A001599 -id:A001599 - cp's OEIS frontend

A328945 Numbers m that are neither arithmetic (A003601) nor harmonic (A001599).

2, 4, 8, 9, 10, 12, 16, 18, 24, 25, 26, 32, 34, 36, 40, 48, 50, 52, 58, 63, 64, 72, 74, 75, 76, 80, 81, 82, 84, 88, 90, 98, 100, 104, 106, 108, 112, 117, 120, 121, 122, 124, 128, 130, 136, 144, 146, 148, 152, 156, 160, 162, 170, 171, 172, 175, 176, 178, 180

Offset: 1

Jaroslav Krizek, Oct 31 2019

nonn

Numbers m such that neither the arithmetic mean of the divisors of m nor the harmonic mean of the divisors of m is an integer.
Numbers m such that neither A(m) = A000203(m)/A000005(m) nor H(m) = m * A000005(m)/A000203(m) is an integer.
Corresponding values of A(m): 3/2, 7/3, 15/4, 13/3, 9/2, 14/3, 31/5, 13/2, 15/2, 31/3, 21/2, 21/2, 27/2, ...
Corresponding values of H(m): 4/3, 12/7, 32/15, 27/13, 20/9, 18/7, 80/31, 36/13, 16/5, 75/31, 52/21, 64/21, ...

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000005, A000203, A001599, A003601, A007340, A328944.

[m: m in [1..10^5] | not IsIntegral(m * NumberOfDivisors(m) / SumOfDivisors(m)) and not IsIntegral(SumOfDivisors(m) / NumberOfDivisors(m))]

filter:= proc(n) local D,d,t;
  D:=numtheory:-divisors(n);
  d:= nops(D);
  convert(D,`+`) mod d <> 0 and not ((d/add(1/t,t=D))::integer)
end proc:
select(filter, [$1..200]); # Robert Israel, Dec 14 2023

Select[Range[180], !Divisible[DivisorSigma[1, #], DivisorSigma[0, #]] && !Divisible[# * DivisorSigma[0, #], DivisorSigma[1, #]] &] (* Amiram Eldar, Nov 01 2019 *)

isok(m) = my(f = factor(m), prd = sigma(f)/numdiv(f)); (denominator(prd) != 1) && (denominator(m/prd) != 1); \\ Michel Marcus, Nov 05 2019

A335318 Harmonic numbers (A001599) k with a record abundancy index sigma(k)/k.

Original entry on oeis.org

1, 6, 140, 270, 672, 8190, 30240, 332640, 14303520, 17428320, 27027000, 191711520, 2144862720, 3506025600, 5943057120, 14378364000, 45578332800, 288662774400, 505159855200, 2020639420800, 10680522652800, 54557264361600

Offset: 1

Amiram Eldar, May 31 2020

The corresponding record values are 1, 2, 2.4, 2.666..., 3, 3.2, 4, 4.363..., ...

The terms 1, 6, 672 and 30240 are multiply perfect numbers (A007691) with abundancy indices 1, 2, 3, and 4, respectively. There is no 5-multiperfect number (A046060) in this sequence since A046060(1) = 14182439040 is larger than the harmonic number 5943057120 which is 5-abundant, having an abundancy index 5.067...

			The first 7 harmonic numbers are 1, 6, 28, 140, 270, 496 and 672. Their abundancy indices are 1, 2, 2, 2.4, 2.666..., 2 and 3. The record values, 1, 2, 2.4, 2.666... and 3 occur at 1, 6, 140, 270 and 672, the first 5 terms of this sequence.

Cf. A000203, A001599, A335316, A335317.

rm = 0; s = {}; Do[h = DivisorSigma[0, n]/(r = DivisorSigma[1, n]/n); If[IntegerQ[h] && r > rm, rm = r; AppendTo[s, n]], {n, 1, 10^6}]; s

A153789 First differences of harmonic (or Ore) numbers A001599.

Original entry on oeis.org

5, 22, 112, 130, 226, 176, 966, 1332, 3230, 1928, 62, 10410, 20, 9226, 2394, 2520, 23100, 49804, 12136, 49600, 6200, 63910, 4550, 90580, 27720, 179040, 156120, 30660, 27300, 197496, 138294, 332010, 118440, 638820, 50960, 60760, 166740

Offset: 1

Omar E. Pol, Jan 20 2009

nonn

Muniru A Asiru, Table of n, a(n) for n = 1..71

Cf. A001599.

