This title space is too small! The topic is Hardy-Ramanujan numbers. #5

Python code (by no means efficient, but only takes a second or two on my computer):

bound = 0.00000001
# error bound for checking if a number is a cube,
# since cube root may not be totally accurate

n = int(input("Enter number: "))

numbers = []
a = 1
while a < (n//2)**(1/3) + bound:
    # test if b=n-a^3 is a cube number
    b = (n - (a**3))**(1/3)
    k = b%1 # distance from nearest integer
    b = round(b)
    if k < bound or k > 1 - bound:
        # found a cube
        print(a,b)
        numbers.append(a)
        numbers.append(b)
    a += 1

print("Sum: " + str(sum(numbers)))

Limit on cubes to be checked: 365903

Pairs found: $\left( \begin{array}{cc} 38787 & 365757 \\ 107839 & 362753 \\ 205292 & 342952 \\ 221424 & 336588 \\ 231518 & 331954 \\ \end{array} \right)$

Sum: 2544864

The check for a cube being an integer is exact as failures are returned as radicals.

The program runs in about 15 milliseconds.

The two $ (dollar sign) should be replaced by backslashes. This is necessary because of errors in Brilliant's code block handling.

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s = 48988659276962496;
q = Timing[n = Ceiling[N[Power[s, (3)^-1]]];
   Print[n];
   c = Table[i^3, {i, n}];
   ParallelTable[j = Power[s - i^3, (3)^-1];
    If[j $[Element] Integers $[And] i <= j, {i, j}, Nothing], {i, n}]
   ];
Print[q];
Print[Plus @@ Flatten[q[[2]]]]