Sequence of consecutive integers containing a relatively prime number. Is there a published result that says the following?
Let $p$ be an prime number. Then for all integer $n>p$ there is at least one number in the consecutive sequence $2n,2n+1,2n+2,\ldots,2n+2p-1$ that is relatively prime to all prime numbers less than or equal to $p$.
For example, if $p=5$ and $n=17$, the sequence $34,35,36,37,38,39,40,41,42,43$ has several numbers not divisible by $2,3,5$. Now this just needs to be true for any $n>p$ and any prime $p$. 
I have already asked my colleague who received his Ph.D. in Number Theory. He suggested I pose it to a wider audience, hence why I'm here. Any suggestions on results would be greatly appreciated.
 A: The conjecture is false. 
Let $p_1, p_2, p_3, \dots$ be the primes. Define the primorial Jacobsthal function $h(k)$ to be the smallest length $\ell$ such that every consecutive sequence of integers of length $\ell$ contains a number relatively prime to $p_1, \dots p_k$. Your conjecture is equivalent to the assertion that $h(k) \le 2 p_k$. 
(The condition is invariant upon adding or subtracting multiples of $P = \prod_{i=1}^k p_i$, so the condition that $n > p$ is irrelevant, and for fixed $p$ the conjecture can be verified by a finite calculation; it suffices to examine consecutive sequences of integers between $0$ and $P - 1$ inclusive.)
According to the table of values of $h(k)$ in the linked paper we have
$$p_{14} = 43, h(14) = 90 > 86$$
so there is a consecutive sequence of $89$ integers none of which is relatively prime to the primes up to $43$. This is the smallest counterexample.
You can show that $h(k) \ge 2p_{k-1}$ by constructing a consecutive sequence of integers of length $2p_{k-1} - 1$ none of which is relatively prime to $p_1, \dots p_k$. This can be done by considering the integers $n - a, n - a + 1, \dots n + a - 1, n + a$ where $n$ is a solution to the system of congruences
$$n \equiv 0 \bmod p_1 \dots p_{k-2}$$
$$n-1 \equiv 0 \bmod p_{k-1}$$
$$n+1 \equiv 0 \bmod p_k$$
and $a = p_{k-1} - 1$. This construction is actually best possible for $p_k \le 19$; for example when $p_k = 19$ it produces the sequence $60044, \dots 60076$. It first stops being best possible when $p_k = 23$, where this construction gives a sequence of length $37$ but the longest sequence has length $39$ (namely $20332472, \dots 20332510$). 
For completeness' sake here is the (very sloppy, I'm sure) Python script I used to test the conjecture; it allowed me to compute enough values of $h(k)$ that I could look it up on the OEIS, and looking at the longest sequences led me to the construction above. Note that it computes $h(k) - 1$, the length of the longest sequence of integers not relatively prime to $p_1, \dots p_k$. 
from sympy import sieve
from math import gcd
from operator import mul
import functools as ft

def longest(p):
    primes = list(sieve.primerange(1, p+1))
    product = ft.reduce(mul, primes, 1)

    biglist = list(range(0, primordial(p)))

    def isrelprime(n):
        answer = True
        for p in primes:
            if gcd(n,p) != 1:
                answer = False
        return answer

    best = 0
    bestlen = 0
    curlen = 0
    for i in biglist:
        if not isrelprime(i):
            curlen += 1
        else:
            curlen = 0
        if curlen > bestlen:
            best = i
            bestlen = curlen

    print("\nPrime : " + str(p))
    print("Best : " + str(best))
    print("Length : " + str(bestlen))

    for i in range(0, bestlen):
        print(str(best - i) + " has gcd " + str(gcd(best - i, product)))

primes = list(sieve.primerange(1, 20))
for p in primes:
    longest(p)

And here's a little bit of extra code for computing the gcd of a particular sequence of consecutive integers with the product of the primes up to some prime.
def check(n, p, len):
    primes = list(sieve.primerange(1, p+1))
    product = ft.reduce(mul, primes, 1)
    for i in range(0, len):
        print(str(n - i) + " has gcd " + str(gcd(n - i, product)))

check(60076, 19, 33)
check(20332510, 23, 39)

