As stated in the last post in this series I thought I was done with this number theory but, "every time I try to get out, they keep pulling me back in!" I sent my ABC Factorial Proof off for publication at a respected number theory journal back in January, and it got through desk rejection and they are searching for a referee to peer review it.
Number fact: The number 63,700,992 = 2^18 x 3^5 is a cute little (two primes) π-complete number stuck between 63,700,991 and 63,700,993 which are both prime. This sequence of very special numbers starts with 6 = 2 x 3, which is stuck between primes 5, 7.
Because I found a way to map n(n+1) to m(m+k) and proved a π-complete limit for n(n+1) it means I can reverse the map to show that if some k made m(m+k) unbounded, then that contradicts my proof for n(n+1). So that proves there is a π-complete limit for all m(m+k). The value of that limit is k times the highest n for which n(n+1) is missing only the prime factors of k. Q.E.D.
I found another special case because 949,025 x 949,026 is missing prime 19. So for n(n+k) the π-complete limit with be 949,025 * k for all k that are multiples of 19.
I now have a proof that n(n+k) is π-complete limited for all positive integers, k. In most cases, that limit will be where n = 633,555 * k, however, there are interesting special cases where k is a multiple of 5, and the limit goes up to 709,631 * k. This is because 709,631 * 709,632 could not be a solution for n(n+1) because the product is missing the single prime factor 5. When k is a multiple of 5, it puts that missing prime factor back in, so it extends the limit. I am currently looking for any other special cases where a missing prime factor can be "filled in" but still in all cases, there will be a π-complete limit.
Number fact: The number 63,700,992 = 2^18 x 3^5 is a cute little (two primes) π-complete number stuck between 63,700,991 and 63,700,993 which are both prime. This sequence of very special numbers starts with 6 = 2 x 3, which is stuck between primes 5, 7.
Because I found a way to map n(n+1) to m(m+k) and proved a π-complete limit for n(n+1) it means I can reverse the map to show that if some k made m(m+k) unbounded, then that contradicts my proof for n(n+1). So that proves there is a π-complete limit for all m(m+k). The value of that limit is k times the highest n for which n(n+1) is missing only the prime factors of k. Q.E.D.
I found another special case because 949,025 x 949,026 is missing prime 19. So for n(n+k) the π-complete limit with be 949,025 * k for all k that are multiples of 19.
Also:
2,697,695 x 2,697,696 is missing prime 23
3,897,165 x 3,897,166 is missing prime 11
5,142,500 x 5,142,501 is missing prime 19
5,909,760 x 5,909,761 is missing prime 7
11,859,210 x 11,859,211 is missing prime 17
14,753,024 x 14,753,025 is missing prime 23
16,092,999 x 16,093,000 is missing prime 23
18,085,704 x 18,085,705 is missing prime 19
80,061,344 x 80,061,345 is missing prime 17
I now have a proof that n(n+k) is π-complete limited for all positive integers, k. In most cases, that limit will be where n = 633,555 * k, however, there are interesting special cases where k is a multiple of 5, and the limit goes up to 709,631 * k. This is because 709,631 * 709,632 could not be a solution for n(n+1) because the product is missing the single prime factor 5. When k is a multiple of 5, it puts that missing prime factor back in, so it extends the limit. I am currently looking for any other special cases where a missing prime factor can be "filled in" but still in all cases, there will be a π-complete limit.