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Numbers n with half-integral abundancy index, sigma(n)/n = k+1/2 with integer k.
+10
25
2, 24, 4320, 4680, 26208, 8910720, 17428320, 20427264, 91963648, 197064960, 8583644160, 10200236032, 21857648640, 57575890944, 57629644800, 206166804480, 17116004505600, 1416963251404800, 15338300494970880
COMMENTS
Obviously, all terms must be even (cf. formula), but e.g. a(9) and a(12) are not divisible by 3. See A007691 for numbers with integral abundancy. Odd numbers and higher powers of 2 cannot be in the sequence; 6 is in A000396 and thus in A007691, and n=10,12,14,18,20,22 don't have integral 2*sigma(n)/n. Conjecture: with number 1, multiply-anti-perfect numbers m: m divides antisigma(m) = A024816(m). Sequence of fractions antisigma(m) / m: {0, 0, 10, 2157, 2337, 13101, 4455356, ...}. - Jaroslav Krizek, Jul 21 2011 The above conjecture is equivalent to the conjecture that there are no odd multiply perfect numbers ( A007691) greater than 1. Proof: (sigma(n)+antisigma(n))/n = (n+1)/2 for all n. If n is even then sigma(n)/n is a half-integer if and only if antisigma(n)/n is an integer. Since all members of this sequence are known to be even, the only way the conjecture can fail is if antisigma(n)/n is an integer, in which case sigma(n)/n is an integer as well. - Nathaniel Johnston, Jul 23 2011 These numbers are called hemiperfect numbers. See Numericana & Wikipedia links. - Michel Marcus, Nov 19 2017
EXAMPLE
a(1) = 2 since sigma(2)/2 = (1+2)/2 = 3/2 is of the form k+1/2 with integer k=1.
a(2) = 24 is in the sequence since sigma(24)/24 = (1+2+3+4+6+8+12+24)/24 = (24+12+24)/24 = k+1/2 with integer k=2.
PROG
(PARI) isok(n) = denominator(sigma(n, -1)) == 2; \\ Michel Marcus, Sep 19 2015 (PARI) forfactored(n=1, 10^7, if(denominator(sigma(n, -1))==2, print1(n[1]", "))) \\ Charles R Greathouse IV, May 09 2017 (Python)
from fractions import Fraction
from sympy import divisor_sigma as sigma
def aupto(limit):
for k in range(1, limit):
if Fraction(int(sigma(k, 1)), k).denominator == 2:
print(k, end=", ")
Numbers n such that sigma(n) = 7n/2.
+10
13
4320, 4680, 26208, 20427264, 197064960, 21857648640, 57575890944, 88898072401645056, 301183421949935616, 9083288595228991885541376, 22290964134962716779872256, 230361837156847526055247872
COMMENTS
This sequence also contains the term 230361837156847526055247872. - Yasutoshi Kohmoto, May 10 2008 301183421949935616 also belongs to this sequence. - Avinoam Kalma (a.kalma(AT)gmail.com), May 04 2009
22290964134962716779872256 is in this sequence. [From Gerard P. Michon, May 10 2009]
EXAMPLE
Sigma(4320)=15120=7*4320/2, so 4320 is in the sequence.
MATHEMATICA
Do[If[DivisorSigma[1, m]==3.5*m, Print[m]], {m, 2*10^8}]
EXTENSIONS
a(8), a(10), a(12) [and above] found by Michel Marcus, added by Gerard P. Michon, Jun 04 2009
Numbers n such that sigma(n)/n = 5/2.
+10
9
24, 91963648, 10200236032
COMMENTS
There are no other terms through 2^34. - Walter Nissen, Apr 17 2009
MATHEMATICA
Select[Range[100000], DivisorSigma[1, #]/# == 5/2 &] (* Robert Price, Apr 03 2019 *)
PROG
(PARI) isok(n) = sigma(n, -1) == 5/2; \\ Michel Marcus, Apr 04 2019
Numbers n whose abundancy is equal to 13/2; sigma(n)/n = 13/2.
