cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A335267 Composite numbers whose harmonic mean of their divisors that are larger than 1 is an integer.

Original entry on oeis.org

6, 15, 28, 30, 91, 117, 135, 252, 270, 496, 703, 864, 936, 1891, 1989, 2295, 2701, 4284, 4590, 5733, 8128, 8432, 12403, 18721, 19872, 21528, 38503, 41580, 49141, 51319, 56896, 79003, 88831, 104653, 121920, 146611, 188191, 218791, 226801, 235053, 269011, 286903
Offset: 1

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Author

Amiram Eldar, May 29 2020

Keywords

Comments

The primes are excluded from this sequence since they are trivial terms.
The corresponding harmonic means are 3, 5, 5, 5, 13, 9, 9, 9, 9, 9, 37, ...
Equivalently, composite numbers m such that (sigma(m)-m) | m*(tau(m)-1), or A001065(m) | A168014(m).
The semiprimes terms of this sequence are of the form p*q where p and q = 2*p - 1 are primes (A129521).
If m is a k-perfect numbers, k = 2, 3, ... (i.e., sigma(m) = k*m), then sigma(m)-m = (k-1)*m. If (k-1)*m | m*(tau(m)-1) then (k-1) | (tau(m)-1). If k is odd then tau(m) is also odd, so m is a square, and sigma(m) is odd. Since m | sigma(m) this means that m is also odd. Since there is no known odd multiply-perfect number except for 1 (A007691), there are no known k-perfect numbers with odd k in this sequence.
The perfect numbers (k=2, A000396) are terms: if m is a perfect number then sigma(m)-m = m.
The 4-perfect number (k=4, A027687) m are terms if 3 | (tau(m)-1). Of the first 36 terms of A027687 there are 8 such terms, the first is A027687(26).
The 6-perfect number (k=6, A046061) m are terms if 5 | (tau(m)-1). Of the first 245 terms of A046061 there are 20 such terms, the first is A046061(19).
Hemiperfect numbers that are terms of this sequence include A055153(i) for i = 10, 18 and 20, A141645(21), and A159271(i) for i = 97 and 103.

Examples

			6 is a term since its divisors other than 1 are 2, 3 and 6, and their harmonic mean, 3/(1/2 + 1/3 + 1/6) = 3, is an integer.
		

Crossrefs

A000396 and A129521 are subsequences.
Similar sequences: A001599, A247077, A247078.
Cf. A000005 (tau), A000203 (sigma).

Programs

  • Mathematica
    Select[Range[10^6], CompositeQ[#] && Divisible[# * (DivisorSigma[0, #] - 1), DivisorSigma[1, #] - #] &]
    Select[Range[287000],CompositeQ[#]&&IntegerQ[HarmonicMean[ Rest[ Divisors[ #]]]]&] (* Harvey P. Dale, Jan 21 2021 *)

A336745 Numbers m that divide the product phi(m) * sigma(m) * tau(m), where phi is the Euler totient function (A000010), sigma is the sum of divisors function (A000203) and tau is the number of divisors function (A000005).

Original entry on oeis.org

1, 2, 6, 8, 9, 12, 18, 24, 28, 32, 36, 40, 54, 72, 80, 84, 96, 108, 117, 120, 128, 135, 144, 162, 196, 200, 216, 224, 234, 240, 243, 252, 270, 288, 324, 360, 384, 400, 405, 448, 468, 486, 496, 512, 540, 576, 588, 600, 625, 640, 648, 672, 675, 720, 756, 768, 775, 810, 819
Offset: 1

Views

Author

Bernard Schott, Aug 02 2020

Keywords

Comments

If s and t are terms with gcd(s, t) = 1, then s*t is another term as phi, sigma and tau are multiplicative functions.
The only prime term is 2 because prime p must divide 2*(p-1)*(p+1) to be a term.

Examples

			For 24, phi(24) = 8, sigma(24) = 60 and tau(24) = 8, then 8*60*8 / 24 = 160, hence 24 is a term.
		

Crossrefs

Subsequences: A000396 (perfect numbers), A005820 (tri-perfect), A027687 (4-perfect), A046060 (5-multiperfect), A046061 (6-multiperfect), A007691 (multiply-perfect numbers), A336715 (m divides phi(m)*tau(m)), A004171, A005010.

Programs

  • Maple
    with(numtheory):
    filter:= m -> irem(tau(m)*phi(m)*sigma(m), m) =0:
    select(filter,[$1..850]);
  • Mathematica
    Select[Range[1000], Divisible[Times @@ DivisorSigma[{0, 1}, #] * EulerPhi[#], #] &] (* Amiram Eldar, Aug 02 2020 *)
  • PARI
    isok(m) = !(eulerphi(m)*sigma(m)*numdiv(m) % m); \\ Michel Marcus, Aug 05 2020

A379492 Multiperfect numbers k for which gcd(k,A003961(k))*gcd(sigma(k),A276086(k)) is not equal to gcd(k,A276086(k))*gcd(sigma(k),A003961(k)), where A003961(n) is fully multiplicative with a(prime(i)) = prime(i+1), and A276086 is the primorial base exp-function.

Original entry on oeis.org

120, 672, 523776, 1476304896, 14182439040, 31998395520, 518666803200, 13661860101120, 30823866178560, 740344994887680, 796928461056000, 212517062615531520, 69357059049509038080, 87934476737668055040, 154345556085770649600, 170206605192656148480, 1161492388333469337600, 1802582780370364661760, 9186050031556349952000
Offset: 1

Views

Author

Antti Karttunen, Jan 02 2025

Keywords

Crossrefs

Intersection of A007691 and A000027\A379486.
Cf. A046061 (seems to be a subsequence), A323653.

Programs

A227882 Known number of n_multiperfect numbers that can produce an hemiperfect of abundancy (2*n-1)/2.

Original entry on oeis.org

1, 3, 19, 0, 87, 117, 0, 30, 0, 0
Offset: 2

Views

Author

Michel Marcus, Oct 25 2013

Keywords

Comments

The hemiperfect that are obtained are coprime to p = 2*n-1.
When p=2*n-1 is prime, if m is a n-multiperfect is such that valuation(m, p) = 1, then let's define k = m/p, sigma(k) = sigma(m/p) = sigma(m)/sigma(p) = (n*m)/(p+1) = (n*m)/(2*n) = m/2. So sigma(k)/k = m/(2*k) = (k*p)/(2*k) = p/2 = (2*n-1)/2.

Examples

			a(2) = 1, since the only perfect number multiple of 3 is 6, and 6/3=2 has abundancy 3/2.
a(3) = 3, since the 3 known hemiperfect of abundancy 5/2 are coprime to 5.
a(5) = a(8) = a(11) = 0, since for those n, 2*n-1 is not prime.
a(10) is also 0, since all known 10-multiperfect are at least divisible by 19^2.
		

Crossrefs

Cf. A000396 (2), A005820 (3), A027687 (4), A046060 (5), A046061 (6), A007691 (integer abundancy).
Cf. A141643 (5/2), A055153 (7/2), A141645 (9/2), A159271 (11/2), A160678 (13/2), A159907 (half-integer abundancy).
Cf. A006254.
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