## The Programmatic Side of Mathematica V: Segue into Primes

You might recall in the last post the expansion of a general binomial $(a + b)^z$:

binexpand[z_] := Expand[(a + b)^z]
TableForm[Array[binexpand,6]]

You can also declare a loop in the way Mathematica does so that it gives a series of binomial expansions. Below is the output of the second statement from above, the first 6 expansions of a general binomial:

And if you want to mess around with the Map command, you can put square roots on everything:

You don’t always need to use Array[] to generate a list. You can also use Range[hi] which generates a list from 1 to “hi”:

Prime[Range[100]]

Generates the first 100 primes:

{2, 3, 5, 7, 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, 127, 131, 137, 139, 149, 151, \
157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, \
239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, \
331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, \
421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, \
509, 521, 523, 541}

In earlier posts I had been curious about the distribution of prime numbers along the number line, and took my research to very large numbers, but I fear not large enough since I don’t have a supercomputer handy. My considered opinon first-hand, is that the primes are initially densely distributed, then gets gradually sparser in a way that is hard to predict. Mathematica claims it can easily access primes as high as on the order of $10^8$ in magnitude, and likely does the higher primes more slowly due to the fact that it uses primes up to that magnitude to search for still higher primes. If a natural number n has no prime numbers as factors, then n itself is prime.

The Mersenne primes, those Mersenne primes which are still being sought after, numbers large enough to fill every page in a 200-page novel, are notoriously hard to prove indivisible, since I would fear that the primes up to $10^8$ are not good enough. You need higher primes still, to many more orders of magnitude. Knowing the previous Mersenne prime is unreliable, since who knows how many primes lay between the current prime being checked and the previous one? There are likely to be hundreds, or even thousands of undiscovered primes which lay between $2^{p_{n-2}} - 1$ and $2^{p_{n-1}} - 1$, assuming $2^{p_{n-2}}$ and $2^{p_{n-1}}$ are prime (p, the exponent, is prime as well). The method of last resort, as I understand it, is to try every number up to $\lfloor \sqrt{2^{p}} \rfloor$ past $10^8$. While trying primes below $10^8$ weeds out several numbers, I think would still take a very long time for certain numbers that remain unfactored past that, possibly days to months even on a fast computer processor — to weed out one number or to successfully prove it as prime.

All other prime numbers which lay between two consecutive Mersenne primes (especially the big numbers) are nearly impossible to find, and I am not aware of anyone who bothers trying to find them (that doesn’t mean no one is trying, though). To find the next prime up from a Mersenne prime means looking to the next consecutive odd number, and the next, and the next, until you find one. A lot of time — months of time — is possibly wasted on a lot of composite numbers. Mersenne numbers narrow the search down, but only to a certain kind of prime number.

Mathematica has, for what it’s worth, a function called Prime[n], which returns the nth prime. Prime[1] returns 2; Prime[2] returns 3, and so on. The 100th prime is 541:

Prime[100]
541

And if we subtracted

Prime[100] - Prime[1]
539

we can say that 539 integers lay between the first and the 100th prime. We can now do this:

Prime[1000100] - Prime[1000000]
1606

which tells us there is a difference of 1606 between the millionth prime and the 1,000,100th prime.

Prime[1000000100] - Prime[1000000000]
1974

The air gets thinner, but only slightly, when counting between the billionth prime and the 1,000,000,100th prime. By the trillionth prime, we get a difference of 3546 by the 1,000,000,000,100th prime. Or you can say 100 primes per 3546 consecutive integers. The trillionth prime is 29,996,224,275,833, or just under 30 trillion. Mathematica’s Prime[] function chokes when getting to numbers on the order of a quadrillion, even though the documentation claims arguments to Prime[] are allowed to go into the quintillions: 1,152,921,504,606,846,976 or 260.

I could say that this still provides an incredible list of primes for the possible writing of a prime factorization function, however, Mathematica has beat me to it, wtih the function FactorInteger[]. Even the Mathematica website at Wolfram Math World has defined a function that modifies the output of this function further:

FactorForm[n_?NumberQ,fac_:Automatic]:=
Times@@(HoldForm[Power[##]]&@@@FactorInteger[n,fac])

So, a call to FactorForm[] can be:

I don’t know what “Times@@” does, but I don’t seem to get anything other than the default terminal font. At any rate, the ordered pairs returned by FactorInteger[] are transformed into base integers and exponents.

## In Pursuit of Trends in Primes

INTRODUCTION

The prime numbers are numbers which are divisible only by itself and 1. This means that all primes have two factors, so for that reason, “1” is not prime.

OBSERVATIONS

The first 100000 integers seems to have the greatest density of prime numbers. 9592 primes were found there, meaning on average, nearly one in 10 of the first 100000 integers were prime.

Dissecting this interval further, for the first 1000 contiguous integers there are 168 primes:

 INTERVAL    # OF PRIMES  FRACTION
1-100          25        0.25
101-200          21        0.21
201-300          16        0.16
301-400          16        0.16
401-500          17        0.17
501-600          14        0.14
601-700          16        0.16
701-800          14        0.14
801-900          15        0.15
901-1000         14        0.14
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
TOTAL           168        0.168

we see we’re already in trouble. While the first 200 integers clearly show dominance in having the most primes, with the first 100 integers having the most of all, the remainder fall into some kind of pseudo-random torpor, with some intervals higher, and some lower. 16 primes occur in three of the intervals, and 14 occur in another three.