+10
8
170974031122008628879954060917200710847692800, 1893010442758976546037991125738431754692198400, 54361481238923605327597493185154939181072384000
COMMENTS
This sequence includes many terms but it is conjectured to be finite.
EXAMPLE
a(1) = 2^23 3^9 5^2 7^5 11^5 13^2 17 19^3 31 37 43 61^2 97 181 241.
As the "sum of divisors" function (sigma) is a multiplicative function, sigma(a(1)) is the product of the values of sigma at the above prime powers, respectively given as follows, in factorized form:
sigma(a(1)) = (3^2 5 7 13 17 241) (2^2 11^2 61) (31) (2^3 3 19 43) (2^2 3^2 7 19 37) (3 61) (2 3^2) (2^3 5 181) (2^5) (2 19) (2^2 11) (3 13 97) (2 7 13) (2 7^2) (2 11^2).
a(1) belongs to the sequence because the latter product boils down to 13/2 times the former.
CROSSREFS
Cf. A000203 (sigma function, sum of divisors), A141643 (abundancy = 5/2), A055153 (abundancy = 7/2), A141645 (abundancy = 9/2), A159271 (abundancy = 11/2), A159907 (half-integral abundancy, "hemiperfect numbers"), A088912 (least numbers of given half-integer abundancy). A007691 (multiperfect numbers, abundancy is an integer), A000396 (perfect numbers, abundancy = 2), A005101 (abundant numbers, abundancy is greater than 2), A005100 (deficient numbers, abundancy is less than 2).
Numbers m such that m divides sigma(2*m).
+10
6
1, 3, 12, 14, 60, 248, 336, 2160, 2340, 4064, 13104, 15120, 16380, 261888, 1089270, 4455360, 8714160, 10213632, 11784960, 16775168, 22766400, 45981824, 71495424, 98532480, 229909120, 689727360, 738152448, 4291822080, 4294934528, 5100118016, 7091219520
COMMENTS
If m belongs to the sequence, then sigma(2*m)/m is an integer, so sigma(2*m)/(2*m) is either an integer or half of an integer, so 2*m is either perfect, multiperfect or hemiperfect. - Michel Marcus, Jul 09 2013
Numbers such that numerator(sigma(n)/n) is odd and denominator(sigma(n)/n) is even.
+10
5
2, 4, 8, 16, 18, 20, 24, 32, 36, 40, 48, 50, 52, 64, 68, 72, 80, 88, 96, 98, 100, 104, 112, 116, 128, 136, 144, 148, 152, 160, 162, 164, 176, 180, 192, 196, 200, 208, 212, 224, 232, 240, 242, 244, 256, 272, 288, 292, 296, 304, 320, 324, 328, 338, 344, 352
EXAMPLE
sigma(2)/2 = 3/2 (odd/even).
MATHEMATICA
Select[Range[1000], OddQ[Numerator[DivisorSigma[1, #]/#]] && EvenQ[Denominator[DivisorSigma[1, #]/#]] &] (* Vincenzo Librandi, Jun 24 2014 *)
PROG
(PARI) oeab(n) = {for (i=1, n, ab = sigma(i)/i; if ((numerator(ab) % 2 == 1) && (denominator(ab) % 2 == 0), print1(i, ", ")); ); }
a(n) = smallest m such that sigma(m) = (n+1/2)*m.
+10
4
2, 24, 4320, 8910720, 17116004505600, 170974031122008628879954060917200710847692800
COMMENTS
2 is the only number m such that sigma(m)=1.5*m.
A direct consequence of Robin's theorem is that a(6)>5E16, a(7)>1.898E29, a(8)>2.144E51, a(9)>9.877E89 and a(10)>6.023E157. - Washington Bomfim, Oct 30 2008 If the Riemann hypothesis (RH) is true then Robin's theorem (Guy Robin, 1984) implies that the n-th term of this sequence is greater than exp(exp((n+1/2)/exp(gamma))) where gamma=0.5772156649... is the Euler-Mascheroni constant ( A001620). For the 6th term (which is actually 1.7*10^44) this lower bound is 5.0*10^16. Similarly, if RH is true, the next term (7th term) is at least 1.9*10^29 (and is probably more than 10^90 or so). - Gerard P. Michon, Jun 10 2009 An upper bound for a(7) is provided by a 97-digit integer of abundancy 15/2 (5.71379...10^96) discovered by Michel Marcus on July 4, 2009. The factorization of that number is: 2^53 3^15 5^6 7^6 11^3 13 17 19^3 23 29 31 37 41 43 61 73 79 97 181 193 199 257 263 4733 11939 19531 21803 87211 262657.
