+0

Row 1: Starting with 1 and 3 each successive number is the sum of the two previous numbers

+5
199
2

Row 1: Starting with 1 and 3 each successive number is the sum of the two previous numbers

Row 2: Take 1 from the number above

Row 3: List the natural numbers. Those in bold are factors of the number above.

 1 3 4 7 11 18 29 47 76 123 199 322 521 0 2 3 6 10 17 28 46 75 122 198 321 520 1 2 3 4 5 6 7 8 9 10 11 12 13

The bold blue numbers are all prime numbers. If the table is continued in row 3 the number above 705 in row 2 is 2169133972532938006110694904080729167368737086736963884235248637362562310877666927155150078519441454973795318130267004238028943442676926535761270635 which is exactly divisible by 705 but 705 is clearly not a prime number. It can be called a "pseudo prime". Up to this stage all the 126 prime numbers have been correctly found.

The  questions is, despite the occasional "pseudo prime", does this method produce all prime numbers (it does up to 100 000).

Guest Mar 3, 2017
#1
0

Very interesting observation!! Did you come up with this scheme yourself? It would have been much more interesting if Row 3 produced ONLY primes, instead of listing all the counting numbers from 1 to infinity, such as this famous observation by Euler: x^2 + x + 41. If you start with 1 and sub it into the equation, it will produce only prime numbers up to 41. Beyond that, it was proven by two mathematicians, that up to 10,000,000 the equation gives 50% primes and 50% composites!.

Another observation. If 705 in Row 3 produces a number in Rows 1 and 2 that is 148 digits long, how big a number would the 1,000,000th number in Row 3 produce in Row 1 and 2?

Is this some new observation or has been known to number theorists for a while? With such large numbers in Rows 1 and 2, maybe number theorists can prove or disprove it in a much more clever ways.

Guest Mar 3, 2017
#2
+26753
+5

For anyone who wants to investigate this further, the following might be of interest:

Alan  Mar 3, 2017