%I #45 Mar 22 2024 11:04:04
%S 1,3,4,6,8,10,12,18,20,24,30,36,48,60,72,84,90,96,108,120,168,180,240,
%T 336,360,420,480,504,540,600,630,660,672,720,840,1080,1260,1440,1680,
%U 2160,2520,3360,3780,3960,4200,4320,4620,4680,5040,7560,9240,10080,12600
%N Numbers n such that (pod(n)/tau(n)) > (pod(k)/tau(k)) for all k < n.
%C pod(n) = the product of the divisors of n (A007955), tau(n) = the number of the divisors of n (A000005).
%C Contains all members of A002182 except 2. - _Robert Israel_, Nov 09 2017
%C Is this the same as A034288 except for 3? - _Georg Fischer_, Oct 09 2018
%C From _David A. Corneth_, Oct 11 2018: (Start)
%C Various methods exist to find terms for this sequence, possibly combinable:
%C - Brute force; checking every positive integer up to some bound.
%C - Finding terms based on the prime signature.
%C - Relating to that, the number of divisors.
%C - Finding terms based on the GCD of some earlier found terms.
%C - ... (?)
%C There seems to be a method that helps finding terms < 10^150 for the similar A034287. (End)
%H David A. Corneth, <a href="/A291899/b291899.txt">Table of n, a(n) for n = 1..200</a> (First 126 terms by Robert Israel)
%H David A. Corneth, <a href="/A291899/a291899_1.gp.txt">Conjectured first 1171 terms</a>
%F Numbers n such that (A007955(n)/A000005(n)) > (A007955(k)/A000005(k)) for all k < n.
%F Numbers n such that (A291186(n)/A137927(n)) > (A291186(k)/A137927(k)) for all k < n.
%e 6 is a term because pod(6)/tau(6) = 36/4 = 9 > pod(k)/tau(k) for all k < 6.
%p f:= proc(n) local t; t:= numtheory:-tau(n); simplify(n^(t/2))/t end proc:
%p N:= 20000: # to get all terms <= N
%p Res:= NULL: m:= 0:
%p for n from 1 to N do
%p v:= f(n);
%p if v > m then Res:= Res, n; m:= v fi
%p od:
%p Res; # _Robert Israel_, Nov 09 2017
%t With[{s = Array[Times @@ Divisors@ # &, 12600]}, Select[Range@ Length@ s, Function[m, AllTrue[Range[# - 1], m > s[[#]]/DivisorSigma[0, #] &]][s[[#]]/DivisorSigma[0, #]] &]] (* _Michael De Vlieger_, Oct 10 2017 *)
%t DeleteDuplicates[Table[{n,Times@@Divisors[n]/DivisorSigma[0,n]},{n,13000}],GreaterEqual[ #1[[2]],#2[[2]]]&][[;;,1]] (* _Harvey P. Dale_, Mar 03 2024 *)
%o (Magma) a:=1; S:=[a]; for n in [2..60] do k:=0; flag:= true; while flag do k+:=1; if &*[d: d in Divisors(a)] / #[d: d in Divisors(a)] lt &*[d: d in Divisors(k)] / #[d: d in Divisors(k)] then Append(~S, k); a:=k; flag:=false; end if; end while; end for; S;
%Y Cf. A000005, A002182, A034287, A034288, A007955, A120736, A137927, A291186, A067128.
%K nonn
%O 1,2
%A _Jaroslav Krizek_, Oct 10 2017