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This question Method for coming up with consecutive integers not relatively prime to $(100!)$

asks, how many consecutive numbers can have a prime factor below $100$

In general, let $d(p)$ be the maximum possible number of consecutive numbers that have a prime factor not exceeding $p$. A link to OEIS with the optimal values for small numbers $p$ is given, but I did not find a concrete method.

How can I calculate $d(p)$ for a given prime $p$ ?

My only idea is that the number of numbers out of $k$ numbers having a prime factor not exceeding $p$ can be bounded from above by $$\left\lceil \frac{k}{2}\right\rceil+\left\lceil \frac{k}{2\cdot 3}\right\rceil+ \left\lceil \frac{k}{2\cdot 5} \right\rceil +\cdots+\left\lceil\frac{k}{2p}\right\rceil$$ This approach allows to get a rough upper bound, but it seems to be difficult to calculate the exact value in general.

Any ideas ?

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    What do you mean by "gap between consecutive numbers"??? This gap is $1$ by definition, isn't it? – 2017-01-02
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    So perhaps you should rephrase the entire question to "what is the maximum amount of consecutive numbers which are not coprime to $p\#$?". – 2017-01-02
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    Well, I answered your question, then you changed it completely! – 2017-01-02
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    Then, I made a mistake in my formulation, sorry. – 2017-01-02
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    To clarify : At most $5$ consecutive numbers, (For example $2,3,4,5,6$ ), have a prime factor not exceeding $5$, so $d(5)=5$ – 2017-01-02

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