formula to show last digits consecutive occurrences discrepancy of prime numbers

Hi,

Here's some more data using primorial 210 instead of primorial 30 equations showing distribution of prime last digits:

There are too many permutations to show for all 210 equations that show the last digits of consecutive primes, but here are some highlight results of a subset of the 48 primorial210 equations:

y=210x+b where b = {1 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 121 127 131 137 139 143 149 151 157 163 167 169 173 179 181 187 191 193 197 199 209}

Here are some of the overall set of procurance permutations for the above equations:

primeA, primeA+1, count

97 101 581 37 41 578 17 19 572 113 121 571 73 79 568 29 31 566 31 37 562 193 197 560 ... 151 173 131 19 41 131 187 209 130 127 149 130 113 139 128 169 191 127 121 143 127 67 89 125 31 53 124 101 121 124 ... 59 103 31 149 181 31 101 137 30 61 101 30 53 89 30 97 131 30 167 197 29 181 13 29 131 167 29 ... 53 103 6 197 43 6 97 139 6 97 143 6 19 67 6 143 191 6 23 67 6 89 139 6 19 61 6 31 83 6 ... 83 157 1 11 73 1 23 107 1 173 43 1 169 47 1 131 191 1 23 83 1 83 151 1 103 179 1 67 121 1 181 31 1

That is a huge discrepancy of consecutive last digits occurrences that can be explained with simple statistics, based on how far ahead in number gap the next prime is, ie the top occurring last digit grouping "97 to 101" occurred 581 times, and this is a 4 digit spacing, while the lower occurring last digit groupings, ie "181 to 31" at the bottom, only occurred 1 time. This is a (209-181)+31=59 number gap in the primes, so it makes sense that it occurred less often than the closer spaced primes.

The pairs like "17 to 17" didn't even show up once in the test I did since that is 210 digits apart for consecutive primes within a primorial210 block.

The same thing can be done for larger primorial equation sets, ie primorial 2310 with 480 equations and 480^2 permutation of the consecutive last digit pairings will show a huge discrepancy in the last digits (up to last 4 digits) of consecutive prime numbers.

Here is the C# code excerpt used to do this test:

foreach (int prime in Primes) { foreach (int i in new[] { 1, 11, 13, 17, 19, 23,

29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 121, 127, 131, 137, 139, 143, 149, 151, 157, 163, 167, 169, 173, 179, 181, 187, 191, 193, 197, 199, 209 }) { if (((prime - i) % 210) == 0) {

x210plusSORTED.Add(i); } } }

Dictionary pairCounts = new Dictionary(); ChirpletPairCounts chirpletPairCounts = new ChirpletPairCounts();

pairCounts = chirpletPairCounts.GetPairCounts(x210plusSORTED);

var pairCountsSorted = pairCounts.Values.OrderByDescending(x => x);

class ChirpletPairCounts {

public ChirpletPairCounts() {

}

public Dictionary GetPairCounts(IList sequence) { Dictionary pairCounts = new Dictionary(); for (int i = 0; i < sequence.Count - 1; i++) { var Key = Tuple.Create(sequence[i], sequence[i + 1]); if (pairCounts.ContainsKey(Key)) { pairCounts[Key]++; } else { pairCounts.Add(Key, 1); } } return pairCounts; }

}

cheers, Jamie

Reply to
Jamie Morken
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On Mon, 21 Mar 2016 13:49:11 -0700, Jamie Morken Gave us:

Don't you mean "some more OFF TOPIC CRAP" you keep adding this group to in your inane cross posts?

Reply to
DecadentLinuxUserNumeroUno

Also for primorial 6 it is the simplest pattern:

two equations:

6x+1 6x+5

gives 2*2 = 4 permutations of last consecutive digits in a primorial 6 block of numbers.

