A265714 a(n) = the smallest number k such that floor(Sum_{d|k} 1/sigma(d)) = n.

1, 60, 110880, 4658179125600, 950542574818669103079134726400, 204614292026733833316841991529248485168966921782532186656980932752000

Offset: 1

Author

Jaroslav Krizek, Dec 24 2015

Keywords

• nonn

Comments

If a(4) exists, it must be bigger than 5*10^6.

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

Conjecture: a(n) is also the smallest number k such that Sum_{d|k} 1/sigma(d) >= n.

Sequence of numbers n such that floor(Sum_{d|n} 1/sigma(d)) = k for k = 1, 2, 3:

k = 1: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, ... (A265711);

k = 2: 60, 72, 84, 90, 120, 144, 168, 180, 210, 216, 240, 252, ... (A265712);

k = 3: 110880, 166320, 221760, 277200, 327600, 332640, 360360, ... (A265713).

From Robert Israel, Dec 24 2015: (Start)

Note that g(k) = Sum_{d | k} 1/sigma(d) is multiplicative, with g(p) = 1 + 1/(p+1) for prime p.

Since Product_p (1+1/(p+1)) diverges, there are certainly numbers k with floor(g(k)) = n, including some squarefree numbers.

Conjectured values: a(4) = 4658179125600,

a(5) = 1188178218523336378848918408000,

a(6) = 5354073974699535305124032111682002028587967786642925550857667740344000.

These do have the correct value of floor(g(k)), but may not be the lowest possible.

(End)

Probably, a(7) = 1058687979...2471360000 = 349# * 23# * 7# * 5# * 3#^2 * 2#^3. - Hiroaki Yamanouchi, Dec 31 2015

Examples

			For n = 2; a(2) = 60 because 60 is the smallest number with floor (Sum_{d|60} 1/sigma(d)) = floor(155/72) = 2.
		

Crossrefs

Cf. A069934, A000203, A265708, A265709, A265710, A265711, A265712, A265713, A266227, A266228.

Programs

Magma

a:=1; S:=[a]; for n in [2..3] do k:=0; flag:= true; while flag do k+:=1; if &+[1/SumOfDivisors(d): d in Divisors(k)] ge n then Append(~S, k); a:=k; flag:=false; end if; end while; end for; S;

Extensions

a(4)-a(6) from Hiroaki Yamanouchi, Dec 31 2015