Calculate the lowest number where the sum of the sequence of numbers exceeds a given value Given you have an infinite sequence of numbers defined as follows:

1: 1 = 1
2: 1 + 2 = 3
3: 1 + 3 = 4
4: 1 + 2 + 4 = 7
5: 1 + 5 = 6
6: 1 + 2 + 3 + 6 = 12
7: 1 + 7 = 8
...


The sequence is the sum of the divisors of n, including 1 and n.

Given a positive integer x as input, calculate the lowest number n which will produce a result greater than x. 

Test cases

f(100) = 48, ∑ = 124
f(25000) = 7200, ∑ = 25389
f(5000000) = 1164240, ∑ = 5088960


Expected Output

Your program should return both n and the sum of its divisors, like so:

$ ./challenge 100
48,124


Rules

This is code-golf so the shortest code in bytes, in each language wins.
Jelly, 18 12 11 10 bytes

1Æs&gt;¥#ḢṄÆs


Try it online!

-1 byte thanks to Mr. Xcoder!

How it works

1Æs&gt;¥#ḢṄÆs - Main link. Argument: n (integer)
1   ¥#     - Find the first n integers where...
 Æs        -   the divisor sum
   &gt;       -   is greater than the input
       Ṅ   - Print...
      Ḣ    -   the first element
        Æs - then print the divisor sum