| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | |||
|---|---|---|---|---|---|---|---|---|---|---|
| 104 | 29 | 3 | 5 | 9 | 0 | 1 | 42 | 80562 | ||
| 132 | 31 | 9 | 4 | 3 | 1 | 2054 | 0 | 190783 | ||
| 89 | 21 | 10 | 2 | 5 | 22 | 0 | 1 | 224241 | ||
| 135 | 24 | 6 | 2 | 3 | 0 | 1 | 4 | 211696 | ||
- Least k such that the interval 100k to 100k+99 has exactly n primes ending in 1
- Least k such that the interval 100k to 100k+99 has exactly n primes ending in 3
- Least k such that the interval 100k to 100k+99 has exactly n primes ending in 7
- Least k such that the interval 100k to 100k+99 has exactly n primes ending in 9
It is easy to see that there can never be more than seven primes ending in the same digit between multiples of 100, as minimally three numbers must divide by 3. When I looked more closely it seemed clear that there could be six primes in a century ending in the same digit, but I was unsure that there could be seven. Brian Kehrig, in a LinkedIn conservation, said there could be and gave the smallest example after the first century (all factorisations courtesy of Alpertron, although I have expanded repeated factors rather than use exponents):
8056203 = 3 × 2 685401
8056207 = 73 × 110359
8056209 = 3 × 7 × 19 × 61 × 331
2. 8056211 is prime
8056213 = 11 × 41 × 17863
8056217 = 13 × 467 × 1327
8056219 is prime
8056221 = 3 × 2685407
8056223 = 7 × 47 × 47 × 521
8056227 = 3 × 2685409
8056229 = 29 × 83 × 3347
3. 8056231 is prime
8056233 = 3 × 3 × 3 × 23 × 12973
8056237 = 7 × 7 × 164413
8056239 = 3 × 571 × 4703
4. 8056241 is prime
8056243 = 13 × 619711
8056247 = 19 × 167 × 2539
8056249 = 17 × 31 × 15287
8056251 = 3 × 3 × 7 × 127877
8056253 = 179 × 45007
8056257 = 3 × 11 × 244129
8056259 is prime
5. 8056261 is prime
8056263 = 3 × 1609 × 1669
8056267 is prime
8056269 = 3 × 3 × 13 × 37 × 1861
6. 8056271 is prime
8056273 = 59 × 136547
8056277 is prime
8056279 = 7 × 11 × 23 × 4549
8056281 = 3 × 67 × 149 × 269
8056283 = 17 × 473899
8056287 = 3 × 3 × 3 × 29 × 10289
8056289 is prime
7. 8056291 is prime
8056293 = 3 × 7 × 383633
8056297 is prime
It’s notable that a(7) for all these sequences lies in the “core” of the “eight-digit gap” where the geometric distance between prime-rich centuries is at a maximum. Within the numbers encompassing these centuries there is no century containing so many as fifteen total primes. The 80,563rd century is the 263rd with thirteen total primes, while the 190,784th, 211,697th, and 224,242nd all have too few primes to be similarly counted on OEIS.
The possibility of these sequences I observed many years ago when I noticed that all the following numbers were composite:
- 8907 = 3 × 2969
- 8917 = 37 × 241
- 8927 = 79 × 113
- 8937 = 3 × 3 × 3 × 331
- 8947 = 23 × 389
- 8957 = 13 × 13 × 53
- 8967 = 3 × 7 × 7 × 61
- 8977 = 47 × 191
- 8987 = 11 × 19 × 43
- 8997 = 3 × 2999
I also noticed at a fairly early stage the unusual factorisations of the 105th century:
10401 = 3 × 3467
10403 = 101 × 103
10407 = 3 × 3469
10409 = 7 × 1487
10411 = 29 × 359
10413 = 3 × 3 × 13 × 89
10417 = 11 × 947
10419 = 3 × 23 × 151
10421 = 17 × 613
10423 = 7 × 1489
1. 10427 is prime
2. 10429 is prime
10431 = 3 × 3 × 19 × 61
3. 10433 is prime
10437 = 3 × 7 × 7 × 71
10439 = 11 × 13 × 73
10441 = 53 × 197
10443 = 3 × 59 × 59
10447 = 31 × 337
10449 = 3 × 3 × 3 × 3 × 3 × 43
10451 = 7 × 1493
4. 10453 is prime
5. 10457 is prime
6. 10459 is prime
10461 = 3 × 11 × 317
7. 10463 is prime
10467 = 3 × 3 × 1163
10469 = 19 × 19 × 29
10471 = 37 × 283
10473 = 3 × 3491
8. 10477 is prime
10479 = 3 × 7 × 499
10481 = 47 × 223
10483 = 11 × 953
9. 10487 is prime
10489 = 17 × 617
10491 = 3 × 13 × 269
10493 = 7 × 1499
10497 = 3 × 3499
10. 10499 is prime
