Sphenic number
As Sphenische numbers ( altgr. σφήν sphene " wedge" ) are used in the mathematical theory of numbers denotes the natural numbers that are the product of exactly three different prime numbers. For example, the number 30 is a sphenische number because it can be represented by a product of the primes 2, 3 and 5 ( prime factorization ). 60, however, is not sphenische number: Although these exactly three prime numbers can be represented by the product (60 = 22 ⋅ 3 ⋅ 5), but in the prime factorization takes the two double-click. The sphenischen numbers are so fast primes of order 3
In the Oeconomischen Encyclopedia of Johann Georg Krünitz from the late 18th and early 19th centuries a sphenische number is defined as
Thus, a number that can be represented as the product of three different integers, but need not be prime numbers ( in the example given, 24 = 2 ⋅ 3 ⋅ 4 2 and 3, although primes but not 4 ).
All sphenischen numbers n = pqr (p < q < r ) have exactly 8 splitter ( namely 1, p, q, r, pq, pr, qr and pqr ). In general, if n is square-free and product of k primes, then n has exactly divider ( 1 and n included). Sphenische numbers are, by definition, is square-free (that have dividers). The Möbius function is obtained for each sphenische number -1. A famous sphenische number is the Hardy - Ramanujan number 1729 = 71319th The sphenische number 1001 = 71113 use is uncommon in Teilbarkeitsüberlegungen.
The first sphenischen figures are: 30, 42, 66, 70, 78, 102, 105, 110, 114, 130, 138, 154, 165, ...
The current (2009) sphenische largest known number is, the product of the three largest known primes.
Imperfection
The sum of the divider, and including 1 and n, sec = (p 1) (q 1 ) (r 1). Sphenische numbers are not perfect, because otherwise s would = 2pqr. The left-hand side of ( p 1) (q 1 ) (r 1 ) = 2pqr is divisible by 4 (for q 1 and r 1 are straight), so 4 divides the right side, so that p = must be 2 (because p < q < r). Therefore divides 2 1 = 3, the left, and then the right side, from which q = 3 follows. s = 2pqr therefore reduces to the contradictory equation 12 (r 1 ) = 12r.
All odd numbers sphenischen are deficient because
So (p 1) (q 1 ) (r 1) < 2pqr. Among the straight sphenischen numbers are only 23r (with any prime factor r > 3) and 257 = 70 abundant (all other deficient ). Sphenische numbers of the form 23r are pseudo- complete (see perfect number ) because it is the sum of at least some divisor (namely r, 2r, 3r ) can be prepared. 70 is the only strange (ie abundant, but not pseudo- perfect ) sphenische number.
The formula for the sum of all divisors of square -free numbers is capable of generalization ( eg proof by induction on k). The are different primes. for
Applies
It follows that n is not perfectly even for k > 3 ( indirect proof above can be easily 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 quite perfect).
Twin figures
The numbers 230 (2 ⋅ 5 ⋅ 23) and 231 ( 3 ⋅ 7 ⋅ 11) form the first pair of two directly consecutive sphenischer numbers; we call such a pair of numbers sphenische twin numbers.
For sphenische twin numbers abc < pqr pqr is true - abc = 1, r and c are thus prime solutions of the Diophantine equation PQX - aby = 1 By a theorem of elementary number theory have all the solutions of the form x = u and y = v dependent pqh, while u and v are positive minimal solutions and h is an integer. Are Looking for twins you need to prime numbers a, b , p, q to determine only an hour, so that x and y are prime numbers. An example 27x - 35y = 1 15y = - 14x or simplified 1 All solutions have the form x = 14 and y = 13 15 h 14 h For h = 5 x and y are prime numbers ( ie 89 and 83 ), so give 3583 = 1245 2789 = 1246 sphenische twin numbers. And for h = 9, 11, 15 and 29 prime can be found solutions and consequently sphenische twins (for example, 6285 and 6286 for h = 29).
Drilling numbers
The first triplet of consecutive sphenischen numbers form 1309 (7 ⋅ 11 ⋅ 17), 1310 (2 ⋅ 5 ⋅ 131) and 1311 ( 3 ⋅ 19 ⋅ 23). In such figures, the drilling sphenischen medium required by a factor of 2 (not the two outer, because even numbers of two adjacent a is divisible by 4 ). A sequence of four or more consecutive straight sphenischen figures are not available, because every fourth integer is divisible by 4 and hence not square-free.
The search for triplets is illustrated by two examples.
Example 1: The two Diophantine equations 217x - 311Y = 1 and 57s - 217t = 1 the solutions x = 1 and y = 1 33 h 34 h and s = 1 and t = 1 k 34 35 k ( h and k integers ). Thus the two solutions arise Drilling numbers, must x = t, ie 33h = 35k. Solutions of this equation are special diophantine h = 35j and k = 33j ( j integer ). Therefore: x = 1 1155 j, y = 1 1190 j, s = 1 1122 j ( and t = x by construction ). x, y are prime numbers and s for j = 44 (namely, 50821, 52361 and 49369 ). So are the numbers 31152361 = 1727913, 21750821 and 5749369 = 1727914 = 1727915 sphenische triplets.
Example 2: 25x - 711y = 1 and 319s - 25t = 1 or calculated: 10x - 77y = 1 and 57s - 10t = 1 The solution set can be described by x = 54 77 h, y = 7 10 h, s = 3 10 k, t k 57 = 17 (h and k are integers ). From the condition x = t followed by a third Diophantine equation: 57k - 77h = 37 are your solutions h = 1 57 j and k = 2 77 j ( j integer ). If they are used in the formulas for x, y and s, we obtain x = 131 4389 j, y = 17 570 j and s = 23 770 j (t = x by construction ). This sphenische Drilling numbers arise, have x, y and s be primes. That is, for j = 0 is already the case (you get the well-known triplet 1309, 1310, 1311 ). Also j = 10 generates prime numbers ( x = 44021, y = 5717 and s = 7723 ). The corresponding triplet is: 440209, 440210, 440211th
Triplets are rare. Nevertheless, it seems likely that one can construct any number. But the proof that there are infinitely many triplets is likely to be difficult. Also the set of Lejeune Dirichlet does not help, because it indicates only that in every sequence x = 131 4389 j but is ever so infinitely many prime numbers exist for itself, required for all three episodes each the same j.