Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$2013$ is not a prime: $3 \times 11 \times 61$. I was born in a prime year, and if I live as expected according to the statistics for U.S. males, I will just reach another prime year, $2027$. That will encompass $11$ prime years in my lifetime, which I assume is high (because birth and probable-expiration hit primes directly).

What is the expected number of prime years in a lifetime of length $x$ years, starting at year $n$? I am aware that the Second Hardy-Littlewood Conjecture is likely to be false for large $n$, but does that conjectured relationship, $\pi(x+y) \le \pi(x) + \pi(y)$, still yield the best interval estimate for small $n$?

share|improve this question
    
In addition to the very nice answer from Eric N, change the first occurrence of i in the WA Routine to "current_age + i", where a is your current age and it will provide your age versus year versus if it is a prime year for the next 200 years. For example, it is currently set to age 15 years old. Live long and prosper! Regards –  Amzoti Jan 1 '13 at 2:02
    
are you going to live to 2027 but skip 2017? –  Jonathan Jan 1 '13 at 4:31
    
@Jonathan: No, I was counting 2017: ..., 2011, 2017, 2027. –  Joseph O'Rourke Jan 1 '13 at 13:07

1 Answer 1

up vote 14 down vote accepted

What is the expected number of prime years in a lifetime of length $x$ years, starting at year $n$?

The exact number will be $\pi(n+x)-\pi(n)$, where $\pi(y)=\sum_{p\leq y}1 $ is the prime counting function, but how large do we expect this to be?

The primes are distributed around $n$ with density $\frac{1}{\log n}$, so the expected number would be between $\frac{x}{\log n}$ and $\frac{x}{\log (n+x)}$. Provided that $n$ is much larger than $x$, this gives $\frac{x}{\log n}$ as the expected number of prime years. Supposing that a man born in $2000$ lives to the ripe old age of $100$, this estimate gives approximately $13$ primes in their lifetime. In reality there are $14$ primes between $2000$ and $2100$, which is not far off.

Note however, that if we take $n$ to be very large, it is possible that a person may live to $100$ and never experience a prime year. Indeed, suppose that an individual was born in the year $K=101!+1$. Then even if they live a long life, and die at $100$ years old, they will never have experienced a prime year, since each of $101!+2$, $101!+3$, $101!+4$,...,$101!+101$ are composite.

In the opposite direction, the Brun-Titchmarsh theorem tells us that $$\pi(x+n)-\pi(n)\leq \frac{2x}{\log x},$$ which gives us an upper bound on the number of prime years one can experience. Even if one were to live to be $200$ years old, they would not see more than $75$ primes, regardless of when they were born.

Added: The density of the primes, which is $\frac{1}{\log n}$ around $n$, goes to zero as $n\rightarrow\infty$, so unless life expectancy increases over time, the expected number of primes experienced in the average lifetime will converge to zero. A person born around the year $1$ million A.D. would expect to see only 7 primes if she were to live to $100$. To achieve the same expected number of primes as a woman born near the year $2000$, she would have to live to $190$.

This can also be used to give a good idea of how slowly $\log x$ grows. For an individual's expected number of primes to be less than $1$ in their lifetime, they would have to be born past the year $10^{44}$ A.D., and considering that best estimates put the death of the sun at around $4\times 10^9$ A.D., this is very far away.

In fact, for the next $368000$ years, every individual who lives to be $100$ will experience at least one prime year. However, there is no prime over a $114$ year period from $370261$ A.D. to $370365$ A.D., so for the unfortunate individual born in $370261$ A.D., they will experience no prime years unless they live past 114 years old.

share|improve this answer
1  
Aside from this being a comprehensive answer, it's nice that in the end you included the opposite question (and answer) as well. –  user54147 Jan 1 '13 at 2:06
1  
This answer is worth well more than the question---Thanks, Eric! And may you see many primes. :-) –  Joseph O'Rourke Jan 1 '13 at 2:14
    
If you are born in the year $101!$, then "if they live a long life" might not equal "$100$ years old". A complete answer should also take into account the increase in life expectancy over the years :) –  TMM Jan 1 '13 at 2:32
    
You upper bound estimate of $75$ primes seems to be a very conservative estimate. Can we obtain a better estimate using the sieve method for the number of primes one would see over a period of $200$ ? i.e. going upto the prime $13$ can we say that within $200$ consecutive number approximately $$200 (1-1/2)(1-1/3)(1-1/5)(1-1/7)(1-/11)(1-/13) \approx 38$$ are likely to be prime? –  user17762 Jan 1 '13 at 2:46
2  
The year $100! + 1$ will likely be long after the heat death of the universe, so I don't think Bob will be too worried about not having any prime-years during his life :) –  BlueRaja - Danny Pflughoeft Jan 1 '13 at 8:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.