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%I #6 Apr 07 2023 17:42:30
%S 12,18,20,88,104,464,1888,1952,29056,29312,29824,30592,30848,32128,
%T 127744,128768,130304,521728,522752,8341504,8353792,8378368,8382464,
%U 134029312,134045696,134094848,134193152,2146926592,2146992128,8586723328,8587902976,8589082624
%N Practical numbers (A005153) that are abundant and have a record low value of abundancy index.
%C The abundancy index of an integer k is sigma(k)/k, where sigma is the sum-of-divisors function (A000203).
%C All the perfect numbers (A000396) are practical, and their abundancy index is 2.
%C If k is a deficient practical number, then sigma(k) = 2*k - 1 (i.e., k is an almost-perfect number, and the only known such numbers are the powers of 2, A000079), so the abundancy index of these numbers approaches to the limit 2 from below.
%C All the terms are either of the form 2^m*p, where p < 2^(m+1) - 1 is a prime, or of the form 2^m*p^2, where p = 2^(m+1) - 1 is a prime.
%C This sequence is infinite since the abundancy index of practical numbers can be arbitrarily close to 2 from above: if k = 2^m*p, and p < 2^(m+1) - 1 then k is practical, and its abundancy index is (2-1/2^m)*(1+1/p) < 2 + 2/p. Therefore, for all eps > 0, taking a prime p and m such that 2/eps < p < 2^(m+1) - 1 will yield a practical number k = 2^m*p with 2 < sigma(k)/k < 2 + eps.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Practical_number">Practical number</a>.
%e The abundancy indices of the first terms are 7/3 > 13/6 > 21/10 > 45/22 > 105/52 > 465/232 > 945/472 > ... > 2.
%t f[p_, e_] := (p^(e + 1) - 1)/(p - 1); pracQ[fct_] := Position[fct[[;; , 1]]/(1 + FoldList[Times, 1, f @@@ Most@ fct]), _?(# > 1 &)] == {};
%t seq = {}; rm = 3; Do[fct = FactorInteger[n]; r = Times @@ (((First /@ fct)^(1 + Last /@ fct) - 1)/(First /@ fct - 1))/n; If[2 < r < rm && pracQ[fct], rm = r; AppendTo[seq, n]], {n, 3, 10^6}]; seq
%o (PARI) lista(kmax) = {my(f, r, rm = 3, prd, prac); forstep(k = 2, kmax, 2, f = factor(k); r = sigma(f, -1); if(r > 2 && r < rm, prd = 1; prac = 1; for(i=2, #f~, prd *= sigma(f[i-1, 1]^f[i-1, 2]); if(f[i, 1] > 1 + prd, prac = 0; break)); if(prac, rm = r; print1(k, ", ")))); }
%Y Subsequence of A005101 and A005153.
%Y Cf. A000079, A000203, A000396, A071927, A335029, A335030.
%K nonn
%O 1,1
%A _Amiram Eldar_, Apr 06 2023
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