Theorem of value restriction of sufficiently large potencies:
n is a natural number with n >= 2
Then applies
W (n^{k}) = 0 for all k > = ln n / ln 2  1 if n is even
W (n^{k}) = 1 for all k > = (n1) / 2 if n is odd
Proof:
We prove this theorem specifically for n = 2 and n = 3 and n = 4 and in general for n > = 5.
W(2^{k}) = 0 for all k = 1, 2, 3, 4, ....(Powers of 4 included)
W(3^{k}) = 1 for all k = 1, 2, 3, 4, ....
Let n > = 5 be a natural number and k a natural number.
First case: n is even
n = 2 * (n / 2) and n^{k} = 2^{k} * (n/2)^{k}. In the worst case, n / 2 is a prime number.
But if n / 2 <= 2^{k}, you can divide each away n / 2 in the first k steps and get the Nullwertzahl 2^{k}. A sufficient condition for n^{k} to be a Nullwertzahl is therefore 2^{k} > = n / 2 or k > = ln (n / 2) / ln 2 = ln n / ln 2  1.
Second case: n is odd
n^{2}  1 = (n + 1)(n  1) = 2^{2} * (n + 1) / 2 * (n  1) / 2
n^{4}  1 = (n^{2} + 1)(n^{2}  1) = (n^{2} + 1)(n + 1)(n  1)
In general
n^{(2^j)}  1 = (n^{(2^(j1))} + 1)(n^{(2^(j1))}  1) =
2^{(j+1)} * (n^{(2^(j1))} + 1) / 2 * (n^{(2^(j2))} + 1) / 2 * ... * (n^{(2^1)} + 1) / 2 * (n + 1) / 2 * (n  1) / 2
Let us first take the case that k = 2^{j} is a power of 2. Then we subtract one from n^{k} and get n^{k}  1, which costs us a move. From the above factorization of n^{(2^j)}  1 we can now divide away all the odd factors one after the other, first the largest, then the second largest, and so on, because of n > = 5 it the largest factor in each case is still smaller than the product of remaining factors. The last thing left is (n  1) / 2, which in the worst case is a prime number. So n^{(2^j)}  1 is a Nullwertzahl if 2^{(j+1)} >= (n  1) / 2, from which 2^{j} > = (n  1) / 4 follows.
In the general case that k is not a power of 2, we divide n from n^{k} away until the remaining exponent is a power of 2 that is 2^{j}. The original exponent k must shrink by half at most. So if k > = (n  1) / 2, then 2^{j} > = (n  1) / 4. So it follows that W (n^{k}) = 1 for all k > = (n  1) / 2.
With this the Theorem of value restriction of sufficiently large potencies is completely proved!
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This theorem has farreaching consequences in Minimum theory.
We denote the set of all Nullwertzahlen with W_{0}, the set of all Einswertzahlen with W_{1}, the set of all Zweiwertzahlen with W_{2}, the set of all Dreiwertzahlen with W_{3}, the set of all Vierwertzahlen with W_{4}, and the set of all kvalued numbers with W_{k}.
If every sufficiently large power of EVERY even natural number is a Nullwertzahl and every sufficiently large power of EVERY odd natural number is a Einswertzahl, then the conjecture that the density of the set of all Numbers W_{k} is 0 is false!
Sometimes we call the set of Nullwertzahlen und Einswertzahlen together as cheap numbers and those numbers with W (n) > = 3 as expensive numbers. So far we have only carried out the total analysis up to 10^{12}, but we took a random look at the distribution of the numbers above 10^{12}. The following table shows the number of Nullwertzahlen up to the Vierwertzahlen in the first milliard above 10^{12}, 10^{13} and 10^{14}.
The distribution is as our theoretical estimates predict. But there is still a long way to go to the smallest Fünfwertzahl, which we suspect to be on the order of 10^{18}.
W_{0}(10^{12} + 10^{9})  W_{0}(10^{12}): 22129699
W_{1}(10^{12} + 10^{9})  W_{1}(10^{12}): 538366808
W_{2}(10^{12} + 10^{9})  W_{2}(10^{12}): 429800315
W_{3}(10^{12} + 10^{9})  W_{3}(10^{12}): 9703057
W_{4}(10^{12} + 10^{9})  W_{4}(10^{12}): 121
W_{0}(10^{13} + 10^{9})  W_{0}(10^{13}): 20455311
W_{1}(10^{13} + 10^{9})  W_{1}(10^{13}): 531664281
W_{2}(10^{13} + 10^{9})  W_{2}(10^{13}): 438266115
W_{3}(10^{13} + 10^{9})  W_{3}(10^{13}): 9614183
W_{4}(10^{13} + 10^{9})  W_{4}(10^{13}): 110
W_{0}(10^{14} + 10^{9})  W_{0}(10^{14}): 19015775
W_{1}(10^{14} + 10^{9})  W_{1}(10^{14}): 525804423
W_{2}(10^{14} + 10^{9})  W_{2}(10^{14}): 445685370
W_{3}(10^{14} + 10^{9})  W_{3}(10^{14}): 9494338
W_{4}(10^{14} + 10^{9})  W_{4}(10^{14}): 94
In the next Posting we will compare the Nullwertzahlen with the prime numbers. The comparison of the functions W_{0}(n) and pi (n) is very interesting.
