A265391 -id:A265391 - cp's OEIS frontend

A265392 a(n) = denominator of Sum_{d|n} 1 / tau(d).

1, 2, 2, 6, 2, 4, 2, 12, 6, 4, 2, 4, 2, 4, 4, 60, 2, 4, 2, 4, 4, 4, 2, 8, 6, 4, 12, 4, 2, 8, 2, 20, 4, 4, 4, 36, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 2, 40, 6, 4, 4, 4, 2, 8, 4, 8, 4, 4, 2, 8, 2, 4, 4, 140, 4, 8, 2, 4, 4, 8, 2, 72, 2, 4, 4, 4, 4, 8, 2, 40, 60, 4, 2

Offset: 1

Jaroslav Krizek, Dec 08 2015

a(n) = denominator of Sum_{d|n} 1 / A000005(d).

			For n = 6; divisors d of 6: {1, 2, 3, 6}; tau(d): {1, 2, 2, 4}; Sum_{d|6} 1 / tau(d) = 1/1 + 1/2 + 1/2 + 1/4 = 9 / 4; a(n) = 4 (denominator).

G. C. Greubel, Table of n, a(n) for n = 1..5000

Cf. A000005, A253139, A265390, A265391 (numerator), A265392, A265393.

[Denominator(&+[1/NumberOfDivisors(d): d in Divisors(n)]): n in [1..1000]]

Table[Denominator[Sum[1/DivisorSigma[0, d], {d, Divisors@ n}]], {n, 83}] (* Michael De Vlieger, Dec 09 2015 *)

a(n) = denominator(sumdiv(n, d, 1/numdiv(d))); \\ Michel Marcus, Dec 09 2015

a(n) = A265391(n) / [Sum_{d|n} 1 / tau(d)] = A265391(n) * A253139(n) / A265390(n).

a(1) = 1; a(p) = 2 for p = prime; a(n) = n for numbers 1, 2, 36, 72, ...

A265390 a(n) = lcm_{d|n} tau(d) * Sum_{d|n} 1/tau(d), where tau(d) represents the number of divisors of d (A000005(d)).

Original entry on oeis.org

1, 3, 3, 11, 3, 9, 3, 25, 11, 9, 3, 33, 3, 9, 9, 137, 3, 33, 3, 33, 9, 9, 3, 75, 11, 9, 25, 33, 3, 27, 3, 147, 9, 9, 9, 121, 3, 9, 9, 75, 3, 27, 3, 33, 33, 9, 3, 411, 11, 33, 9, 33, 3, 75, 9, 75, 9, 9, 3, 99, 3, 9, 33, 1089, 9, 27, 3, 33, 9, 27, 3, 275, 3, 9, 33, 33, 9, 27, 3, 411, 137, 9, 3, 99, 9, 9, 9, 75, 3, 99, 9, 33

Offset: 1

Jaroslav Krizek, Dec 08 2015

			For n = 6; divisors d of 6: {1, 2, 3, 6}; tau(d): {1, 2, 2, 4}; LCM_{d|6} tau(d) = 4; a(6) = 4/1 + 4/2 + 4/2 + 4/4 = 9.

Antti Karttunen, Table of n, a(n) for n = 1..16384
Index entries for sequences computed from exponents in factorization of n

Cf. A000005, A025529, A253139, A265391, A265392, A265393.

[&+[LCM([NumberOfDivisors(d): d in Divisors(n)]) / NumberOfDivisors(d): d in Divisors(n) ]: n in [1..100]]

Table[LCM @@ DivisorSigma[0, Divisors@ n] Sum[1/DivisorSigma[0, d], {d, Divisors@ n}], {n, 74}] (* Michael De Vlieger, Dec 09 2015 *)

A253139(n) = my(d = divisors(n)); lcm(vector(#d, k, numdiv(d[k]))); \\ This function from Michel Marcus, Jan 23 2015
A265390(n) = (A253139(n) * sumdiv(n,d,(1/numdiv(d)))); \\ Antti Karttunen, Nov 24 2017

a(n) = A253139(n) * Sum_{d|n} 1/A000005(d) = A265391(n) * A253139(n) / A265392(n).

Multiplicative with a(p^e) = A025529(e+1) = (1/1 + 1/2 + 1/3 + ... + 1/(e+1)) * lcm{1, 2, 3, ..., e+1}.

More terms from Antti Karttunen, Nov 24 2017

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

Original entry on oeis.org

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

Jaroslav Krizek, Dec 08 2015

nonn

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, ...

			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.

