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2020-07-07, 18:08   #868
sweety439

Nov 2016

13·173 Posts

Quote:
 Originally Posted by sweety439 Checking whether a k-value makes a full covering set with algebraic factors not always very easy. The way I do it is to look for patterns in the factors of the various n-values for specific k-values. If there are algebraic factors, it's most common for them to be in a pattern of f*(f+2), i.e.: 11*13 179*181 etc. In other cases there may be a consistent steady increase in the differences of their factors, which is especially tricky to find but indicates the existence of algebraic factors. e.g. for the case R15 k=47 n-value : factors 1 : 2^5 · 11 2 : 17 · 311 3 : 2^4 · 4957 4 : 31 · 38377 6 : 11 · 43 · 565919 8 : 199 · 1627 · 186019 10 : 17 · 61 · 13067776451 12 : 37 · 82406457849451 20 : 15061 · 236863181 · 2190492030407 Analysis: For n=1 & 3 (and all odd n), all values are divisible by 2 so we only consider even n's. For n=4, the two prime factors does not close. For n=6 & 10, multiplying the 2 lower prime factors together does not come close to the higher prime factor so little chance of algebraic factors. For n=12, the large lowest prime factor that bears no relation to the other prime factor means that there is unlikely to be a pattern to the occurrences of large prime factors so there must be a prime at some point. R33 k=257: n-value : factors 1 : 5 · 53 2 : 2 · 4373 3 : 397 · 727 4 : 2^2 · 2381107 5 : 5^3 · 7 · 359207 7 : 11027 · 31040117 15 : 13337 · 706661 · 51076716238627 19 : 38231 · 14932493857679888742000509 For n=15 & 19 same explanation as R15 k=47 R36 k=1555: n-value : factors 1 : 11 · 727 2 : 31 · 37 · 251 3 : 67 · 154691 4 : 37 · 127 · 271 · 293 7 : 4943 · 3521755879 9 : 59 · 382386761790283 For n=7 & 9 same explanation as R15 k=47 The prime factors for n=12, n=15, and n=7 respectively make it clear to me that these k-values should all yield primes at some point so you are correct to include them as remaining. The higher-math folks may be able to chime in and answer why there are an abnormally large # of k's that are perfect squares that end up remaining even though they don't have known algebraic factors for most bases. IMHO, it's because there ARE algebraic factors for a subset of the universe of n-values on them but not for all of the n-values. Hence they are frequently lower weight than the other k's but NOT zero weight and so should eventually yield a prime.
In fact, (k*b^n+1)/gcd(k+1,b-1) has algebra factors if and only if k*b^n is either perfect odd power or of the form 4*m^4, and (k*b^n-1)/gcd(k-1,b-1) has algebra factors if and only if k*b^n is perfect power, thus forms like (257*33^n-1)/32 cannot have algebra factors, since there is no n such that 257*33^n is perfect power (after all, the exponent of the prime 257 in the prime factorization of 257*33^n is 1 for all n, but any prime factor in the prime factorization of a perfect power cannot have exponent 1).

Last fiddled with by sweety439 on 2020-07-07 at 18:09