A380448 Least primes which can be represented as the sum of distinct positive cubes in exactly n ways, or 0 if no such prime exists.

73, 757, 953, 2521, 1801, 3257, 2953, 4013, 4139, 4789, 5347, 4481, 5669, 4663, 6427, 6659, 5867, 6301, 6841, 7867, 7687, 7741, 10169, 7057, 7723, 7561, 9631, 8443, 8191, 8387, 9883, 10079, 10313, 10891, 10729, 10009, 9109, 10711, 9829, 11161, 10457, 12547, 11699, 10513, 10333, 11159, 13007

Offset: 1

Author

Zhining Yang, Jun 21 2025

Keywords

• nonn

Comments

From David A. Corneth, Jun 22 2025: (Start)

a(234) = 0. We have if a(234) > 0 then a(234) > 10^6.

Proof:

Let Q(k, u) be the number of ways to write k as a sum of distinct cubes c where c <= u^3. Then for all m we have Q(k, m+1) >= Q(k, m).

for 39031 <= k <= 633371 we have Q(k, 40) >= 235.

Those are 633371 - 39031 + 1 = 594341 > 41^3 consecutive values for k.

Therefore Q(k, 41) >= 235 for 39031 <= k <= 633371 + 26^3.

We have n^3 + (n+1)^3 > (n+2)^3 for n >= 6. So via induction we have Q(k, u) > 234 for some u for any k >= 39031 which completes the proof.

Similarly Q(k, 40) > 10000 for 100000 <= k <= 500000 which is more than enough to confirm zero's found in the first 10000 terms in a search up to 10^6. (End)

Examples

			a(4) = 2521 because 2521 = 4^3 + 9^3 + 12^3 = 1^3 + 4^3 + 5^3 + 10^3 + 11^3= 1^3 +4^3+ 6^3 + 8^3 + 12^3 =4^3 + 6^3 + 8^3 + 9^3 + 10^3 and 2521 is the least prime that can be written as the sum of distinct positive cubes in 4 different ways.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000 (first 233 terms from Zhining Yang)

Crossrefs

Cf. A003997, A275154, A385094.

Programs

Mathematica

s = CoefficientList[Series[Product[(1 + x^(r^3)), {r, 20}], {x, 0, 8000}], x];
Table[SelectFirst[Flatten@Position[s, k] - 1, PrimeQ], {k, 20}]