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- 1 rows
- _____________ _____________
- | 1 | | total |
- 1 |100 1 | 2 cols | % unique |
- |_____________|_____________ |_____________|
- | 3 | 15 |
- 2 | 67 2 | 33 5 | 3
- |_____________|_____________|_____________
- | 7 | 63 | 511 |
- 3 | 57 4 | 30 19 | 17 85 | 4
- |_____________|_____________|_____________|_____________
- | 15 | 255 | 4095 | 65535 |
- 4 | 47 7 | 24 62 | 22 898 | 12 7625 | 5
- |_____________|_____________|_____________|_____________|_____________
- | 31 | 1023 | 31267 | 1048575 | 33554431 |
- 5 | 42 13 | 22 226 | 23 7334 | 22 233185 | 12 3956995 |
- |_____________|_____________|_____________|_____________|_____________|
- | 63 | 4095 | 262143 | |
- 6 | 37 23 | 20 830 | 22 57738 | |
- |_____________|_____________|_____________|_____________|
- | 127 | 16383 | 2097151 |
- 7 | 34 43 | 20 3214 | 22 460474 |
- |_____________|_____________|_____________|
- | 255 | 65535 | |
- 8 | 31 79 | 19 12542 | |
- |_____________|_____________|_____________|
- | 511 | 262143 |
- 9 | 30 151 | 19 49726 |
- |_____________|_____________|
- | 1023 |
- 10 | 28 287 |
- |_____________|
- | 2047 |
- 11 | 27 559 |
- |_____________|
- | 4095 |
- 12 | 27 1087 |
- |_____________|
- | 8191 |
- 13 | 26 2143 |
- |_____________|
- | 16383 |
- 14 | 26 4223 |
- |_____________|
- | 32767 |
- 15 | 26 8383 |
- |_____________|
- | 65535 |
- 16 | 25 16639 |
- |_____________|
- | 131071 |
- 17 | 25 33151 |
- |_____________|
- | 262143 |
- 18 | 25 66047 |
- |_____________|
-
- Let U equal the number of unique patterns in a 1*n rectangle.
- By representing patterns as binary numbers we can eliminate translations
- (they are simply all the even numbers) and reflections (half the asymmetric
- odd numbers) and so derive:
-
- U = 2^(n-2) + 2^(n/2) - 1 for even n >= 2
- = 2^(n-2) + 3*2^((n-3)/2) - 1 for odd n >= 1
-
- For a 1*n rectangle the unique/total ratio approaches 25% as n increases.
- Proof: 2^(n-2) = 1/4 of 2^n, and since the total number of patterns = (2^n)-1
- then U/total approaches 1/4 because 2^n grows more rapidly than the other terms.
-
- The above expressions for U have been verified by LifeLab for n = 1 to 18.
- Conjecture: U is either prime or the product of two primes!?
- NO; see n = 28 and 29.
-
- n U
- 1 1 prime
- 2 2 prime
- 3 4 2 * 2
- 4 7 prime
- 5 13 prime
- 6 23 prime
- 7 43 prime
- 8 79 prime
- 9 151 prime
- 10 287 7 * 41
- 11 559 13 * 43
- 12 1087 prime
- 13 2143 prime
- 14 4223 41 * 103
- 15 8383 83 * 101
- 16 16639 7 * 2377
- 17 33151 prime
- 18 66047 prime
- 19 131839 prime
- 20 263167 prime
- 21 525823 191 * 2753
- 22 1050623 7 * 150089
- 23 2100223 59 * 35597
- 24 4198399 1399 * 3001
- 25 8394751 19 * 441829
- 26 16785407 prime
- 27 33566719 prime
- 28 67125247 7 * 7 * 23 * 59561
- 29 134242303 13 * 179 * 57689
- 30 268468223 15913 * 16871
- 31 536920063 2371 * 226453
- 32 1073807359 prime
-
- Dean Hickerson came up with the explanation for why primes occur
- so frequently in the above table:
-
- By the prime number theorem, the probability
- that a number near U(1x32) (about 10^9) is prime is about 1/ln(10^9),
- which is about 1/20. But it's easy to see that U(1xn) is never divisible
- by 2, 3, or 5 for n>3. That raises the probability of its being prime by a
- factor of 2/1 * 3/2 * 5/4 = 15/4, so the probability is now about 1/5. For
- smaller numbers, the probability is even higher, so the large number of
- primes in your list isn't surprising.
-
- In fact we can go even further. Consider the case when n is even. Then
- U(1xn) = x^2 - 2, where x = 1 + 2^(n/2 - 1). So if p is a prime divisor
- of U(1xn), then 2 must be a quadratic residue of n, which means that
- p is congruent to 1 or 7 mod 8. And not even all such primes can occur;
- e.g. 17 does not, since if x^2 is congruent to 2 mod 17 then x is 6 or 11
- mod 17 and x - 1 = 2^(n/2 - 1) is 5 or 10 mod 17. But the residues mod 17
- of powers of 2 run through the cycle 1, 2, 4, 8, 16, 15, 13, 9, 1, 2, ...
- So with so many possible prime divisors excluded, the probability of
- primality for U(1xn) must be much larger than for a random integer about
- the same size. I don't see how to estimate what percentage of primes are
- excluded from being divisors, so I don't know how to give even a heuristic
- asymptotic formula for the number of primes among U(1x2), U(1x4), ...,
- U(1 x 2n).
-
- Similarly, if n is odd, then 8 * U(1xn) = x^2 - 17, where
- x = 3 + 2^((n+1)/2). So if p is a prime divisor then 17 is a quadratic
- residue of p; hence p is congruent to 1, 9, 13, 15, 19, 21, 25, or 33
- mod 34. So again half of all primes are excluded immediately and even
- more when we use the fact that x-3 is a power of 2.
-