A265393 a(n) = the smallest number k such that floor(Sum_{d|k} 1/tau(d)) = n.

1, 6, 24, 60, 180, 420, 840, 2520, 4620, 9240, 13860, 27720, 60060, 55440, 110880, 166320, 180180, 480480, 360360, 900900, 720720, 1441440, 1801800, 2162160, 3063060, 4084080, 7207200, 12612600, 6126120, 27027000, 12252240, 18378360, 43243200, 24504480

Offset: 1

Author

Jaroslav Krizek, Dec 08 2015

Keywords

• nonn

Comments

Further known terms: a(29) = 6126120, a(31) = 12252240.

Are there numbers n > 1 such that Sum_{d|n} 1/tau(d) is an integer?

Sequences of numbers n such that floor(Sum_{d|n} 1/tau(d)) = k for k = 1..6:

k=1: 1, 2, 3, 4, 5, 7, 9, 11, 13, 17, 19, 23, 25, 29, 31, 37, 41, ... (A166684);

k=2: 6, 8, 10, 12, 14, 15, 16, 18, 20, 21, 22, 26, 27, 28, 32, 33, 34, 35, ...;

k=3: 24, 30, 36, 40, 42, 48, 54, 56, 66, 70, 72, 78, 80, 88, 96, 100, ...;

k=4: 60, 84, 90, 120, 126, 132, 140, 144, 150, 156, 168, 198, 204, 216, ...;

k=5: 180, 210, 240, 252, 300, 330, 336, 360, 390, 396, 450, 462, 468, ...;

k=6: 420, 630, 660, 720, 780, 900, 924, 990, 1008, 1020, 1050, 1080, ....

Values of function F = Sum_{d|n} 1/tau(d) for some numbers according to their prime signature: F{} = 1; F{1} = 3/2; F{2} = 11/6; F{1, 1} = 9/4; F{3} = 25/12; F{2, 1} = 11/4; F{4} = 137/60; F{3, 1} = 25/8, ...

Examples

			For n = 2; a(2) = 6 because 6 is the smallest number with floor(Sum_{d|6} 1/tau(d)) = floor(1/1 + 1/2 + 1/2 + 1/4) = floor(9/4) = 2.
		

Links

• Hiroaki Yamanouchi, Table of n, a(n) for n = 1..5000

Crossrefs

Cf. A000005, A253139, A265390, A265391, A265392.

Cf. A237350 (a(n) = the smallest number k such that Sum_{d|k} 1/tau(d) >= n).

Programs

Magma

a:=1; S:=[a]; for n in [2..14] do k:=0; flag:= true; while flag do k+:=1; if Floor(&+[1/NumberOfDivisors(d): d in Divisors(k)]) eq n then Append(~S, k); a:=k; flag:=false; end if; end while; end for; S;

Mathematica

Table[k = 1; While[Floor@ Sum[1/DivisorSigma[0, d], {d, Divisors@ k}] != n, k++]; k, {n, 17}] (* Michael De Vlieger, Dec 09 2015 *)

PARI

a(n) = {k=1; while(k, if(floor(sumdiv(k, d, 1/numdiv(d))) == n, return(k)); k++)} \\ Altug Alkan, Dec 09 2015

Extensions

More terms from Michel Marcus, Dec 23 2015

a(33)-a(34) from Hiroaki Yamanouchi, Dec 31 2015