H:=Concatenation([1],Filtered([2,4..3*10^6],n->IsInt(n*Tau(n)/Sigma(n))));;
a:=List([1..Length(H)-1],i->H[i+1]-H[i]);; Print(a); # Muniru A Asiru, Dec 23 2018

a := Select[Range[1000000], IntegerQ[HarmonicMean[Divisors[#]]] &]; Table[Part[a,n + 1] - Part[a,n], {n,1,30}] (* G. C. Greubel, Aug 29 2016 *)

from numpy import diff
from sympy import divisor_sigma as sigma
def ore(n): return (n*sigma(n, 0))%sigma(n, 1) == 0
print(diff(list(filter(ore, range(1, 9000))))) # Michael S. Branicky, May 22 2021

More terms from R. J. Mathar, Feb 19 2009

A157850 Numbers k such that are harmonic means of divisors of harmonic (Ore) numbers (harmonic numbers is A001599) and equal to one of the divisors of this harmonic numbers.

Original entry on oeis.org

1, 2, 5, 6, 8, 9, 11, 10, 15, 15, 14, 17, 24, 24, 21, 13, 19, 27, 25, 29, 26, 44, 44, 29, 46, 39, 46, 42, 47, 47, 35, 41, 60, 51, 37, 48, 45, 49, 50, 49, 53, 77, 86, 86, 51, 96, 75, 70, 80, 99, 110, 81, 84, 102, 82, 96, 114, 53, 108, 115, 105, 116, 91, 85, 105

Offset: 1

Jaroslav Krizek, Mar 07 2009, Apr 12 2009

nonn

			a(3) = 5, because 5 is harmonic mean of divisors of A007340(3)=140 and also is divisor of 140.

Amiram Eldar, Table of n, a(n) for n = 1..847 (Calculated from the b-file at A007340)

Cf. A007340, A000005, A000203, A001599.

a(n) = f(A007340(n)), where f(k) = k * tau(k)/sigma(k) = k * A000005(k)/A000203(k).

More terms from Amiram Eldar, Jul 09 2019

A174633 Let H(p) = ptau(p)/sigma(p). Numbers 2^(H(p) - 1)(2^H(p) - 1) where H(p) is an integer (i.e., p in A001599).

Original entry on oeis.org

1, 6, 28, 496, 2016, 496, 32640, 130816, 2096128, 523776, 8128, 536854528, 536854528, 134209536, 8589869056, 140737479966720, 140737479966720, 2199022206976, 33550336, 137438691328, 9007199187632128, 562949936644096

Offset: 1

Michel Lagneau, Mar 24 2010

nonn

We prove that perfect numbers A000396 are included in this sequence. A number p is perfect if sigma(p)=2p where sigma(p) is the sum of the divisors of p (A000203). So, p is perfect if and only if H(p) = p*tau(p)/sigma(p) = p*tau(p)/2p = tau(p)/2. Because the numbers of the form 2^(q-1)*(2^q - 1) are perfect, where q is a prime such that 2^q - 1 is also prime, tau(p) = (q-1+1)*2 = 2q, and then H(p) = q. When p is perfect, we have p = 2^(H(p) - 1)*(2^H(p) - 1). Now, we prove that n = 2^(H(p) - 1)*(2^H(p) - 1) => n is an even perfect number. We have H(p) = p*tau(p)/sigma(p) and H is multiplicative. Because gcd(2^(H(p) - 1), 2^H(p) - 1) = 1, we obtain sigma(2^H(p) - 1)/tau(2^H(p) - 1) = 2^H(p) - 1 = 2^H(p)/2. Now, the equation sigma(m)/tau(m) = (m+1)/2 with m odd is possible only if m is prime. Thus, 2^H(p) - 1 is prime.

			For p = 1, H(1) = 1 and n=1; for p=2, H(2) = 2*tau(2)/sigma(2) = 2*2/3 = 4/3 (not integer). For p = 6, H(6) = 6*tau(6)/sigma(6) = 6*4/12 = 2, n = 2^(2-1)*(2^2 - 1) = 2*3 = 6 (first perfect number). Other perfect numbers: 28 (for p=28), 496 (for p=140), 8128 (for p = 8128).

T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, p. 4.
S. Bezuszka, Perfect Numbers, (Booklet 3, Motivated Math. Project Activities) Boston College Press, Chestnut Hill MA 1980.
J.M. De Koninck, A. Mercier, 1001 problèmes en théorie classique des nombres, Ellipses 2004, p. 73.

Muniru A Asiru, Table of n, a(n) for n = 1..725
C. K. Caldwell, Perfect number
J. J. O'Connor and E. F. Robertson, Perfect Numbers

Cf. A000396.

