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) = ???