A318277 -id:A318277 - cp's OEIS frontend

1, 6, 24, 60, 180, 420, 840, 2520, 4620, 9240, 13860, 27720, 55440, 55440, 110880, 166320, 180180, 360360, 360360, 720720, 720720, 1441440, 1801800, 2162160, 3063060, 4084080, 6126120, 6126120, 6126120, 12252240, 12252240, 18378360, 24504480, 24504480, 30630600, 36756720

Offset: 1

Jaroslav Krizek, Dec 13 2015

nonn

Are there numbers n > 1 such that Sum_{d|n} 1/tau(d) is an integer?
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, ...
All terms are of the form Product_{j=1..k} prime(j)^e(j) where e(j+1)<= e(j), and thus products of (not necessarily distinct) primorials. - Robert Israel, Dec 21 2015
From David A. Corneth, Nov 05 2019: (Start)
Instead of checking all divisors of A025487(n), one could use A318277 to see how often each prime signature occurs as a divisor.
Knowing the lcm of the terms below some m drastically improves the possibility of finding terms. In hindsight, knowing the lcm of the terms below 10^25 yields having to consider 1056 terms of A025487 instead of 222124. Is there some way to accurately predict the lcm to improve computation? (End)

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

David A. Corneth, Table of n, a(n) for n = 1..3338 (first 131 terms from Robert Israel, terms <= 10^25)
David A. Corneth, m, Sum_{d|a(m)} 1/tau(d) and the prime signature of a(m)

Cf. A000005, A002110, A025487, A253139, A265390, A265391, A265392, A318277.

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

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

N:= 10^9: # to get all entries <= N
Primorials:= NULL:
p:= 2: P:= p:
while P <= N do
  Primorials:= Primorials, P;
  p:= nextprime(p);
  P:= P*p;
od:
Primorials:= [Primorials]:
S:= {1}:
for i from 1 to nops(Primorials) do
  S:= {seq(seq(s*Primorials[i]^j,
       j = 0 .. floor(log[Primorials[i]](N/s))),s=S)}
od:
A:= NULL:
S:= sort(convert(S,list)):
xmax:= 0:
for s in S do
  x:= floor(add(1/numtheory:-tau(d),d=numtheory:-divisors(s)));
  if x > xmax then
     A:= A, s$(x-xmax);
     xmax:= x
  fi
od:
A; # Robert Israel, Dec 21 2015

s[1] = 1; s[n_] := DivisorSum[n, 1/DivisorSigma[0, #] &]; n = 1; k = 1; seq = {}; Do[While[s[k] < n, k++]; AppendTo[seq, k]; n++, {j, 1, 20}]; seq (* Amiram Eldar, Jan 30 2019 *)

a(n) = {my(k=1); while(sumdiv(k, d, 1/numdiv(d)) < n, k++); k;} \\ Michel Marcus, Dec 20 2015

a(24)-a(30) from Michel Marcus, Dec 20 2015
a(31)-a(35) from Robert Israel, Dec 21 2015
Missing a(31) = 12252240 inserted in data section by Georg Fischer, Nov 05 2019

cp's OEIS Frontend

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

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