Prime numbers and primality testing Yahoo Group Gaps Between Consecutive Odds not Divisible by 3 =============================================== w_sindelar@juno.com Message 1 of 5 Jun 27, 2003 ----------------------------------------------- Let "N" represent an ODD COMPOSITE INTEGER NOT DIVISIBLE by 3. Let A < B represent a pair of CONSECUTIVE N's, meaning that there is no N greater than A and less than B. For example, here is a short list of the first few consecutive N's; 25, 35, 49, 55, 65, 77, 85. Any 2 adjacent integers A and B in the list are called a pair of consecutive odd composite integers not divisible by 3. Let G= (B - A) call it the GAP. The hunt for huge gaps between consecutive PRIMES seems to imply that there is no limit to the size of the gap. As a lark I thought I would see how large of a gap "G" I could find between the 2 consecutive composites "A" and "B" and how many primes exist between them. The largest gap I could find was 20 with 6 primes between!! Can it be possible that is the largest gap possible? Would anyone care to have a go at this with a program other than Ubasic? Thanks folks and regards. Bill Sindelar =============================================== Mark Underwood Message 2 of 5 Jun 27, 2003 ----------------------------------------------- Hi Bill et all The composite N's have factors of the form 6t+1 or 6t-1. The largest gap possible would be of the form 5*(6*(t+1) -1) - 5*(6t +1) which is 20, which you have found. Speaking of multiples of 5, while gazing at a sheet of primes the other day (hehe) I noticed four consecutive primes ending in the same number: 22963, 22973, 22993 and 23003. If consecutive primes were independant events then I guess the odds of four consecutive primes ending in the same number in the set of primes would be 4 * 1/4 * 1/4 * 1/4 * 1/4 which is 1/64. Yet on inspection four consecutive primes of this nature appear to be rarer than this. Perhaps up higher in the list of primes it approaches more this probability. For the fun of it, can anyone produce a longer string of such primes? Mark =============================================== Mikael Klasson Message 3 of 5 Jun 27, 2003 ----------------------------------------------- Hi, > Speaking of multiples of 5, while gazing at a sheet of primes the > other day (hehe) I noticed four consecutive primes ending in the same > number: 22963, 22973, 22993 and 23003. If consecutive primes were > independant events then I guess the odds of four consecutive primes > ending in the same number in the set of primes would be > 4 * 1/4 * 1/4 * 1/4 * 1/4 > which is 1/64. Yet on inspection four consecutive primes of this > nature appear to be rarer than this. Perhaps up higher in the list > of primes it approaches more this probability. > > For the fun of it, can anyone produce a longer string of such primes? For our mutual amusement, I humbly present: 452942837,452942857,452942927,452942947,452942977,452943047,452943097,452943107, 452943137,452943157 :) Mikael =============================================== Mark Underwood Message 4 of 5 Jun 28, 2003 ----------------------------------------------- Hi all Phil Carmody has generated two data sets, each tallying the occurance of consecutive primes ending in the same number. In the first data set we have Length of consecutive Number of Occurances prime string 7 112783 8 17612 9 2745 10 438 11 59 12 7 13 0 14 1 In the second data set we have Length of consecutive Number of Occurances prime string 11 2323 12 360 13 62 14 10 15 2 What surprised me was that the number of occurances does not decrease by a factor of 4 with each unit increase of string length. Rather it appears to decrease by a factor of roughly 6.4 when the occurances are high enough to stabilize the ratio. If anyone would like to try their hand at explaining why it is about 6.4 rather than 4, I would like to hear it! It appears that consecutive primes separated by a multiple of 10 deviate considerably from being considered independent events, but that's about all I can conjure up myself right now. Phil has a graph of the first occurances of these prime strings at http://fatphil.org/maths/trivia/terminal.html Mark =============================================== Mark Underwood Message 5 of 5 Jun 29, 2003 ----------------------------------------------- Hi all, I thought I should clarify somethings about the last post which were probably confusing. Phil Carmody has compiled two data sets from two different test intervals. The first interval went from the prime 3,873,011 to the prime 159,163,479,499. In this first interval Strings of exactly length 7 occur 112783 times. Strings of exactly length 8 occur 17612 times. Strings of exactly length 9 occur 2745 times. Strings of exactly length 10 occur 438 times. Strings of exactly length 11 occur 59 times. Strings of exactly length 12 occur 7 times. Strings of exactly length 13 occur 0 times. Strings of exactly length 14 occur 1 times. The second test interval went from the prime 34,239,812,903 to the prime 4,053,462,479,317 and hence overlaps the first test interval. In this second interval Strings of exactly length 11 occur 2323 times. Strings of exactly length 12 occur 360 times. Strings of exactly length 13 occur 62 times. Strings of exactly length 14 occur 10 times. Strings of exactly length 15 occur 2 times. What surprised me at first was that the number of occurances does not decrease by a factor of 4 with each unit increase of string length. Since primes end in only four digits - 1,3,7 and 9 - if these digits were distributed randomly and equally one would expect that there would be a one in four chance that two consecutive primes would end in the same number. Similarly one would expect strings of n consecutive primes ending in the same digit to be 4 times as abundant as strings of n+1 consecutive primes ending in the same digit. Yet Phil's data shows that strings of n consecutive primes ending in the same digit appear to be about 6.4 times as abundant as strings of n+1 consecutive primes ending in the same digit. That is, when occurances are large enough to stabilize the ratio. I suppose that the progression of the average size of prime gaps in an interval may play a part, and moreso the fact that certain spacings between primes are more likely. But can anyone produce the details of how the 6.4 ratio comes about? Mark PS In addition to this, and for interests sake, Phil Carmody has compiled a listing of the first occurences of these prime strings. At http://fatphil.org/maths/trivia/terminal.html Phil's uses the convention a(n) to represent the first prime of the first string to contain exactly n consecutive primes having the same last digit. For instance, since 139,149 are the first two consecutive primes that have the same last digit, a(2) = 139. Since 1627,1637 and 1657 are the first three consecutive primes that have the same last digit, a(3) = 1627. Here is Phil's compilation: a(1) = 2 a(2) = 139 a(3) = 1,627 a(4) = 18,839 a(5) = 123,229 a(6) = 776,257 a(7) = 3,873,011 a(8) = 23,884,639 a(9) = 36,539,311 a(10) = 196,943,081 a(11) = 452,942,827 a(12) = 73,712,513,057 a(13) = 177,746,482,483 a(14) = 154,351,758,091 a(15) = 4,010,803,176,619 a(16) = ??? =============================================== Cached by Georg Fischer at Nov 14 2019 12:46 with clean_yahoo.pl V1.4