# Large prime factors in a sequence of consecutive numbers

When trying to solve a number theory problem I came across this other problem which sounds interesting. Let $$n$$ be a positive integer, and consider $$n$$ consecutive positive integers $$a_1, \ldots, a_n$$ that are at most $$n^2$$.

What is an upper bound for the number of integers in this kind of list that have a prime factor greater than $$n$$?

What's interesting is that for any such prime factor, it appears only once as a factor in the list, and there can only be at most $$n$$ such primes. I'm guessing that $$n$$ is too large an upper bound and cannot be reached, i.e. there is always at least one number with prime factors all less than or equal to $$n$$.

I don't have any results besides checking some lists of numbers, and I don't really know how to approach this. Any ideas?

At least one of the numbers will be divisible by $$n$$. It cannot have a prime factor greater than $$n$$ because then it would be at least $$n(n+1)$$, so at most $$n-1$$ of them can have a prime factor larger than $$n$$. An example that meets this is $$n=5$$ and the numbers $$19-23$$ where only $$20$$ does not have a prime factor greater than $$5$$.
• Silly me, I should have thought of that. Now I guess the next natural question is whether the bound $n-1$ can be achieved infinitely often. Jan 18, 2020 at 4:05
Denote this upper bound by $$u(n)$$. Let $$l(k)$$ be the largest $$n$$ where $$u(n)=n-k$$. Then some pairs $$k:l(k)$$ are as follows: 1:10 2:15 3:16 4:21 5:27 6:30 7:40 8:37 9:42 10:46 11:57 12:58 13:65 14:70 15:72 16:78 17:79 18:81 19:87 20: 96 21:97 22:100 23:102 24:106 25:107 26:108 27:126 28:127 29:129 30:133 31:136 32:147 33:148 34:155 35:156 36:166 37: 167 38:172 39:165 40:174 41:178 42:185 43:190 44:196 45:198 46:202 47:204 48:207 49:209 50:214 51:221 52:222 53:226 54:227 55:233 56:238 57:243 58:247 59:250 60:251 61:256 62:268 63:262 64:270 65:272 66:275 67:280 68:281 69:282 70: 289 71:292 72:296 73:300 74:310 75:311 76:312 77:316 78:328 79:330 80:332 81:338 82:344 83:347 84:348 85:350 86:352 87:355 88:365 89:366 90:369 91:372 92:378 93:379 94:385 95:388 96:395 97:397 98:408 99:411 100:418 101:430 103:432 104:433 105:435 106:438 108:442 and so on.
In particular, to answer Paul Cusson's comment to Ross Millikan's answer, $$u(n)=n-1$$ last when $$n=10$$. (Some positive integers, such as 102, appear not to be values of $$k=n-u(n)$$.)
Though $$u$$ is not always an increasing function of $$n$$, and $$l$$, where defined, is not always an increasing function of $$k$$, it certainly appears that $$\liminf_{n\to\infty} u(n)=\infty$$ and $$\liminf_{k\to\infty} l(k)=\infty$$ (where the latter limit is taken over values of $$k$$ where $$l(k)$$ is defined).