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

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

Cf. A237350 (a(n) = the smallest number k such that 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 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;

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

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

More terms from Michel Marcus, Dec 23 2015

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

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

Original entry on oeis.org

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

A266226 a(n) = floor(Sum_{d|n} 1 / tau(d)).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 3, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 2, 3, 2, 2, 1, 4, 1, 2, 2, 2, 2, 3, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 2, 3, 1, 3, 2, 2, 1, 4, 2, 2, 2, 3

Offset: 1

Jaroslav Krizek, Dec 24 2015

a(n) = floor(Sum_{d|n} 1 / A000005(d)).
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, ....
See A265393 - the smallest number n such that a(n) = k for k>= 1.

			For n = 6; a(6) = floor(Sum_{d|6} 1/tau(d)) = floor(1/1 + 1/2 + 1/2 + 1/4) = floor(9/4) = 2.

G. C. Greubel, Table of n, a(n) for n = 1..5000

Cf. A000005, A237350, A253139, A265390, A265391, A265392, A265393

[Floor(&+[1/NumberOfDivisors(d): d in Divisors(n)]): n in [1..100]];

Table[Floor[Sum[1/DivisorSigma[0, d], {d, Divisors[ n]}]], {n, 1, 100}] (* G. C. Greubel, Dec 24 2015 *)

A069165 a(n) are the integers corresponding to A069164(n).

Original entry on oeis.org

1, 3, 33, 55, 75, 77, 121, 125, 121, 143, 175, 187, 275, 209, 274, 275, 253, 325, 319, 341, 425, 407, 475, 588, 451, 473, 575, 517, 583, 725, 649, 1233, 671, 775, 737, 781, 803, 925, 869, 1025, 913, 1075, 979, 1175, 1067, 1111, 1133, 1325, 1177, 2055, 1199

Offset: 1

Benoit Cloitre, Apr 09 2002

The previous name was "Integers of the form k * Sum_{d|k} 1/tau(d)", but this did not take into account the occurrence of multiple terms and the non-ascending order.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000005, A069164, A265391, A265392.

Select[Table[n * DivisorSum[n, 1/DivisorSigma[0, #] &], {n, 1, 500}], IntegerQ] (* Amiram Eldar, Apr 29 2025 *)

a(n) = f(A069164(n)), where f(n) = n*A265391(n)/A265392(n). - Amiram Eldar, Apr 29 2025

Name corrected by Hugo Pfoertner, Jul 07 2024

A265719 Numbers n such that Sum_{d|n} 1/tau(d) > Sum_{d|m} 1/tau(d) for all m < n.

Original entry on oeis.org

1, 2, 4, 6, 12, 24, 30, 48, 60, 120, 180, 210, 240, 360, 420, 720, 840, 1260, 1680, 2520, 4620, 5040, 7560, 9240, 13860, 18480, 27720, 55440, 83160, 110880, 120120, 166320, 180180, 240240, 360360, 720720, 1081080, 1441440, 1801800, 2042040, 2162160, 3063060, 3603600, 4084080

Offset: 1

Jaroslav Krizek, Dec 14 2015

nonn

Where record values of Sum_{d|n} 1/tau(d) occur.

Terms a(n) are the smallest number from sequences numbers with following prime signatures: {}, {1}, {2}, {1, 1}, {2, 1}, {3, 1}, {1, 1, 1}, {4, 1}, {2, 1, 1}, {3, 1, 1}, {2, 2, 1}, {1, 1, 1, 1}, {4, 1, 1}, {3, 2, 1}, ...

			For n = 4; a(4) = 6 because 6 is the smallest number such that Sum_{d|a(4)} 1/tau(d) = Sum_{d|6} 1/tau(d) = 9/4 > Sum_{d|a(3)} 1/tau(d) = Sum_{d|4} 1/tau(d) = 11/6.

Cf. A000005, A237350, A253139, A265390, A265391, A265392, A265393.

a:=1; S:=[a]; for n in [2..25] do k:=0; flag:= true; while flag do k+:=1; if &+[1/NumberOfDivisors(d): d in Divisors(a)] lt &+[1/NumberOfDivisors(d): d in Divisors(k)] then Append(~S, k); a:=k; flag:=false; end if; end while; end for; S;

lista(nn) = {m = 0; for (n=1, nn, if ((mm = sumdiv(n, d, 1/numdiv(d))) > m, print1(n, ", "); m = mm););} \\ Michel Marcus, Dec 22 2015

More terms from Michel Marcus, Dec 22 2015

cp's OEIS Frontend

A265392 a(n) = denominator of Sum_{d|n} 1 / tau(d).

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A265390 a(n) = lcm_{d|n} tau(d) * Sum_{d|n} 1/tau(d), where tau(d) represents the number of divisors of d (A000005(d)).

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

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

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A266226 a(n) = floor(Sum_{d|n} 1 / tau(d)).

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A069165 a(n) are the integers corresponding to A069164(n).

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A265719 Numbers n such that Sum_{d|n} 1/tau(d) > Sum_{d|m} 1/tau(d) for all m < n.

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