- #1
ktoz
- 171
- 12
Hi
I was playing around with ways to store numbers more compactly in bases other than 2 and, just for giggles, I tried a "base" of consecutive positive integers that add up to "n". When I applied it to primes, like so
• 2 = 2
1 2 = 3
• 2 3 = 5
1 2 • 4 = 7
1 2 3 • 5 = 11
1 • 3 4 5 = 13
1 2 3 • 5 6 = 17
1 • 3 4 5 6 = 19
1 2 3 4 • 6 7 = 23
1 2 3 4 5 6 • 8 = 29
1 2 3 4 • 6 7 8 = 31
1 2 3 4 5 6 7 • 9 = 37
1 2 3 • 5 6 7 8 9 = 41
1 • 3 4 5 6 7 8 9 = 43
1 2 3 4 5 6 7 • 9 10 = 47
1 • 3 4 5 6 7 8 9 10 = 53
1 2 3 4 5 6 • 8 9 10 11 = 59
1 2 3 4 • 6 7 8 9 10 11 = 61
1 2 3 4 5 6 7 8 9 10 • 12 = 67
1 2 3 4 5 6 • 8 9 10 11 12 = 71
1 2 3 4 • 6 7 8 9 10 11 12 = 73
1 2 3 4 5 6 7 8 9 10 11 • 13 = 79
1 2 3 4 5 6 7 • 9 10 11 12 13 = 83
1 • 3 4 5 6 7 8 9 10 11 12 13 = 89
1 2 3 4 5 6 7 • 9 10 11 12 13 14 = 97
1 2 3 • 5 6 7 8 9 10 11 12 13 14 = 101
1 • 3 4 5 6 7 8 9 10 11 12 13 14 = 103
1 2 3 4 5 6 7 8 9 10 11 12 • 14 15 = 107
1 2 3 4 5 6 7 8 9 10 • 12 13 14 15 = 109
1 2 3 4 5 6 • 8 9 10 11 12 13 14 15 = 113
1 2 3 4 5 6 7 8 • 10 11 12 13 14 15 16 = 127
1 2 3 4 • 6 7 8 9 10 11 12 13 14 15 16 = 131
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 • 17 = 137
1 2 3 4 5 6 7 8 9 10 11 12 13 • 15 16 17 = 139
1 2 3 • 5 6 7 8 9 10 11 12 13 14 15 16 17 = 149
1 • 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 = 151
1 2 3 4 5 6 7 8 9 10 11 12 13 • 15 16 17 18 = 157
1 2 3 4 5 6 7 • 9 10 11 12 13 14 15 16 17 18 = 163
1 2 3 • 5 6 7 8 9 10 11 12 13 14 15 16 17 18 = 167
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 • 18 19 = 173
1 2 3 4 5 6 7 8 9 10 • 12 13 14 15 16 17 18 19 = 179
1 2 3 4 5 6 7 8 • 10 11 12 13 14 15 16 17 18 19 = 181
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 • 20 = 191
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 • 18 19 20 = 193
1 2 3 4 5 6 7 8 9 10 11 12 • 14 15 16 17 18 19 20 = 197
1 2 3 4 5 6 7 8 9 10 • 12 13 14 15 16 17 18 19 20 = 199
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 • 21 = 211
1 2 3 4 5 6 7 • 9 10 11 12 13 14 15 16 17 18 19 20 21 = 223
1 2 3 • 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 = 227
1 • 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 = 229
A potentially interesting pattern emerged. If you look down the columns, you notice that for primes greater than 7 (m = 4) there are no primes missing their 3, 6, 10, 15, 21 terms. 3, 6, 10 etc are exact sums of n(n + 1) / 2
Given:
m, n element of positive integers
p element of primes
Where:
m > 4
m > b
a = m(m+1) / 2
b = n(n + 1) / 2
Then:
a - b != p
I wrote a quick and dirty program and tested this for all primes up to p(300,000) and it seems to hold. How would I go about proving this? It's been years since I did proofs so I don't even know where to start.
Any help appreciated
P.S. is there some command that allows the display of nice neat tables within a post?
