Sphenic number

As sphenische numbers ( Greek σφήν Sphen "wedge") in the be mathematical number theory the natural numbers indicated that the product exactly three different prime numbers are. For example, the number 30 is a sphenic number because it ( prime factorization ) can be represented by a product of the prime numbers 2, 3 and 5. 60, on the other hand, is not a sphenic number: Although this can also be represented by the product of exactly three prime numbers , the 2 appears twice in the prime factorization. The sphenic numbers are therefore almost prime numbers of order 3. ${\displaystyle (60=2^{2}\cdot 3\cdot 5),}$

In the Oeconomischen Encyclopedia of Johann Georg Krünitz from the late 18th and early 19th century sphenic number is defined as

[...] a body number that has three unequal sides, e.g. B. four and twenty, the sides of which are two, three and four.

a number that can be represented as the product of three different whole numbers, but which do not have to be prime numbers (in the given example , 2 and 3 are prime numbers, but 4 are not). ${\displaystyle 24=2\cdot 3\cdot 4}$

All sphenic numbers have exactly 8 divisors (namely 1, p, q, r, pq, pr, qr and pqr). In general, if n is square-free and the product of k prime numbers, then n has exactly divisors (including 1 and n). Sphenic numbers are by definition free of squares (i.e. have divisors). The Möbius function gives −1 for every sphenic number. A famous sphenic number is the Hardy Ramanujan number 1729 = The sphenic number is occasionally useful when considering divisibility. ${\displaystyle n=pqr(p<q<r)}$ ${\displaystyle 2^{k}}$ ${\displaystyle 2^{3}}$ ${\displaystyle 7\cdot 13\cdot 19.}$ ${\displaystyle 1001=7\cdot 11\cdot 13}$

The first sphenic numbers are: 30, 42, 66, 70, 78, 102, 105, 110, 114, 130, 138, 154, 165, ...

The currently (2018) largest known sphenic number is the product of the three largest known prime numbers. ${\displaystyle (2^{74.207.281}-1)\cdot (2^{77.232.917}-1)\cdot (2^{82.589.933}-1),}$

imperfection

The sum of all divisors, including 1 and n, is Sphenic numbers are not perfect , otherwise s = 2pqr. The left side of is divisible by 4 (because and are even), therefore 4 also divides the right side, so that must be (because of ). Hence divides the left, and then also the right side, which follows. is therefore reduced to the contradicting equation ${\displaystyle s=(p+1)(q+1)(r+1).}$ ${\displaystyle (p+1)(q+1)(r+1)=2pqr}$ ${\displaystyle q+1}$ ${\displaystyle r+1}$ ${\displaystyle p=2}$ ${\displaystyle p<q<r}$ ${\displaystyle 2+1=3}$ ${\displaystyle q=3}$ ${\displaystyle s=2pqr}$ ${\displaystyle 12(r+1)=12r.}$

All odd sphenic numbers are deficient because

${\displaystyle {\frac {(p+1)}{p}}\cdot {\frac {(q+1)}{q}}\cdot {\frac {(r+1)}{r}}\leq {\frac {4}{3}}\cdot {\frac {6}{5}}\cdot {\frac {8}{7}}<2,}$

so among the even sphenic numbers are only (with any prime factor ) and abundant (all others are deficient). Sphenic numbers of the form are pseudo-perfect (see perfect number ) because they can be represented as the sum of at least some divisors (namely r, 2r, 3r). 70 is the only odd (i.e., abundant, but not pseudo-perfect) Sphenic number. ${\displaystyle (p+1)(q+1)(r+1)<2pqr.}$ ${\displaystyle 2\cdot 3\cdot r}$ ${\displaystyle r>3}$ ${\displaystyle 2\cdot 5\cdot 7=70}$ ${\displaystyle 2\cdot 3\cdot r}$

The formula for the sum of all divisors of square-free numbers can be generalized (proof e.g. with complete induction over k). They are different prime numbers. For ${\displaystyle p_{i}}$

${\displaystyle n=\prod _{i=1}^{k}p_{i}}$

applies

${\displaystyle s=\prod _{i=1}^{k}(p_{i}+1)}$.