In this interval, it is difficult to figure out if the numbers are meandering up or down. Let’s look at the second thousand:

1001-1100    16    0.16
1101-1200    12    0.12
1201-1300    15    0.15
1301-1400    11    0.11
1401-1500    17    0.17
1501-1600    12    0.12
1601-1700    15    0.15
1701-1800    12    0.12
1801-1900    12    0.12
1901-2000    13    0.13
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
TOTAL:      135    0.135

Is this a downward trend? We observe the first 10 thoudsands:

   1-1000    168    0.168
1001-2000    135    0.135
2001-3000    127    0.127
3001-4000    120    0.120
4001-5000    119    0.119
5001-6000    114    0.114
6001-7000    117    0.117
7001-8000    107    0.107
8001-9000    100    0.100
9001-10000   112    0.112
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
TOTAL:      1219    0.1219

A major jump upward is observed in the last interval, but it does not get to the level of the first 4 intervals. This is noticed nearly universally, and is reproduced here. If you look at the number of primes in regularly-spaced intervals of contiguous sequences of integers with each interval the same size, you tend to see a general declining trend, followed by a seemingly random meanindering of numbers of primes going up and down in number. This observation appears to be independent of
interval size.

In the interval 99001 to 100000, the last 1000 numbers in the interval, we observe that there are 86 primes. Clearly, this is below what we have just observed in the first 10 intervals of 1000, but in the interval 97001-98000, we observe only 82 primes, where we had expected a slightly greater number than 86. The number of primes once again are returning to a similar pseudo-random torpor observed earlier.

I define “pseudo” randomness as frequencies which go on a downward trend in an unpredictably meandering way. True randomness would have frequencies going all over the place.

So, if intervals of size “p” were used, the first p integers would contain the most primes; while there would be a downward trend for integers from p+1 to 2p; 2p+1 to 3p, and so on. By some middle interval kp+1 to (k+1)p, there would occur a relatively low number of primes in some pseudo random torpor. By the time we arrived at interval np+1 to (n+1)p, the frequency of primes become noticeably fewer. I am afraid I am not yet in a position to define the word “noticeably”. You just have to notice it.

But the word “noticeably” might imply that there is an upper limit of frequencies which it does not attain. That is, if “e” was the upper limit of a predicted frequency by interval np+1 to (n+1)p, we need to set the number so as to guarantee to ourselves that it does not rise above the limit. As a first approximation, I am willing to set e, if 2 divides n, as being the same for the interval (n/2)p+1 to ((n/2)+1)p, or the interval half-way to the last one. You obviously can be more restrictive than that, but at least you can see from my numbers that this clearly works out.

Contiguous Intervals of 100000

Taking the whole interval of the first 100000 positive integers contiguously, we find there are 9592 primes, far and away more than any other such interval observed for 100000 integers. The second interval from 100001 to 200000, has 8392 primes. Going from 1 billion, the interval has 4832 primes. While it is not clear if we are returning to torpor, we can see at least that the number of primes are decreasing.

The strongest proof I have observed of a decline returning to a chaos of up-and-down numbers of primes, was to see that these observations were even consistent when I counted 100,000 contiguous integers, and jumping 10^50 integers and counting another interval of 100,000 from there. I kept this up under Maple 9.5 up to 10^2000. The last intervals from 10^1700 took the longest to count (well, _you_ try checking 100000 integers which are each 1700 digits long, and see how long it takes you!). It is still cranking away, and there are 4 intervals left. In all, it will probably take 9 hours on a dual-core processor, so on the optimistic side, I have about 3 hours remaining.

The results so far have been enough to more than indicating a trend:

100000 FROM   # OF PRIMES     EQUAL OR GREATER(*)
10^0            9592
10^50            895
10^100           407
10^150           274
10^200           216
10^250           165
10^300           143
10^350           129
10^400           115
10^450            96
10^500            81
10^550            71
10^600            78                   *
10^650            65
10^700            68                   *
10^750            59
10^800            52
10^850            57                   *
10^900            57                   *
10^950            45
10^1000           39
10^1050           38
10^1100           43                   *
10^1150           33
10^1200           38                   *
10^1250           43                   *
10^1300           21
10^1350           32                   *
10^1400           28
10^1450           28                   *
10^1500           26
10^1550           35                   *
10^1600           32
10^1650           28
10^1700           28                   *
10^1750           29                   *
10^1800           27
10^1850           23

As has been stated by other writers, prime numbers appear to have a pattern, but the pattern has eluded us. It seems the pattern of decreasing numbers of primes as integers increase by many orders of magnitude seems to be elusive indeed, appearing pseudo-random, but with a general downward trend. One might desire to regress the numbers of primes per 100,000 integers to a curve, but I am not sure that would tell us anything meaningful by way of a pattern.