MHB Set of 2015 Consecutive Positive Ints with 15 Primes

lfdahl
Gold Member
MHB
Messages
747
Reaction score
0
Is there a set of $2015$ consecutive positive integers containing exactly $15$ prime numbers?
 
Mathematics news on Phys.org
lfdahl said:
Is there a set of $2015$ consecutive positive integers containing exactly $15$ prime numbers?
[sp]For each natural number $n$ let $S_n$ be the set of $2015$ consecutive positive integers starting at $n$, and let $f(S_n)$ be the number of primes in that set.

To get from $S_n$ to $S_{n+1}$, you have to remove $n$ from $S_n$, and add $n+2015$ to the set. If either both or neither of those two numbers are prime then $f(S_{n+1}) = f(S_n)$. If one of them is prime and the other is not then $f(S_{n+1})$ and $f(S_n)$ will differ by $1$.

When $n=1$, $S_1$ is greater than $15$ (in fact, I think that $S_1 = 305$). When $n = 2016! + 2$, $S_n = 0$, because $2016! + k$ is divisible by $k$ whenever $2\leqslant k \leqslant 2016$.

So as $n$ increases from $1$ to $2016! + 2$, $f(S_n)$ changes by at most $1$ at each step, and has to go from $305$ to $0$. By a sort of integer-valued "intermediate value theorem", it must take the value $15$ at some point.[/sp]
 
Opalg said:
[sp]For each natural number $n$ let $S_n$ be the set of $2015$ consecutive positive integers starting at $n$, and let $f(S_n)$ be the number of primes in that set.

To get from $S_n$ to $S_{n+1}$, you have to remove $n$ from $S_n$, and add $n+2015$ to the set. If either both or neither of those two numbers are prime then $f(S_{n+1}) = f(S_n)$. If one of them is prime and the other is not then $f(S_{n+1})$ and $f(S_n)$ will differ by $1$.

When $n=1$, $S_1$ is greater than $15$ (in fact, I think that $S_1 = 305$). When $n = 2016! + 2$, $S_n = 0$, because $2016! + k$ is divisible by $k$ whenever $2\leqslant k \leqslant 2016$.

So as $n$ increases from $1$ to $2016! + 2$, $f(S_n)$ changes by at most $1$ at each step, and has to go from $305$ to $0$. By a sort of integer-valued "intermediate value theorem", it must take the value $15$ at some point.[/sp]

Thankyou so much, Opalg, for your excellent solution and participation!
 
lfdahl said:
Thankyou so much, Opalg, for your excellent solution and participation!

Yes, we can always count on Chris (Opalg) to post a robust, lucid solution. (Yes)
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top