Here are the last digit occurances for the above equations:

primeA, primeA+1, count

5, 1, 22837 1, 5, 22837 5, 5, 16427 1, 1, 16394

In primorial6 blocks, here are the prime gaps for the above four consecutive last digits:

primeA, primeA+1, primegapFormula, primegap

5, 1, (6-5)+1, 2 1, 5, (5-1), 4 5, 5, (6-5)+5, 6 1, 1, (6-1)+1, 6

Equivalently stated by adding 6 to the second prime IF it is in the subsequent primorial6 block:

primeA, primeA+1, primegapFormula, primegap

5, 1, (7-5), 2 1, 5, (5-1), 4 5, 5, (11-5), 6 1, 1, (7-1), 6

Prime gap of 2 and prime gap of 4 have the same occurrence exactly in primorial 6 number blocks it looks like, both

22837 in this case.

cheers, Jamie

Reply to
Jamie Morken

Here are some excerpts for the primorial 2310 number blocks for the last digits of consecutive primes. There are 480 equations, and 480*480 permutations of prime last digits up to the last four digits up to 2309, so primes ending in 2309 can occur consecutively but would be very rare as they would have a 2310 digit spacing in a primorial 2310 block.

primeA, primeA+1, count

17 19 75 347 349 74 1279 1283 74 1291 1297 74 2137 2141 74 727 731 74 71 73 73 ... 1 17 36 197 211 36 299 311 36 2099 2113 36 19 29 36 179 191 36 521 527 36 481 493 36 853 859 36 ... 1517 1543 9 2053 2077 9 1739 1763 9 1973 1997 9 2143 2171 9 851 871 9 1459 1487 9 1489 1513 9 1663 1693 9 ... 1957 2017 2 347 377 2 1297 1333 2 1933 1993 2 569 607 2 731 769 2 1559 1619 2 1487 1537 2 109 151 2 ... 839 883 1 1319 1367 1 1427 1469 1 379 433 1 229 289 1 883 929 1 221 263 1 461 529 1

Most of the possible permutations won't show up unless very long input sequences of primes are used, but their probability of occurrences is straight forward to calculate.

17 to 19 digit last pair was most common overall.

cheers, Jamie

Reply to
Jamie Morken

Here is a summary of last digit counts for different primorial number blocks for the most common and less common consecutive last digits

primeA, primeA+1, count

primorial6: (2 values, primorial 6 is a special case as just 2 formulas produce primes ending in 1,3,7,9) most common:

5, 1, 22837 1, 5, 22837

least common:

5, 5, 16427 1, 1, 16394

primorial30: (8 values) most common:

7,11,2945 11,13,2916 17,19,2914 29,1,2904 23,29,2893 13,17,2892 19,23,2887 1,7,2849

least common:

13,13,274 11,11,266 17,17,265 19,19,261 7,7,260 1,1,241 23,23,226 29,29,221

primorial210: (48 values) most common:

97,101,581 37,41,578 17,19,572 113,121,571 73,79,568 29,31,566 31,37,562 193,197,560

*less common:

41,101,1 83,139,1 149,13,1 41,109,1 1,67,1 127,1,1 113,197,1 11,67,1

primorial2310: (48 values) most common:

17,19,75 347,349,74 1279,1283,74 1291,1297,74 2137,2141,74 727,731,74 71,73,73 1123,1129,73

*less common:

1471,1523,1 793,841,1 1201,1249,1 601,643,1 1289,1343,1 2203,2267,1 641,697,1 1889,1933,1

primorial30030: (5760 values) most common:

4127,4129,14 1447,1451,13 15557,15559,13 2243,2251,12 6121,6131,12 8831,8837,12 19751,19753,12 9197,9199,12

*less common:

8399,8441,1 8461,8501,1 8509,8531,1 8573,8609,1 8623,8633,1 8681,8693,1 8783,8819,1 8819,8863,1

*less common instead of least common as full set of permutations was not calculated for primorial 210 and higher, but last digit groups with same values should always have the lowest occurrence in a given primorial block

Reply to
Jamie Morken

Here is another way to view the data sorted by the first prime in the pair of consecutive primes:

primeA, primeA+1, count

primorial6:

