#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
#include <iostream>

using namespace std;

bool IsPrime(unsigned long long n) {
	bool r = true;

	for (unsigned long long i = 3; i < sqrt((double) n) + 1; i+= 2)
	{
		if (n % i ==0) {
			r = false;
			break;
		}
	}

	return r;
}

bool IsPalindrome(unsigned long long n) {
	bool r = true;
	char s[30];
	int l = sprintf(s, "%llu", n);

	if (l == 1 && n != 1) {
		r = true;
	} else	{
		for (int i = 0; i < l/2; i++) {
			if (s[i] != s[l-i-1]) {
				r = false;
				break;
			}
		}
	}

	return r;
}

/*
 * usage: palprime [lbound] [ubound]
 */
int main(int argc, char** argv) {
	unsigned long long beginNum = 3;
	unsigned long long endNum = 3;

	if (argc == 2) { // lbound default to 3
#ifdef _WIN32
		endNum = _strtoui64(argv[1], NULL, 10);
#else
		endNum = strtoull(argv[1], NULL, 10);
#endif

	} else if (argc == 3) {
#ifdef _WIN32
		beginNum = _strtoui64(argv[1], NULL, 10);
		endNum = _strtoui64(argv[2], NULL, 10);
#else
		beginNum = strtoull(argv[1], NULL, 10);
		endNum = strtoull(argv[2], NULL, 10);
#endif
	}

        unsigned long long i = beginNum;

        while (i < endNum) {
                char s[30];
                int l = sprintf(s, "%llu", i);

		//length cannot be even as even length palindrome numbers
		//can be divided by 11.
                if (l % 2 == 0) {
                    i = ((unsigned long long) (i / 10)) * 100 + 1;
                    continue;
                }

		if (IsPalindrome(i)) {
			if (IsPrime(i)) {
				cout << i << endl;
			}
		}

                i+=2;

                if (s[0] % 2 == 0) {
                    i+=pow(10, l-1);

		    //leading/ending number cannot be 5
                    if (((int) (s[0] - '0')) + 1 == 5) {
                        i += 2 * pow(10, l-1);
                    }
                }
	}
	return (EXIT_SUCCESS);
}

At first, I was trying to find all the palprimes that can be represented by 64 bit integers. But soon I realized that it would take months to do so using the code above with a quad-core PC (using 4 processes with different ranges). Anyway, here’s the last few palindromic primes less than 10,000,000,000,000:

9999899989999
9999901099999
9999907099999
9999913199999
9999919199999
9999938399999
9999961699999
9999970799999
9999980899999
9999987899999

And here are a few interesting ones:

11357975311
1112345432111
1300000000031
1700000000071
1900000000091
7900000000097
9200000000029
1357900097531

Alternatively, if we “bin” the prime numbers according to the differences (gaps) between two consecutive prime numbers, we would yield another logarithmic distribution: the histogram of such differences will be logarithmic as well.

So, take the following sequence of prime numbers for example:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29…

The distances between the consecutive numbers in the above sequence are:

3-2 = 1
5-3 = 2
7-5 = 2
11-7 = 4
19-17 = 2
23-19 = 4
29-23 = 6

We then calculate the histogram based on the gaps.

Denote the histogram bin index as
\[i = \lfloor\frac{d}{2}\rfloor\]

For the example above, we get the following distributions for H[i]:
\[H[0] = 1, H[1] = 3 H[2]=2, H[3]=1\]

Now we calculate the distribution (H[i]) described above using prime numbers within the following interval (for all prime numbers up to 4,294,967,291):
\[[2^0, 2^{32}]\]

We can obtain the following plot for the histogram (H[i]):

Prime number distribution

To illustrate that the distribution is indeed around a logarithm curve, we take the logrithms of the values for each H[i] and get the following figure:

Prime number distribution

As you can see, the distribution is around a straight line.