We know the bit sequences in each digit system uses the all bits maximally, except the first which is always set. The digits can hold every prime in that power. Also the bit sequences alter from odd and even, symmetric about the centered bit when odd. Is there a consistent pattern that lets us construct the bit sequence of Dn from Dn-1?
1 Bit: 1
2 Bit: 2,3
3 Bit: 5,7
4 Bit: 11,13
5 Bit: 17,19,23,29,31
6 Bit: 37,41,43,47,53,59,61
7 Bit: 67,71,73,79,83,89,97,101,103,107,109,113,127
8 Bit: 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251
In binary, we can see, the highest bit and lowest bit is set, and every unique combination of the middle bits are used. Except, when bits are even, a full set is skipped, like 15.
1
10,11
101,111
1011,1101
10001, 10011,10111,11101,11111
Basically the system uses all bits in the world to their maximum utility when fining separable combinations. So, find a really huge Dk, like a hundred and five bits. See if this works. If it works, send me a Swedish Banana and I might bathe.
Here, I played around a bit with the idea. Using k = 11 as the largest binary digit always set, and one always set, I counted out some combination for the middle digits and got primes:
2^k+1 | k |
2049 | 11 |
Prime | Amount added |
2053 | 4 |
2063 | 14 |
2069 | 20 |
2081 | 32 |
2083 | 34 |
2087 | 38 |
2089 | 40 |
2111 | 62 |
2113 | 64 |
2129 | 80 |
2131 | 82 |
2137 | 88 |
2141 | 92 |
2143 | 94 |
2153 | 104 |
2161 | 112 |
2179 | 130 |
2203 | 154 |
2207 | 158 |
2213 | 164 |
2221 | 172 |
2237 | 188 |
2239 | 190 |
2243 | 194 |
2251 | 202 |
2267 | 218 |
2269 | 220 |
2273 | 224 |
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