Thread: Minimum (Number Theory Game) View Single Post 2021-08-12, 13:02 #18 ThomasK   Aug 2021 108 Posts Theorem of value restriction of sufficiently large potencies: n is a natural number with n >= 2 Then applies W (nk) = 0 for all k > = ln n / ln 2 - 1 if n is even W (nk) = 1 for all k > = (n-1) / 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(2k) = 0 for all k = 1, 2, 3, 4, ....(Powers of 4 included) W(3k) = 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 nk = 2k * (n/2)k. In the worst case, n / 2 is a prime number. But if n / 2 <= 2k, you can divide each away n / 2 in the first k steps and get the Nullwertzahl 2k. A sufficient condition for nk to be a Nullwertzahl is therefore 2k > = n / 2 or k > = ln (n / 2) / ln 2 = ln n / ln 2 - 1. Second case: n is odd n2 - 1 = (n + 1)(n - 1) = 22 * (n + 1) / 2 * (n - 1) / 2 n4 - 1 = (n2 + 1)(n2 - 1) = (n2 + 1)(n + 1)(n - 1) In general n(2^j) - 1 = (n(2^(j-1)) + 1)(n(2^(j-1)) - 1) = 2(j+1) * (n(2^(j-1)) + 1) / 2 * (n(2^(j-2)) + 1) / 2 * ... * (n(2^1) + 1) / 2 * (n + 1) / 2 * (n - 1) / 2 Let us first take the case that k = 2j is a power of 2. Then we subtract one from nk and get nk - 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 2j > = (n - 1) / 4 follows. In the general case that k is not a power of 2, we divide n from nk away until the remaining exponent is a power of 2 that is 2j. The original exponent k must shrink by half at most. So if k > = (n - 1) / 2, then 2j > = (n - 1) / 4. So it follows that W (nk) = 1 for all k > = (n - 1) / 2. With this the Theorem of value restriction of sufficiently large potencies is completely proved! __________________________________________________________ This theorem has far-reaching consequences in Minimum theory. We denote the set of all Nullwertzahlen with W0, the set of all Einswertzahlen with W1, the set of all Zweiwertzahlen with W2, the set of all Dreiwertzahlen with W3, the set of all Vierwertzahlen with W4, and the set of all k-valued numbers with Wk. 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 Wk 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 1012, but we took a random look at the distribution of the numbers above 1012. The following table shows the number of Nullwertzahlen up to the Vierwertzahlen in the first milliard above 1012, 1013 and 1014. 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 1018. W0(1012 + 109) - W0(1012): 22129699 W1(1012 + 109) - W1(1012): 538366808 W2(1012 + 109) - W2(1012): 429800315 W3(1012 + 109) - W3(1012): 9703057 W4(1012 + 109) - W4(1012): 121 W0(1013 + 109) - W0(1013): 20455311 W1(1013 + 109) - W1(1013): 531664281 W2(1013 + 109) - W2(1013): 438266115 W3(1013 + 109) - W3(1013): 9614183 W4(1013 + 109) - W4(1013): 110 W0(1014 + 109) - W0(1014): 19015775 W1(1014 + 109) - W1(1014): 525804423 W2(1014 + 109) - W2(1014): 445685370 W3(1014 + 109) - W3(1014): 9494338 W4(1014 + 109) - W4(1014): 94 In the next Posting we will compare the Nullwertzahlen with the prime numbers. The comparison of the functions W0(n) and pi (n) is very interesting.  