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%I #19 Apr 23 2020 23:01:29
%S 1,2,4,6,8,12,24,36,48,72,96,120,144,240,288,360,480,576,720,1152,
%T 1440,2160,2880,4320,5760,8640,11520,17280,25920,30240,34560,51840,
%U 60480,69120,103680,120960,172800,181440,207360,241920,345600,362880,414720,483840,725760
%N Recursively superabundant numbers: numbers m such that A330575(m)/m > A330575(k)/k for all k < m.
%C Fink (2019) defined this sequence. He asked whether 720 is the largest term that is also superabundant number (A004394).
%C He noted that up to 10^6 all the recursively superabundant numbers are also recursively highly composite numbers (A333952), except for 181440 (the next term which is not recursively highly composite is 2177280). He asked whether there are a finite number of numbers that are both recursively highly composite and recursively superabundant (in analogy to A166981).
%C From _David A. Corneth_, Apr 13 2020: (Start)
%C The 76 terms in the b-file are products of primorials (Cf. A025487) and 7-smooth numbers (Cf. A002473). All terms are in A025487.
%C Proof: As A330575(n) = Sum_{d|n} A074206(d) * n/d we have A330575(n) / n = Sum_{d|n} A074206(d)/d which is maximal for some prime signature when n is a product of primorials.
%C Assuming terms below 10^17 are 13-smooth gives the 213 11-smooth numbers in the Corneth a-file. (End)
%H David A. Corneth, <a href="/A333953/b333953.txt">Table of n, a(n) for n = 1..213</a> (first 76 terms from Amiram Eldar)
%H Thomas Fink, <a href="https://arxiv.org/abs/1912.07979">Recursively divisible numbers</a>, arXiv:1912.07979 [math.NT], 2019. See section 5.
%t s[1] = 1; s[n_] := s[n] = n + DivisorSum[n, s[#] &, # < n &]; seq={}; rm = 0; Do[r1 = s[n]/n; If[r1 > rm, rm = r1; AppendTo[seq, n]], {n, 1, 10^4}]; seq
%Y Cf. A002473, A004394, A166981, A025487, A330575, A333952.
%K nonn
%O 1,2
%A _Amiram Eldar_, Apr 11 2020
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