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A number k is abundant if sigma(k) > 2k (cf. A005101), perfect if sigma(k) = 2k (this sequence), or deficient if sigma(k) < 2k (cf. A005100), where sigma(k) is the sum of the divisors of k (A000203).

The numbers 2^(p-1)*(2^p - 1) are perfect, where p is a prime such that 2^p - 1 is also prime (for the list of p's see A000043). There are no other even perfect numbers and it is believed that there are no odd perfect numbers.

Numbers k such that Sum_{d|k} 1/d = 2. - Benoit Cloitre, Apr 07 2002

For number of divisors of a(n) see A061645(n). Number of digits in a(n) is A061193(n). - Lekraj Beedassy, Jun 04 2004

All terms other than the first have digital root 1 (since 4^2 == 4 (mod 6), we have, by induction, 4^k == 4 (mod 6), or 2*2^(2*k) = 8 == 2 (mod 6), implying that Mersenne primes M = 2^p - 1, for odd p, are of the form 6*t+1). Thus perfect numbers N, being M-th triangular, have the form (6*t+1)*(3*t+1), whence the property N mod 9 = 1 for all N after the first. - Lekraj Beedassy, Aug 21 2004

The earliest recorded mention of this sequence is in Euclid's Elements, IX 36, about 300 BC. - Artur Jasinski, Jan 25 2006

Theorem (Euclid, Euler). An even number m is a perfect number if and only if m = 2^(k-1)*(2^k-1), where 2^k-1 is prime. Euler's idea came from Euclid's Proposition 36 of Book IX (see Weil). It follows that every even perfect number is also a triangular number. - Mohammad K. Azarian, Apr 16 2008

Triangular numbers (also generalized hexagonal numbers) A000217 whose indices are Mersenne primes A000668, assuming there are no odd perfect numbers. - Omar E. Pol, May 09 2008, Sep 15 2013

If a(n) is even, then 2*a(n) is in A181595. - Vladimir Shevelev, Nov 07 2010

Except for a(1) = 6, all even terms are of the form 30*k - 2 or 45*k + 1. - Arkadiusz Wesolowski, Mar 11 2012

a(4) = A229381(1) = 8128 is the "Simpsons's perfect number". - Jonathan Sondow, Jan 02 2015

Theorem (Farideh Firoozbakht): If m is an integer and both p and p^k-m-1 are prime numbers then x = p^(k-1)*(p^k-m-1) is a solution to the equation sigma(x) = (p*x+m)/(p-1). For example, if we take m=0 and p=2 we get Euclid's result about perfect numbers. - Farideh Firoozbakht, Mar 01 2015

The cototient of the even perfect numbers is a square; in particular, if 2^p - 1 is a Mersenne prime, cototient(2^(p-1) * (2^p - 1)) = (2^(p-1))^2 (see A152921). So, this sequence is a subsequence of A063752. - Bernard Schott, Jan 11 2019

Euler's (1747) proof that all the even perfect number are of the form 2^(p-1)*(2^p-1) implies that their asymptotic density is 0. Kanold (1954) proved that the asymptotic density of odd perfect numbers is 0. - Amiram Eldar, Feb 13 2021

If k is perfect and semiprime, then k = 6. - Alexandra Hercilia Pereira Silva, Aug 30 2021

This sequence lists the fixed points of A001065. - Alois P. Heinz, Mar 10 2024

sigma(n)/n is in A054030.