H:=[];; for p in [1..240000] do if IsInt(p*Tau(p)/Sigma(p)) then Add(H,p*Tau(p)/Sigma(p)); fi; od; a:=List(H,i->(2^(i-1))*(2^i-1)); # Muniru A Asiru, Nov 28 2018

for p from 1 to 10000000 do
        H := p*numtheory[tau](p)/numtheory[sigma](p) ;
        if type(H,'integer') then
                (2^(H-1))*(2^H-1) ;
                printf("%d,",%) ;
        end if;
end do:

h[p_] := p*DivisorSigma[0,p]/DivisorSigma[1,p]; hp=Select[Table[h[p],{p,1,10^6}],IntegerQ]; (2^(hp-1))*(2^hp-1) (* Jean-François Alcover, Sep 13 2011 *)

A331666 Refactorable numbers (A033950) that are simultaneously arithmetic (A003601) and harmonic (A001599).

Original entry on oeis.org

1, 672, 30240, 23569920, 45532800, 164989440, 447828480, 623397600, 1381161600, 1862023680, 2144862720, 3134799360, 3831421440, 13584130560, 14182439040, 16569653760, 21943595520, 22933532160, 34482792960, 35032757760, 40752391680, 53621568000, 56481384960

Offset: 1

Jaroslav Krizek, Jan 23 2020

nonn

Numbers m such that all values of sigma(m)/tau(m), m/tau(m) and m * tau(m)/sigma(m) are any integers (f, g, and h respectively).
Corresponding values of numbers f, g and h: (1, 84, 1260, 294624, 474300, 1178496, 2946240, 3298400, 5754840, 11784960, ...); (1, 28, 315, 73656, 118575, 257796, 699732, 721525, 1198925, 2909412, 1675674, ...); (1, 8, 24, 80, 96, 140, 152, 189, 240, 158, 260, 266, 220, 380, 384, 296, 392, ...).
Multiply-perfect numbers from this sequence are in A047728.

			For m = 672, f = sigma(m)/tau(m) = 2016/24 = 84; g = m/tau(m) = 672/24 = 28; h = m * tau(m)/sigma(m) = 672*24/2016 = 8.

Amiram Eldar, Table of n, a(n) for n = 1..118 (terms below 10^14)

Intersection of A033950 and A007340.

Cf. A000005, A000203, A001599, A003601, A047728.

[m: m in [1..10^6] | IsIntegral(SumOfDivisors(m) / NumberOfDivisors(m)) and  IsIntegral(m / NumberOfDivisors(m)) and IsIntegral(m * NumberOfDivisors(m) / SumOfDivisors(m))]

Select[Range[3*10^7], Divisible[#, (d = DivisorSigma[0, #])] && Divisible[(s = DivisorSigma[1, #]), d] && Divisible[#*d, s] &] (* Amiram Eldar, Jan 24 2020 *)

is(k) = {my(f = factor(k), s = sigma(f), d = numdiv(f)); !(k % d) && !(s % d) && !((k * d) % s) ;} \\ Amiram Eldar, May 09 2024

A336397 Numbers k such that both k and A122111(k) [its conjugated prime factorization] are Ore's Harmonic numbers (A001599).

Original entry on oeis.org

1, 6, 753480, 429520946400

Offset: 1

Antti Karttunen, Jul 20 2020

The initial four terms were found from the b-file of 937 terms provided in A001599.
There also at least these additional terms present: 439151400324000, 889013810412000, 106713790278732000, that were found from the file of conjectured initial 10000 terms of A001599 provided by David A. Corneth under that entry. However, there might be further terms in the range between.
Note that A122111(889013810412000) = 106713790278732000, and vice versa, which proves that the intersection of A001599 and A088902 (equal to the intersection of A088902 with A336317) is not equal to this sequence, but just a proper subsequence.

Intersection of A001599 and A336317.

Cf. A088902, A122111.

isA001599(n) = !((sigma(n,0)*n)%sigma(n,1));
isA336397(n) = (isA001599(n) && isA001599(A122111(n)));
\\ Or alternatively, by using standalone program given in A336317:
isA336397(n) = (isA001599(n) && isA336317(n));

A144641 Indices of harmonic (or Ore) numbers (A001599) that are also perfect numbers.

Original entry on oeis.org

2, 3, 6, 11, 59, 176, 310

Offset: 1

Omar E. Pol, Jan 19 2009

Cf. A000396, A001599.

a(6)-a(7) from Donovan Johnson, Aug 25 2012

A335388 Harmonic numbers (A001599) with a record number of divisors that are harmonic numbers.

Original entry on oeis.org

1, 6, 140, 672, 30240, 332640, 2178540, 17428320, 163390500, 191711520, 326781000, 1307124000, 2701389600, 8104168800, 14378364000, 40520844000, 1175104476000, 7131668544000, 49921679808000

Offset: 1

Amiram Eldar, Jun 04 2020

The corresponding record values are 1, 2, 3, 4, 7, 9, 12, 15, 17, 19, 21, 25, 29, 34, 35, 43, 54, 75, 84.

