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%I #19 Feb 01 2024 09:41:33
%S 1,2,4,8,12,16,24,32,48,72,96,144,192,288,384,576,864,1152,1728,2304,
%T 3456,4608,5184,6912,10368,13824,20736,27648,41472,55296,62208,82944,
%U 124416,165888,207360,248832,331776,373248,414720,497664,622080,746496,829440,995328
%N Numbers k where Sum_{d | k} 1/rad(d) increases to a record.
%C Let rad(n) be the radical of n, which equals the product of all prime factors of n (A007947). Let g(n) = 1/rad(n) and let f(n) = Sum_{d | n} g(d). This is a multiplicative function whose value on a prime power is f(p^k) = 1 + k/p. Hence f is a weighted divisor-counting function that weights divisors d higher when they have few and small prime divisors themselves. This sequence lists the values where f(n) increases to a record, analogously to highly composite numbers (A002182) or superabundant numbers (A004394). The numbers in this sequence are much smoother than those in the other two sequences, since the definition of f(n) strongly disfavors a lack of smoothness in n.
%H Charlie Neder, <a href="/A322447/b322447.txt">Table of n, a(n) for n = 1..200</a>
%e The divisors of 12 are 1,2,3,4,6,12, so f(12) = 1 + (1/2) + (1/3) + (1/2) + (1/6) + (1/6) = 8/3, which exceeds f(n) for n = 1,...,11. Alternately, since f is multiplicative, f(12) = f(4)*f(3) = (1+2/2)*(1+1/3).
%e f(207360) = f(2^9)*f(3^4)*f(5) = (11/2)*(7/3)*(6/5) = 15.4, which exceeds f(n) for n < 207360. (Note that this is the first value of the sequence that is divisible by 5; earlier values are all 3-smooth.)
%t rad[n_] := Times @@ (First@# & /@ FactorInteger@n); f[n_] := DivisorSum[n, 1/rad[#] &]; fm = 0; s = {}; Do[f1 = f[n]; If[f1 > fm, fm = f1; AppendTo[s, n]], {n, 1, 10^6}]; s (* _Amiram Eldar_, Dec 08 2018 *)
%o (PARI) rad(n) = factorback(factorint(n)[, 1]); \\ A007947
%o lista(nn) = {my(m=0, newm); for (n=1, nn, newm = sumdiv(n, d, 1/rad(d)); if (newm > m, m = newm; print1(n, ", ")););} \\ _Michel Marcus_, Dec 09 2018
%Y Cf. A007947 (radical), A002182, A004394.
%Y Also smooth numbers: A003586, A051037, A002473.
%K nonn
%O 1,2
%A _David S. Metzler_, Dec 08 2018
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