A291899 Numbers n such that (pod(n)/tau(n)) > (pod(k)/tau(k)) for all k < n.

1, 3, 4, 6, 8, 10, 12, 18, 20, 24, 30, 36, 48, 60, 72, 84, 90, 96, 108, 120, 168, 180, 240, 336, 360, 420, 480, 504, 540, 600, 630, 660, 672, 720, 840, 1080, 1260, 1440, 1680, 2160, 2520, 3360, 3780, 3960, 4200, 4320, 4620, 4680, 5040, 7560, 9240, 10080, 12600

Offset: 1

Author

Jaroslav Krizek, Oct 10 2017

Keywords

• nonn

Comments

pod(n) = the product of the divisors of n (A007955), tau(n) = the number of the divisors of n (A000005).

Contains all members of A002182 except 2. - Robert Israel, Nov 09 2017

Is this the same as A034288 except for 3? - Georg Fischer, Oct 09 2018

From David A. Corneth, Oct 11 2018: (Start)

Various methods exist to find terms for this sequence, possibly combinable:

- Brute force; checking every positive integer up to some bound.

- Finding terms based on the prime signature.

- Relating to that, the number of divisors.

- Finding terms based on the GCD of some earlier found terms.

- ... (?)

There seems to be a method that helps finding terms < 10^150 for the similar A034287. (End)

Examples

			6 is a term because pod(6)/tau(6) = 36/4 = 9 > pod(k)/tau(k) for all k < 6.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..200 (First 126 terms by Robert Israel)
• David A. Corneth, Conjectured first 1171 terms

Crossrefs

Cf. A000005, A002182, A034287, A034288, A007955, A120736, A137927, A291186, A067128.

Programs

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;

Maple

f:= proc(n) local t; t:= numtheory:-tau(n); simplify(n^(t/2))/t end proc:
N:= 20000: # to get all terms <= N
Res:= NULL: m:= 0:
for n from 1 to N do
  v:= f(n);
  if v > m then Res:= Res, n; m:= v fi
od:
Res; # Robert Israel, Nov 09 2017

Mathematica

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 *)
DeleteDuplicates[Table[{n,Times@@Divisors[n]/DivisorSigma[0,n]},{n,13000}],GreaterEqual[ #1[[2]],#2[[2]]]&][[;;,1]] (* Harvey P. Dale, Mar 03 2024 *)

Formula

Numbers n such that (A007955(n)/A000005(n)) > (A007955(k)/A000005(k)) for all k < n.

Numbers n such that (A291186(n)/A137927(n)) > (A291186(k)/A137927(k)) for all k < n.