			1 has one divisor which is a harmonic number: 1. 6 has 2 divisors which are harmonic numbers: 1 and 6. The next harmonic number with more than 2 divisors which are harmonic numbers is 140 with 3 such divisors: 1, 28 and 140.

Cf. A001599, A335317.

h = Cases[Import["https://oeis.org/A001599/b001599.txt", "Table"], {, }][[;; , 2]]; hQ[n_] := Divisible[n * DivisorSigma[0, n], DivisorSigma[1, n]]; hsigma[n_] := DivisorSum[n, Boole @ hQ[#] &]; hm = 0; s = {}; Do[h1 = hsigma[h[[k]]]; If[h1 > hm, hm = h1; AppendTo[s, h[[k]]]], {k, 1, Length[h]}]; s

A342358 Balanced numbers (A020492) that are also arithmetic numbers (A003601) and harmonic numbers (A001599).

Original entry on oeis.org

1, 6, 140, 270, 2970, 332640, 14303520, 5297292000

Offset: 1

Bernard Schott, Mar 09 2021

Equivalently, numbers m such that sigma(m)/phi(m), sigma(m)/tau(m) and m*tau(m)/sigma(m) are all integers where phi = A000010, tau = A000005 and sigma = A000203.
Conjecture: 1 would be the only odd term of this sequence, because Oystein Ore conjectured that 1 is the only odd harmonic number (see link), and 1 is an arithmetic and balanced number (A342103).
Proposition: there are no primes in the sequence. Proof: the only prime that is both arithmetic and balanced is 3 (A342103), but 3 is not an harmonic number.
As Hans-Joachim Kanold (1957) proved that the asymptotic density of the harmonic numbers is 0 (see link), the asymptotic density of this sequence is also 0.
a(9) > 6.5*10^14 (verified using list of balanced numbers from Jud McCranie). All the numbers in this range that are both balanced and harmonic numbers are also arithmetic numbers. - Amiram Eldar, Mar 09 2021

			For 6: tau(6) = 4, phi(6) = 2, sigma(6) = 12, 6*tau(6)/sigma(6) = 6*4/12 = 2, sigma(6)/tau(6) = 3 and sigma(6)/phi(6) = 2, hence 6 is a term.

Hans-Joachim Kanold, Über das harmonische Mittel der Teiler einer natürlichen Zahl, Math. Ann., Vol. 133 (1957), pp. 371-374.
Oystein Ore, On the averages of the divisors of a number, Amer. Math. Monthly, Vol. 55, No. 10 (1948), pp. 615-619.
Oystein Ore, On the averages of the divisors of a number (annotated scanned copy).

Intersection of A001599, A003601 and A020492.
Intersection of A001599 and A342103.
Intersection of A007340 and A020492.
Cf. A000005, A000010, A000203.

with(numtheory): filter:= q -> (sigma(q) mod phi(q) = 0) and (sigma(q) mod tau(q) = 0 and (q*tau(q) mod sigma(q) = 0) : select(filter, [$1..300000]);

Select[Range[350000], And @@ Divisible[(s = DivisorSigma[1, #]), {(d = DivisorSigma[0, #]), EulerPhi[#]}] && Divisible[#*d, s] &] (* Amiram Eldar, Mar 09 2021 *)

isok(m) = my(s=sigma(m), t=numdiv(m)); !(s % eulerphi(m)) && !(s % t) && !((m*t) % s); \\ Michel Marcus, Mar 09 2021

a(6)-a(8) from Amiram Eldar, Mar 09 2021

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A328945 Numbers m that are neither arithmetic (A003601) nor harmonic (A001599).

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A335318 Harmonic numbers (A001599) k with a record abundancy index sigma(k)/k.

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A153789 First differences of harmonic (or Ore) numbers A001599.

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A157850 Numbers k such that are harmonic means of divisors of harmonic (Ore) numbers (harmonic numbers is A001599) and equal to one of the divisors of this harmonic numbers.

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A174633 Let H(p) = p*tau(p)/sigma(p). Numbers 2^(H(p) - 1)*(2^H(p) - 1) where H(p) is an integer (i.e., p in A001599).

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A331666 Refactorable numbers (A033950) that are simultaneously arithmetic (A003601) and harmonic (A001599).

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A336397 Numbers k such that both k and A122111(k) [its conjugated prime factorization] are Ore's Harmonic numbers (A001599).

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A144641 Indices of harmonic (or Ore) numbers (A001599) that are also perfect numbers.

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A174633 Let H(p) = ptau(p)/sigma(p). Numbers 2^(H(p) - 1)(2^H(p) - 1) where H(p) is an integer (i.e., p in A001599).