It follows that n is also imperfect (the indirect proof above can easily be extended to the general case). So: All square-free numbers with at least three prime factors are not perfect (6 on the other hand, a square-free number with only two prime factors, is absolutely perfect). ${\displaystyle k>3}$

Twin numbers

The numbers and form the first pair of two consecutive sphenic numbers; Such a pair of numbers is called sphenic twin numbers . ${\displaystyle 230(2\cdot 5\cdot 23)}$ ${\displaystyle 231(3\cdot 7\cdot 11)}$

For sphenic twin numbers , r and c are prime number solutions of the Diophantine equation. According to a theorem of elementary number theory , all solutions have the form and , where u and v are positive minimal solutions and h is an integer. In the search for twins one only needs to determine one h for various prime numbers a, b, p, q, so that x and y become prime numbers. An example or simplified All solutions have the form and For are x and y prime numbers (namely 89 and 83), so that and result in sphenic twins. For and one also finds prime number solutions and consequently sphenic twins (e.g. 6285 and 6286 for ). ${\displaystyle abc<pqr}$ ${\displaystyle pq\cdot r-ab\cdot c=1.}$ ${\displaystyle pq\cdot x-ab\cdot y=1.}$ ${\displaystyle x=u+ab\cdot h}$ ${\displaystyle y=v+pq\cdot h}$ ${\displaystyle 2\cdot 7\cdot x-3\cdot 5\cdot y=1}$ ${\displaystyle 14x-15y=1.}$ ${\displaystyle x=14+15\cdot h}$ ${\displaystyle y=13+14\cdot h.}$ ${\displaystyle h=5}$ ${\displaystyle 3\cdot 5\cdot 83=1245}$ ${\displaystyle 2\cdot 7\cdot 89=1246}$ ${\displaystyle h=9,11,15}$ ${\displaystyle 29}$ ${\displaystyle h=29}$

Triplets

Form the first triplet of consecutive spherical numbers and . With such sphenic triplet numbers, the middle one necessarily has the factor 2 (not the two outer ones, because one of two adjacent even numbers is divisible by 4). There is no sequence of four or more consecutive sphenic numbers, since every fourth whole number is divisible by 4 and is therefore not square-free. ${\displaystyle 1309(7\cdot 11\cdot 17),1310(2\cdot 5\cdot 131)}$ ${\displaystyle 1311(3\cdot 19\cdot 23)}$

The search for triplets will be explained using two examples.

1st example: The two Diophantine equations and have the solutions and or and (h and k integers). So that the two solutions result in triplet numbers, must , so . Solutions to this particular Diophantine equation are and (j integer). Therefore: (and according to construction). x, y and s are prime numbers for (namely 50821, 52361 and 49369). So the numbers and sphenic triplets form. ${\displaystyle 2\cdot 17\cdot x-3\cdot 11\cdot y=1}$ ${\displaystyle 5\cdot 7\cdot s-2\cdot 17\cdot t=1}$ ${\displaystyle x=1+33\cdot h}$ ${\displaystyle y=1+34\cdot h}$ ${\displaystyle s=1+34\cdot k}$ ${\displaystyle t=1+35\cdot k}$ ${\displaystyle x=t}$ ${\displaystyle 33\cdot h=35\cdot k}$ ${\displaystyle h=35\cdot j}$ ${\displaystyle k=33\cdot j}$ ${\displaystyle x=1+1155\cdot j,y=1+1190\cdot j,s=1+1122\cdot j}$ ${\displaystyle t=x}$ ${\displaystyle j=44}$ ${\displaystyle 3\cdot 11\cdot 52361=1727913,2\cdot 17\cdot 50821=1727914}$ ${\displaystyle 5\cdot 7\cdot 49369=1727915}$

2nd example: and or calculated: and The solution set can be described by (h and k whole numbers). A third Diophantine equation follows from the condition : Its solutions are and (j integer). If you put them in the formulas for x, y and s, you get and ( after construction). For sphenic triplet numbers to arise, x, y and s must be prime numbers. This is already the case (you get the well-known triplet 1309, 1310, 1311). Also generates prime numbers ( and ). The corresponding triplet is: 440209, 440210, 440211. ${\displaystyle 2\cdot 5\cdot x-7\cdot 11\cdot y=1}$ ${\displaystyle 3\cdot 19\cdot s-2\cdot 5\cdot t=1}$ ${\displaystyle 10x-77y=1}$ ${\displaystyle 57s-10t=1.}$ ${\displaystyle x=54+77h,y=7+10h,s=3+10k,t=17+57k}$ ${\displaystyle x=t}$ ${\displaystyle 57k-77h=37.}$ ${\displaystyle h=1+57j}$ ${\displaystyle k=2+77j}$ ${\displaystyle x=131+4389j,y=17+570j}$ ${\displaystyle s=23+770j}$ ${\displaystyle t=x}$ ${\displaystyle j=0}$ ${\displaystyle j=10}$ ${\displaystyle x=44021,y=5717}$ ${\displaystyle s=7723}$

Triplets are rare. Nevertheless, it seems reasonable to assume that you can construct as many as you want. But it would be hard to prove that there are an infinite number of triplets. Also, the set of Lejeune Dirichlet further does not help because it only states that in each episode , etc. depending infinite for many primes, but required for all three episodes each the same j. ${\displaystyle x=131+4389j}$

swell

1. ↑ Economic Encyclopedia online
2. ↑ Follow A007304 in OEIS
3. ↑ This can e.g. B. prove with the prime number tester

Web links

• Follow A007304 in OEIS (English)