5,5,16427 5,1,22837 1,5,22837 1,1,16394

primorial30: full list of 64 (8*8) permutations

29,29,221 29,23,363 29,19,519 29,17,774 29,13,1345 29,11,1526 29,7,2153 29,1,2904 23,29,2893 23,23,226 23,19,431 23,17,543 23,13,720 23,11,993 23,7,1824 23,1,2209 19,29,2154 19,23,2887 19,19,261 19,17,431 19,13,517 19,11,957 19,7,1020 19,1,1561 17,29,1551 17,23,2181 17,19,2914 17,17,265 17,13,380 17,11,516 17,7,724 17,1,1278 13,29,1160 13,23,1590 13,19,2163 13,17,2892 13,13,274 13,11,448 13,7,560 13,1,737 11,29,783 11,23,1133 11,19,1551 11,17,2193 11,13,2916 11,11,266 11,7,421 11,1,547 7,29,633 7,23,795 7,19,1138 7,17,1624 7,13,2087 7,11,2945 7,7,260 7,1,330 1,29,410 1,23,665 1,19,811 1,17,1087 1,13,1585 1,11,2159 1,7,2849 1,1,241

primorial210: (only excerpts)

209,73,1 209,67,2 209,61,2 209,59,1 209,53,1 209,47,4 209,43,9 209,41,9 209,37,19 209,31,37 209,29,49 209,23,68 209,19,105 209,17,177 209,13,241 209,11,374 209,1,542 199,209,549 199,71,1 199,67,2 199,61,1 199,59,1 199,47,4 199,43,8 199,41,8 199,37,4 199,31,16 199,29,35 199,23,40 199,19,69 199,17,97 199,13,141 199,11,271 199,1,371 197,209,370 197,199,555 197,53,1 197,47,3 197,43,6 197,41,10

primorial2310:

2309,41,2 2309,37,2 2309,31,5 2309,29,11 2309,23,5 2309,19,13 2309,17,28 2309,13,34 2309,1,66 2297,2309,55 2297,47,1 2297,41,1 2297,37,3 2297,31,4 2297,29,4 2297,23,7 2297,19,4 2297,17,18 2297,13,28 2297,1,37 2293,2309,44 2293,2297,60 2293,29,3 2293,23,5 2293,19,5 2293,17,11 2293,13,17 2293,1,19 2291,2309,23 2291,2297,37 2291,2293,59 2291,37,1 2291,29,2 2291,23,2

primorial30030:

30029,29,2 30029,19,3 30029,17,3 30029,1,9 30013,30029,10 30013,29,1 30013,23,1 30013,19,1 30013,1,2 30011,30029,2 30011,30013,6 30011,17,1 30011,1,4 30007,30029,3 30007,30013,7 30007,30011,5 30007,31,1 30007,1,1 30001,30029,2 30001,30013,1 30001,30011,2 30001,30007,4 30001,19,1 29999,30007,5 29999,30001,4 29999,17,2 29993,30007,6 29993,30001,2 29993,29999,5 29989,30013,1 29989,30011,4 29989,30001,1 29989,29999,2

There is a harmonic sine wave visible in the sorted count values, ie visible easily in the above primorial30, with frequency that corresponds to a given primorial block size. The frequency is just

8 per 30 in the case of primorial 30 as seen above. Each primorial has a beat frequency of EulerPhi(primorial)

ie:

primorial beatFrequency

6 2 30 8 210 48 2310 48 30030 5760

ie in wolfram alpha:

formatting link

These can explain all the harmonics observed in the prime number distributions I think.

For the primorials 210 and above, I didn't run all the permutations, but the partial full wave of oscillation is visible, and with a bit of imagination the full oscillating sine wave of EulerPhi(primorial) can be seen :D

The whole thing could be displayed with a statistical distribution formula instead of manually calculating the primes for it too, as for higher primorials the gaps get too big between some of the permutations to occur enough to show the sine wave plot without a supercomputer. Thats why the pattern becomes less visible for big primorials, but if they are fully plotted they should have nice sine waves of EulerPhi(primorial)

cheers, Jamie

Reply to
Jamie Morken

They're all odd.

Reply to
William Elliot

All but one!

--
Virgil 
"Mit der Dummheit kampfen Gotter selbst vergebens." (Schiller)
Reply to
Virgil

How odd!

Reply to
William Elliot

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