Also numbers such that the sum of the reciprocals of the divisors is an integer. - Harvey P. Dale, Jul 24 2001

Luca's solution of problem 11090, which proves that for k>1 there are an infinite number of n such that n divides sigma_k(n), does not apply to this sequence. However, it is conjectured that this sequence is also infinite. - T. D. Noe, Nov 04 2007

Numbers k such that sigma(k) is divisible by all divisors of k, subsequence of A166070. - Jaroslav Krizek, Oct 06 2009

A017666(a(n)) = 1. - Reinhard Zumkeller, Apr 06 2012

Bach, Miller, & Shallit show that this sequence can be recognized in polynomial time with arbitrarily small error by a probabilistic Turing machine; that is, this sequence is in BPP. - Charles R Greathouse IV, Jun 21 2013

Conjecture: If n is such that 2^n-1 is in A066175 then a(n) is a triangular number. - Ivan N. Ianakiev, Aug 26 2013

Conjecture: Every multiply-perfect number is practical (A005153). I've verified this conjecture for the first 5261 terms with abundancy > 2 using Achim Flammenkamp's data. The even perfect numbers are easily shown to be practical, but every practical number > 1 is even, so a weak form says every even multiply-perfect number is practical. - Jaycob Coleman, Oct 15 2013

Numbers such that A054024(n) = 0. - Michel Marcus, Nov 16 2013

Numbers n such that k(n) = A229110(n) = antisigma(n) mod n = A024816(n) mod n = A000217(n) mod n = (n(n+1)/2) mod n = A142150(n). k(n) = n/2 for even n; k(n) = 0 for odd n (for number 1 and eventually odd multiply-perfect numbers n > 1). - Jaroslav Krizek, May 28 2014

The only terms m > 1 of this sequence that are not in A145551 are m for which sigma(m)/m is not a divisor of m. Conjecture: after 1, A323653 lists all such m (and no other numbers). - Antti Karttunen, Mar 19 2021

These six terms are believed to comprise all 3-perfect numbers. - cf. the MathWorld link. - Daniel Forgues, May 11 2010

If there exists an odd perfect number m (a famous open problem) then 2m would be 3-perfect, since sigma(2m) = sigma(2)*sigma(m) = 3*2m. - Jens Kruse Andersen, Jul 30 2014

According to the previous comment from Jens Kruse Andersen, proving that this sequence is complete would imply that there are no odd perfect numbers. - Farideh Firoozbakht, Sep 09 2014

If 2 were prepended to this sequence, then it would be the sequence of integers k such that numerator(sigma(k)/k) = A017665(k) = 3. - Michel Marcus, Nov 22 2015

From Antti Karttunen, Mar 20 2021, Sep 18 2021, (Start):

Obviously, any odd triperfect numbers k, if they exist, have to be squares for the condition sigma(k) = 3*k to hold, as sigma(k) is odd only for k square or twice a square. The square root would then need to be a term of A097023, because in that case sigma(2*k) = 9*k. (See illustration in A347391).

Conversely to Jens Kruse Andersen's comment above, any 3-perfect number of the form 4k+2 would be twice an odd perfect number. See comment in A347870.

(End)

It is conjectured that there are only finitely many terms. - N. J. A. Sloane, Jul 22 2012

Odd perfect number (unlikely to exist) and infinitely many Mersenne primes will make the sequence infinite - take the product of the OPN and coprime EPNs.

Conjecture: A010888(a(n)) divides a(n). Tested for n up to 36 incl. - Ivan N. Ianakiev, Oct 31 2013

From Farideh Firoozbakht, Dec 26 2014: (Start)

Theorem: If k>1 and p=a(n)/2^(k-2)+1 is prime then for each positive integer m, 2^(k-1)*p^m is a solution to the equation sigma(phi(x))=2*x-2^k, which implies the equation has infinitely many solutions.

Proof: sigma(phi(2^(k-1)*p^m)) = sigma(2^(k-2)*(p-1)*p^(m-1)) = sigma(2^(k-2)*(p-1))*sigma(p^(m-1)) = sigma(a(n))*(p^m-1)/(p-1) = 4*a(n)*(p^m-1)/(p-1) = 2^k*(p^m-1) = 2*(2^(k-1)*p^m)-2^k.