Similarly, an upper bound for a(8) is provided by a 286-digit integer of abundancy 17/2 (3.30181...10^285) equal to x/17, where x is the smallest known number of abundancy 9 (a 287-digit integer discovered by Fred W. Helenius in 1995). This is so because 17 happen to occur with multiplicity 1 in the factorization of x. (End)
A new upper bound for a(7) was found on Aug 15 2009 by Michel Marcus, who broke his own record by finding two "small" multiples of 2^35*3^20*5^5*7^6*11^2*13^2*17 that are of abundancy 15/2. The lower one (1.27494722...10^88) has only 89 digits. - Gerard P. Michon, Aug 15 2009 These are the least hemiperfects of abundancy n + 1/2. - Walter Nissen, Aug 17 2010 On Jul 24 2010, Michel Marcus found a 191-digit integer of abundancy 17/2 (2.7172904...10^190) whose factorization starts with 2^81 3^29 5^9 7^10 11^4 13^3 17^2 19 23^2... This is the best upper bound to a(8) known so far. - Gerard P. Michon, Aug 22 2010
REFERENCES
Guy Robin, Grandes valeurs de la fonction somme des diviseurs et hypothèse de Riemann, J. Math. Pures Appl. 63 (1984), 187-213. [From Gerard P. Michon, Jun 10 2009]
EXAMPLE
a(2)=24 because 1+2+3+4+6+8+12+24=2.5*24 and 24 is the earliest m such that sigma(m)=2.5*m.
MATHEMATICA
a[n_] := (For[m=1, DivisorSigma[1, m]!=(n+1/2)m, m++ ]; m); Do[Print[a[n]], {n, 4}]
Composite numbers whose harmonic mean of their divisors that are larger than 1 is an integer.
+10
3
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
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.
EXAMPLE
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.
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 *)
Known number of n_multiperfect numbers that can produce an hemiperfect of abundancy (2*n-1)/2.
+10
0
1, 3, 19, 0, 87, 117, 0, 30, 0, 0
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.
EXAMPLE
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.
a(n) = smallest m such that sigma(m) = n*m/2.
+10
0
1, 2, 6, 24, 120, 4320, 30240, 8910720, 14182439040, 17116004505600, 154345556085770649600, 170974031122008628879954060917200710847692800, 141310897947438348259849402738485523264343544818565120000
COMMENTS
For even n, a(n) is a multiply perfect number; for odd n it is a hemiperfect number.
Note that 1 is the only number with abundancy 1, and 2 is the only number with abundancy 3/2 (in other words, 1 and 2 are solitary numbers; see A014567). For k >= 4 it is not known whether there are finitely many or infinitely many numbers with abundancy k/2. Also it is not known whether a(n) < a(n+1) always holds. On the Riemann Hypothesis (RH), a(n) > exp(exp(n/(2*exp(gamma)))), where gamma = 0.5772156649... is the Euler-Mascheroni constant ( A001620).
EXAMPLE
a(7) = 4320 since sigma(4320) = 15120 = 7/2*4320 and 4320 is the smallest m such that sigma(m)/m = 7/2.
MATHEMATICA
Nest[Append[#, Block[{m = #1[[-1]] + 1}, While[DivisorSigma[1, m] != #2 m/2, m++]; m]] & @@ {#, Length@ # + 2} &, {1}, 6] (* Michael De Vlieger, Aug 05 2018 *)
PROG
(PARI) for(n=2, 10, for(m=1, 10^12, if(sigma(m)/m==n/2, print1(m, ", "); break())))
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