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%I #6 Sep 19 2023 16:52:25
%S 1,2,4,8,12,32,24,60,48,72,96,120,144,420,384,240,432,360,576,480,864,
%T 840,1200,720,1728,1800,4080,1920,2400,1440,4752,2160,3960,2520,3600,
%U 2880,5280,3360,9504,4320,9240,5760,12240,7200,7920,5040,10800,8640,19800,12600
%N The smallest number k that has exactly n of its divisors in A005153.
%C For any n >= 1, the number 2^(n - 1) has exactly n divisors in A005153.
%e a(1) = 1 because 1 has only one divisor 1 = A005153(1).
%e a(2) = 2 because 2 has exactly two divisors 1 = A005153(1) and 2 = A005153(2).
%e Numbers 3, 5, 7, 9, 11 have only divisor 1 in A005153, 4 has divisors 1, 2, 4 in A005153, numbers 6 and 10 have only two divisors in A005153, and 8 has three divisors in A005153. The number 12 has the divisors {1, 2, 3, 4, 6, 12} and exactly five of them, 1, 2, 4 and 12 are in A005153, so a(5) = 12.
%t f[p_, e_] := (p^(e + 1) - 1)/(p - 1); pracQ[n_] := (ind = Position[(fct = FactorInteger[n])[[;; , 1]]/(1 + FoldList[Times, 1, f @@@ Most@fct]), _?(# > 1 &)]) == {}; d[n_] := DivisorSum[n, 1 &, pracQ[#] &]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = d[n]; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[50, 10^5] (* _Amiram Eldar_, Aug 21 2023 *)
%o (Magma) sk:=func<n,k|&+[Divisors(n)[i]:i in [1..k]]>; pr:=func<n|forall{k:k in [2..#Divisors(n)]|sk(n,k-1) ge Divisors(n)[k]-1}>; a:=[]; for n in [1..50] do k:=1; while #[d:d in Divisors(k)|pr(d)] ne n do k:=k+1; end while; Append(~a,k); end for; a;
%Y Cf. A000005, A005153.
%K nonn
%O 1,2
%A _Marius A. Burtea_, Aug 18 2023
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