10 while P<5000
20 P=nxtprm(P)
30 Diff=10000-P
40 if prmdiv(Diff)=Diff then print P;Diff;" ";:Ct=Ct+1
45 :if Ct @ 5=0 then print
50 wend
60 print:print Ct
finds
59 9941 71 9929 113 9887 149 9851 167 9833
197 9803 233 9767 251 9749 257 9743 281 9719
311 9689 449 9551 461 9539 467 9533 479 9521
503 9497 509 9491 521 9479 563 9437 569 9431
587 9413 659 9341 677 9323 719 9281 743 9257
761 9239 773 9227 797 9203 827 9173 839 9161
863 9137 941 9059 971 9029 1031 8969 1049 8951
1151 8849 1163 8837 1181 8819 1193 8807 1217 8783
1259 8741 1301 8699 1307 8693 1319 8681 1373 8627
1427 8573 1487 8513 1499 8501 1553 8447 1571 8429
1613 8387 1637 8363 1709 8291 1877 8123 1889 8111
1907 8093 1913 8087 1931 8069 2063 7937 2081 7919
2099 7901 2207 7793 2243 7757 2273 7727 2297 7703
2309 7691 2351 7649 2357 7643 2393 7607 2411 7589
2417 7583 2423 7577 2441 7559 2459 7541 2477 7523
2543 7457 2549 7451 2693 7307 2753 7247 2789 7211
2879 7121 2897 7103 2957 7043 2999 7001 3023 6977
3041 6959 3083 6917 3089 6911 3137 6863 3167 6833
3209 6791 3221 6779 3299 6701 3347 6653 3449 6551
3527 6473 3671 6329 3677 6323 3701 6299 3779 6221
3797 6203 3803 6197 3911 6089 3947 6053 3989 6011
4013 5987 4019 5981 4073 5927 4133 5867 4139 5861
4157 5843 4217 5783 4259 5741 4283 5717 4289 5711
4349 5651 4409 5591 4481 5519 4493 5507 4517 5483
4523 5477 4583 5417 4649 5351 4691 5309 4703 5297
4721 5279 4919 5081
for a count of 127. Since order doesn't matter, only those with lower number first are shown and counted.
Approximation method:
The prime number theorem approximates the number of primes below a given number, n, as n / ln(n). For 100,000,000 this would be 100000000/18.42068 ~= 5428681. That also means that about 1/18.42 of the numbers under 100,000,000 are prime. Offhand, then, one would think if you subtracted a prime from 100,000,000 the probability that you'd get another prime would be 1/18.42 and the number of pairs, counting order separately would be 5428681/18.42 or about 294717. That counts both orders of a pair separately as the numerator included all primes, not just those smaller than 100000000/2, so you'd have to halve the number to discount order: 147358.
But how well does that method work when applied to the 10,000 version?
10000/ln(10000) implies about 1086 primes under 10000 and we'd expect 118 pairs counting both orders separately or 59 not considering order.
What went wrong?
As 10000 is even and all primes but the number 2 are odd, you're virtually guaranteed that the difference between the first prime chosen and 10000 is odd (not divisible by 2). You're also guaranteed that the complement of the first prime is not divisible by 5. So out of every 10 numbers in the range, 6 are ruled out automatically and the estimate needs to be multiplied by 10/4 or 5/2, making the estimate for the 10,000 version 59*5/2 ~= 148, in better agreement with the actual count of 127.
The same correction applies to the case of 100,000,000: 147358*5/2 = 368395.
Check:
10 while P<50000000
20 P=nxtprm(P)
30 Diff=100000000-P
40 if prmdiv(Diff)=Diff then print P;Diff;" ";:Ct=Ct+1
45 :if Ct @ 5=0 then print
50 wend
60 print:print Ct
finds the actual count is 291400.
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Posted by Charlie
on 2013-03-15 15:43:03 |
computer aided solutions