It seems that for all such equations there exist such an infinite set of solutions. So I conjecture that the sequence is infinite! (End)

If 3 were prepended to this sequence, then it would be the sequence of integers k such that numerator(sigma(k)/k)=4. - Michel Marcus, Nov 22 2015

Conjectured finite and probably these are the only terms; cf. Flammenkamp's link. [Georgi Guninski, Jul 25 2012]

On the Riemann Hypothesis, a(n) > exp(exp(n / e^gamma)) for n > 3. Unconditionally, a(n) > exp(exp(0.9976n / e^gamma)) for n > 3, where the constant is related to A004394(1000000). - Charles R Greathouse IV, Sep 06 2012

Each of the terms 1, 6, 120, 30240 divides all larger terms <= a(8). See A227765, A227766, ..., A227769. - Jonathan Sondow, Jul 30 2013

Is a(n) < a(n+1)? - Jeppe Stig Nielsen, Jun 16 2015

Equivalently, a(n) is the smallest number k such that sigma(k)/k = n. - Derek Orr, Jun 19 2015

The number of divisors of these terms are: 1, 4, 16, 96, 1920, 110592, 1751777280, 63121588161085440. - Michel Marcus, Jun 20 2015

Given n, let S_n be the sequence of integers k that satisfy numerator(sigma(k)/k) = n. Then a(n) is a member of S_n. In fact a(n) = S_n(i), where the successive values of i are 1, 1, 2, 2, 4, 2, (23, 6, 31, 12, ...), where the terms in parentheses need to be confirmed. - Michel Marcus, Nov 22 2015

The first four terms are the only multiperfect numbers in A025487 among the 1600 initial terms of A007691. Can it be proved that these are the only ones among the whole A007691? See also A349747. - Antti Karttunen, Dec 04 2021

Numbers k > 1 such that nearest common ancestor of k and sigma(k) in Doudna tree is the parent of k, and sigma(k) is not a descendant of k.

Any hypothetical odd term x in A005820 (triperfect numbers) would also be a member of this sequence. This is illustrated in the following diagram which shows how the neighborhood of such x would look like in the Doudna tree (A005940). If m (the parent of x, x = A003961(m), m = A064989(x)) is even, then x is a multiple of 3, while if m is odd, then 3 does not divide x. Because the abundancy index decreases when traversing leftwards in the Doudna tree, m must be a term of A068403. Both x and m would also need to be squares, by necessity.

.

<--A003961-- m ---(*2)--->

.............../ \...............

/ \

/ \

x 2m

/ \ / \

etc.../ \.....2x sigma(x) = 3x..../ \.....4m

/ \ / \ / \

etc. etc. etc. \ / etc.

\ /

6x 9x = sigma(2x)

/ \ / \

etc. \ etc. etc.

\

12x = sigma(3x) if m odd.

.

From the diagram we also see that 2x would then need to be a term of A347392 (as well as that of A159907 and also in A074388, thus sqrt(x) should be a term of A097023), and furthermore, if x is not a multiple of 3 (i.e., when m is odd), then sigma(3*x) = 4*sigma(x) = 4*(3*x), thus 3*x = sigma(x) would be a term of A336702 (particularly, in A027687) and x would be a term of A323653.

Moreover, any odd square x in this sequence (for which sigma(x) would also be odd), would have an abundancy index of at least three (sigma(x)/x >= 3). See comments in A347383.

Note how 401625 = 6375 * 63 = 945 * 425, 46305 = 945 * 49, 9261 = 189 * 49, 6375 = 2125 * 3, 945 = 189 * 5 = 15 * 63 and 9261*2125 = 19679625. It seems that when the multiplicands are coprime, then they are both terms of this sequence, e.g. 2125 and 3, 189 and 5, 2125 and 9261.

From Antti Karttunen, Jul 10 2024: (Start)

Regarding the observation above, for two coprime odd numbers x, y, if both are included here because sigma(x) = 2^a * A064989(x) and sigma(y) = 2^b * A064989(y), then also their product x*y is included because in that case sigma(x*y) = 2^(a+b) * A064989(x*y).

