##3 = 360 YIELDS 1, 3, 5, 6, 5, 3, 1, that is, 1 WAY OF CHOOSING 0 (OR ALL) FACTORS of 360; 3 WAYS OF CHOOSING 1 (OR 5) FACTORS; 5 WAYS OF CHOOSING 2 (or 4) FACTORS; 6 WAYS OF CHOOSING 3 FACTORS; and 1+3+5+6+5+3+1 = 24 = 4!.)
As is well known, Euler's f-function, for functand m, specifies the count of numbers relatively prime to m (do not divide m), by the FORMULA: for k the number of prime factors of m, f(m) = m(1 - 1/p1)(1 - 1/p2)...(1 - 1/pk) = f(m) = mPi=1i=k((pi - 1)/pi).
Whence this f-pattern?
- To find a gnomon for the f-pattern, consider the single prime number p, which is divisible on by numbers 1 & p, its trivial or improper factors. A proper factor would have to be in the range between 1 and p. But it has no proper factors, so all the numbers given by p - 1 are relatively prime to p, nonfactoring p. Then p - 1 is our gnomon, that is, f(p) = p - 1.
- Now, suppose our number is product of two primes: n = p1·p2. Then the numbers p1 - 1 are relatively prime to n; also, the numbers p2 - 1 are relatively prime to n, that is, f(p1) = p1 - 1, and f(p2) = p2 - 1.
- Furthermore, since p1, p2 are MUTUALLY INDEPENDENT PRIMES, it follows that the NUMBER-RANGES p1 - 1, p2 - 1 are COMBINATORIALLY independent. Hence, by THE MULTIPLICATION RULE OF COMBINATORICS, we can MULTIPLY THESE NUMBER-RANGES. So, for n = p1·p2, we find: f(n) = (p1 - 1)·(p2 - 1).
- In general, suppose that n is the product of k primes: n = p1·p2·...·pk. Then, f(n) = (p1 - 1)·(p1 - 1)·...·(pk - 1) = f(p1)·f(p2) ·...·f(pk), as the Euler f-pattern, GENERATED from the general GNOMON, pi - 1.
But how does it transform into the form given above -- with the (1 - pi) terms?
Given Euler's f-function, ##n my partorial function, #n my primorial function, pi the ith prime, and reprise of some results above:
- RULE OF EQUIVALENCE: MULTIPLY & DIVIDE ANY NUMBER m BY THE SAME NONZERO NUMBER, AND IT IS UNCHANGED: m = (k/k)m, if k is NONZERO. So, applying this rule, we have: f(n) = (n/n)(p1 - 1)·(p1 - 1)·...·(pk - 1).
- Moving the denominator n after the "range"-product, we have f(n) = (n)[(p1 - 1)·(p1 - 1)·...·(pk - 1)]/n
.- Remember, the denominator term, n, CAN BE EXPANDED AS THE PRODUCT OF k PRIMES: n = p1·p2·...·pk. And EACH OF THOSE PRIMES IS UNIQUELY ASSOCIATED WITH A RANGE TERM IN THE PRODUCT. So, we CAN MOVE EACH PRIME IN THE GENERAL DENOMINATOR to BECOME DENOMINATOR OF A PRODUCT TERM: f(n) = (n)[(p1 - 1) ·(p1 - 1)·...·(pk - 1)]/[p1·p2·...·pk] = n[((p1 - 1)/p1)·((p1 - 1)/p2) ·...·((pk - 1))/pk].
- Finally, look at a single RATIO on the right-hand side of this, say, (p1 - 1)/p1. We can DIVIDE THROUGH to obtain: (p1 - 1)/p1 = (1 - 1/p1). And similarly with the other terms. So we obtain the original formula, above: f(m) = mPi=1i=k((pi - 1)/pi).
- f=m·Pi=1i=k(1 - 1/pi).(1)
- f=Pi=1i=k((pi - 1)/pi).(2)
- f=Pi=1i=kf(pi).(3)
- ##(n + 1) = ##n·#n
.(4)The "differential" result is proven as a Corollary to a Theorem on ##n, proven by means of two Lemmas on #n adapted from these last (4) results.
LEMMA I: f(#n) = Pi=1i=(pi - 1), from (1), (2).
LEMMA II: f(#n)=Pi=1i=f(pi), from (3).
Theorem:f(##n) = ##(n-1)·f(#n).
Proof:
f(##n) =##n·Pi=1i=n(1 - 1/pi), from (1) and DEFINITIONS.f(##n) =##n·[Pi=1i=n(pi - 1)]/Pi=1i=n(pi), by simple algebra.
f(##n) =[(##n)/Pi=1i=n(pi))]·[Pi=1i=n(pi - 1)], bringing denominator forward.
f(##n) =[(##n/#n)]·[Pi=1i=n(pi - 1)], by DEF. of #n.
f(##n) =##(n - 1)·[Pi=1i=n(pi - 1)], by (4).
f(##n) =##(n - 1)·(#n), by DEF. OF #n.
f(##n) =##(n - 1)·Pi=1i=f(pi), from LEMMA II.
COROLLARY: f IS A DIFFERENTIAL OPERATOR on ##n. (Sends ##n into multiple of ##(n - 1).)
MADSTERS:
- Elsewhere, I've applied the partorial function as a probability function. Can you find other applications?
- Before finding the Aigner criterion, I read somewhere that the criterion for a differential operator is being a homologue of Leibnitz's product-derivative rule, as follows: given u = u(x), v = v(x), Dx(u·v) = u·Dx(v) + v·Dx(u). Can you see how this form might arise in connection with partorials, primorials?
- With or without the Leibnizian homologue, can the above result be expanded into "a calculus"?
Of course, the partorial function can become the basis of a further recursion. As generated on Mathematica, parpartorial[n_] := partorial[n]*parpartorial[n - 1]; parpartorial[1] := 2. Thus, primorial[3] = 12; partorial[3] = 360; parpartorial[3] = 8640.