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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)

Let sigma^m (j) be the result of applying the sum-of-divisors function (A000203) m times to j; call j (m,k)-perfect if sigma^m (j) = k*j; then this is the sequence of (2,k)-perfect numbers.

From Michel Marcus, May 14 2016: (Start)

For these numbers, the quotient k = sigma(sigma(j))/j is an integer (see A098223). Then also k = (sigma(s)/s)*(sigma(j)/j) with s = sigma(j). That is, k = abundancy(s)*abundancy(j).

So looking at the abundancy of these terms may be interesting. Indeed we see that 459818240 and 51001180160 are actually 3-perfect numbers (A005820), and the reason they are here is that they are coprime to 3. So their sums of divisors are 4-perfect numbers (A027687), yielding q=12.

In a similar way, we can see that the 5-perfect numbers (A046060) that are coprime to 5 will be terms of this sequence with q=30. There are 20 such numbers, the smallest being 13188979363639752997731839211623940096. (End)

From Michel Marcus, May 15 2016: (Start)

It is also interesting to note that for a(2)=8, s=sigma(8)=15 is also a term. This happens to be the case for chains of several terms in a row:

8, 15, 24, 60, 168, 480 with k = 3,4,7,8,9,10;

512, 1023, 1536, 4092, 10752, 32736 with k = 3,4,7,8,9,10;

29127, 47360, 116508, 331520, 932064, 2983680 with k = 4,7,8,9,13,14;

1556480, 3932040, 14008320 with k = 9,13,14;

106151936, 251650560, 955367424 with k = 9,13,14;

312792480, 1505806848 with k = 19,20;

6604416000, 30834059256 with k = 19,20;

9623577600, 46566269568 with k = 19,20.

When j is a term, we can test if s=sigma(j) is also a term; this way we get 6 more terms: 572941926400, 845734196736, 1422976331052, 4010593484800, 11383810648416, 36095341363200.

And the corresponding chains are:

173238912000, 845734196736 with k = 19,20;

355744082763, 572941926400, 1422976331052, 4010593484800, 11383810648416, 36095341363200 with k = 4,7,8,9,13,14. (End)

From Altug Alkan, May 17 2016: (Start)

Here are additional chains for the above list:

57120, 217728 with k = 13,14;

343976, 710400 with k = 7,8;

1980342, 5621760 with k = 10,14;

4404480, 14913024 with k = 11,12;

238608384, 775898880 with k = 11,12. (End)

Currently, the coefficient pairs are [1, 1], [3, 4], [4, 7], [7, 8], [8, 9], [9, 10], [9, 13], [10, 14], [11, 12], [13, 14], [16, 17], [16, 21], [17, 18], [19, 20], [23, 24], [25, 26], [25, 31], [27, 28], [29, 30], [31, 32], [32, 33], [37, 38]. It is interesting to note that for some of them, the pair (s,t) also satisfies t=sigma(s). - Michel Marcus, Jul 03 2016; Sep 06 2016

Using these empirical pairs of coefficients in conjunction with the first comment allows us to determine whether some term is the sum of divisors of another yet unknown smaller term. - Michel Marcus, Jul 04 2016

For m in A090748 = A000043 - 1 and c in A205597 (= odd a(n)), c*2^m is in the sequence, unless 2^(m+1)-1 | sigma(c). Indeed, from sigma(x*y) = sigma(x)*sigma(y) for gcd(x,y) = 1, we get sigma(sigma(c*2^m)) = sigma(sigma(c))*2^(m+1), so c*2^m is in the sequence if sigma(sigma(c))/c = k/2 (where k can't be odd: A330598 has no odd c). - M. F. Hasler, Jan 06 2020

Apart from missing 2, this sequence gives all numbers k such that the binary expansion of A156552(k) is a prefix of that of A156552(sigma(k)), that is, for k > 1, numbers k for which sigma(k) is a descendant of k in A005940-tree. This follows because of the two transitions x -> A005843(x) (doubling) and x -> A003961(x) (prime shift) used to generate descendants in A005940-tree, using A003961 at any step of the process will ruin the chances of encountering sigma(k) anywhere further down that subtree.

Proof: Any left child in A005940 (i.e., A003961(k) for k) is larger than sigma(k), for any k > 2 [see A286385 for a proof], and A003961(n) > n for all n > 1. Thus, apart from A003961(2) = 3 = sigma(2), A003961^t(k) > sigma(k), where A003961^t means t-fold application of prime shift, here with t >= 1. On the other hand, sigma(2n) > sigma(n) for all n, thus taking first some doubling steps before a run of one or more prime shift steps will not rescue us, as neither will taking further doubling steps after a bout of prime shifts.

The first terms of A325637 not included in this sequence are 154345556085770649600 and 9186050031556349952000, as they have abundancy index 6.

From Antti Karttunen, Nov 29 2021: (Start)

Odd terms of this sequence are given by the intersection of A349169 and A349174.

A064989 applied to the odd terms of this sequence gives the fixed points of A326042, i.e., the positions of zeros in A348736, and a subset of the positions of ones in A348941.

Odd terms of this sequence form a subsequence of A348943, but should occur neither in A348748 nor in A348749.

(End)

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

a(n) tells about the degree of relatedness between n and sigma(n) in Doudna tree (see the illustration in A005940). It is 0 for those n where sigma(n) is one of the descendants of n, 1 for those n where the nearest common ancestor of n and sigma(n) is the parent of n, 2 for those n where the nearest common ancestor of n and sigma(n) is the grandparent of n, and so on.

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

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

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.