Also, for two coprime odd numbers x, y, if both are included here because sigma(x) = A065119(i) * x and sigma(y) = A065119(j) * y, then also their product x*y is included because sigma(x*y) = A065119(k) * x*y, where A065119(k) = A065119(i)*A065119(j). The existence of such numbers (that would include odd triperfect and odd 6-perfect numbers, see A046061) is so far hypothetical, none is known.

It is not possible that the odd x is in this sequence if sigma(x) = k*A003961^e(x) and e = A061395(k)-2 >= 1.

Note that all odd terms < 2^33 here are some of the exponentially odd divisors of 19679625 (see A374199, also A374463 and A374464).

(End)

Question: from a(6) = 189 onward, are the rest of terms all in A347390?

Conjecture: sequence is finite.

If it exists, a(14) > 2^33.

Multiperfect numbers m such that sigma(m) divides sigma(sigma(m)).

Also k-multiperfect numbers m such that k*m is also multiperfect.

Corresponding values of numbers k(n) = sigma(a(n)) / a(n): 1, 3, 3, 5, 5, 5, 5, 5, ...

Corresponding values of numbers h(n) = sigma(k(n) * a(n)) / (k(n) * a(n)): 1, 4, 4, 6, 6, 6, 6, 6, ...

Number of k-multiperfect numbers m such that sigma(n) is also multiperfect for k = 3..6: 2, 0, 20, 0.

From Antti Karttunen, Mar 20 2021, Feb 18 2022: (Start)

Conjecture 1 (a): This sequence consists of those m for which sigma(m)/m is an integer (thus a term of A007691), and coprime with m. Or expressed in a slightly weaker form (b): {1} followed by those m for which sigma(m)/m is an integer, but not a divisor of m. In a slightly stronger form (c): For m > 1, sigma(m)/m is always the least prime not dividing m. This would imply both (a) and (b) forms.

Conjecture 2: This sequence is finite.

Conjecture 3: This sequence is the intersection of A007691 and A351458.

Conjecture 4: This is a subsequence of A349745, thus also of A351551 and of A351554.

Note that if there existed an odd perfect number x that were not a multiple of 3, then both x and 2*x would be terms in this sequence, as then we would have: sigma(x)/x = 2, sigma(2*x)/(2*x) = 3, sigma(6*x)/(6*x) = 4. See also the diagram in A347392 and A353365.

(End)

From Antti Karttunen, May 16 2022: (Start)

Apparently for all n > 1, A336546(a(n)) = 0. [At least for n=2..23], while A353633(a(n)) = 1, for n=1..23.

The terms a(1) .. a(23) are only cases present among the 5721 known and claimed multiperfect numbers with abundancy <> 2, as published 03 January 2022 under Flammenkamp's site, that satisfy the condition for inclusion in this sequence.

They are also the only 23 cases among that data such that gcd(n, sigma(n)/n) = 1, or in other words, for which the n and its abundancy are relatively prime, with abundancy in all cases being the least prime that does not divide n, A053669(n), which is a sufficient condition for inclusion in A351458.

(End)

The graph supports the conjecture that all numbers except 2 appear only a finite number of times. Sequences A000396, A005820, A027687, A046060 and A046061 give the n for which the abundancy sigma(n)/n is 2, 3, 4, 5 and 6, respectively. See A134639 for the number of n having abundancy greater than 2. - T. D. Noe, Nov 04 2007

Numbers k for which A351546(k) is a unitary divisor of k.

The condition guarantees that A351555(k) = 0, therefore this is a subsequence of A351554.

The condition is also a necessary condition for A349745, therefore it is a subsequence of this sequence.

All six known 3-perfect numbers (A005820) are included in this sequence.

All 65 known 5-multiperfects (A046060) are included in this sequence.

Not all multiperfects (A007691) are present (only 587 of the first 1600 are), but all 23 known terms of A323653 are terms, while none of the (even) terms of A046061 or A336702 are.