I was playing around with ways to store numbers more compactly in bases other than 2 and, just for giggles, I tried a "base" of consecutive positive integers that add up to "n". When I applied it to primes, like so
• 2 = 2
1 2 = 3
• 2 3 = 5
1 2 • 4 = 7
1 2 3 • 5 = 11
1 • 3 4 5 = 13
1 2 3 • 5 6 = 17
1 • 3 4 5 6 = 19
1 2 3 4 • 6 7 = 23
1 2 3 4 5 6 • 8 = 29
1 2 3 4 • 6 7 8 = 31
1 2 3 4 5 6 7 • 9 = 37
1 2 3 • 5 6 7 8 9 = 41
1 • 3 4 5 6 7 8 9 = 43
1 2 3 4 5 6 7 • 9 10 = 47
1 • 3 4 5 6 7 8 9 10 = 53
1 2 3 4 5 6 • 8 9 10 11 = 59
1 2 3 4 • 6 7 8 9 10 11 = 61
1 2 3 4 5 6 7 8 9 10 • 12 = 67
1 2 3 4 5 6 • 8 9 10 11 12 = 71
1 2 3 4 • 6 7 8 9 10 11 12 = 73
1 2 3 4 5 6 7 8 9 10 11 • 13 = 79
1 2 3 4 5 6 7 • 9 10 11 12 13 = 83
1 • 3 4 5 6 7 8 9 10 11 12 13 = 89
1 2 3 4 5 6 7 • 9 10 11 12 13 14 = 97
1 2 3 • 5 6 7 8 9 10 11 12 13 14 = 101
1 • 3 4 5 6 7 8 9 10 11 12 13 14 = 103
1 2 3 4 5 6 7 8 9 10 11 12 • 14 15 = 107
1 2 3 4 5 6 7 8 9 10 • 12 13 14 15 = 109
1 2 3 4 5 6 • 8 9 10 11 12 13 14 15 = 113
1 2 3 4 5 6 7 8 • 10 11 12 13 14 15 16 = 127
1 2 3 4 • 6 7 8 9 10 11 12 13 14 15 16 = 131
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 • 17 = 137
1 2 3 4 5 6 7 8 9 10 11 12 13 • 15 16 17 = 139
1 2 3 • 5 6 7 8 9 10 11 12 13 14 15 16 17 = 149
1 • 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 = 151
1 2 3 4 5 6 7 8 9 10 11 12 13 • 15 16 17 18 = 157
1 2 3 4 5 6 7 • 9 10 11 12 13 14 15 16 17 18 = 163
1 2 3 • 5 6 7 8 9 10 11 12 13 14 15 16 17 18 = 167
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 • 18 19 = 173
1 2 3 4 5 6 7 8 9 10 • 12 13 14 15 16 17 18 19 = 179
1 2 3 4 5 6 7 8 • 10 11 12 13 14 15 16 17 18 19 = 181
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 • 20 = 191
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 • 18 19 20 = 193
1 2 3 4 5 6 7 8 9 10 11 12 • 14 15 16 17 18 19 20 = 197
1 2 3 4 5 6 7 8 9 10 • 12 13 14 15 16 17 18 19 20 = 199
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 • 21 = 211
1 2 3 4 5 6 7 • 9 10 11 12 13 14 15 16 17 18 19 20 21 = 223
1 2 3 • 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 = 227
1 • 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 = 229
A potentially interesting pattern emerged. If you look down the columns, you notice that for primes greater than 7 (m = 4) there are no primes missing their 3, 6, 10, 15, 21 terms. 3, 6, 10 etc are exact sums of n(n + 1) / 2
Given:
m, n element of positive integers
p element of primes
Where:
m > 4
m > b
a = m(m+1) / 2
b = n(n + 1) / 2
Then:
a - b != p
I wrote a quick and dirty program and tested this for all primes up to p(300,000) and it seems to hold. How would I go about proving this? It's been years since I did proofs so I don't even know where to start.
Any help appreciated
P.S. is there some command that allows the display of nice neat tables within a post?
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