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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

For any other number n than those in A326182 we have a(n) < A003961(n).

Fixed points k (for which a(k) = k) satisfy A003973(k) = 2^e * A003961(k) for some exponent e >= 0. Applying A003961 to such numbers gives the odd terms in A336702, of which there are likely to be just a single instance, its initial 1. (Clarified Nov 07 2021).

Conjecture: There are no other fixed points than a(1) = 1. If true, then there are no odd perfect numbers. This condition is equivalent to the condition that if A161942 has no fixed points larger than one, then there are no odd perfect numbers. This follows as whenever k is a fixed point, that is, a(k) = k, then we should also have A003961(a(k)) = A003961(A064989(sigma(A003961(k)))) = A161942(A003961(k)) = A003961(k). Note that A003961 is an injective and surjective mapping from natural numbers to odd numbers, A064989 is its (left) inverse, and composition A003961(A064989(n)) is equivalent to A000265(n).

From Antti Karttunen, Aug 05 2020: (Start)

For any hypothetical odd perfect number x, we would have A003973(k) = 2 * A003961(k), with k = A064989(x) and x = A003961(k). Thus we would have a(k) = A064989(sigma(A003961(k))) = A064989(sigma(x)) = A064989(2*x) = A064989(x) = k. On the other hand, A003973(k) = sigma(A003961(k)) < A003961(A003961(k)) [see A286385 for the reason why], so a necessary condition for this is that x should be one of the terms of A246282. (Clarified Dec 01 2020).

(End)

Are all terms nonnegative? This question is equivalent to the question posed in A285705.

From Antti Karttunen, Aug 05 2020: (Start)

The answer to the above question is yes. Because both A000203 and A003961 are multiplicative sequences, it suffices to prove that for any prime p, and e >= 1, q^e >= sigma(p^e) = ((p^(1+e))-1) / (p-1), where q = A151800(p), i.e., the next larger prime after p. If p is a lesser twin prime, then q = p+2 (and this difference can't be less than 2, apart from case p=2), and it is easy to see that (n+2)^e > ((n^(e+1)) - 1) / (n-1), for all n >= 2, e >= 1.

See comments in A326042.

(End)

This is the inverse Möbius transform of A337549, from which it is even easier to see that all terms are nonnegative. - Antti Karttunen, Sep 22 2020

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.

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.

Numbers k such that A348990(k) [= k/gcd(k, A003961(k))] is equal to A348992(k), which is the odd part of A349162(k), thus all terms must be odd, as A348990 preserves the parity of its argument.

Equally, numbers k for which gcd(A064987(k), A191002(k)) is equal to A000265(gcd(A064987(k), A341529(k))).

Also odd numbers k for which A348993(k) = A319627(k).

Odd terms of A336702 are given by the intersection of this sequence and A349174.

Conjectures:

(1) After 1, all terms are multiples of 3. (Why?)

(2) After 1, all terms are in A104210, in other words, for all n > 1, gcd(a(n), A003961(a(n))) > 1. Note that if we encountered a term k with gcd(k, A003961(k)) = 1, then we would have discovered an odd multiperfect number.

(3) Apart from 1, 15, 105, 3003, 13923, 264537, all other terms are abundant. [These apparently are also the only terms that are not Zumkeller, A083207. Note added Dec 05 2024]

(4) After 1, all terms are in A248150. (Cf. also A386430).

(5) After 1, all terms are in A348748.

(6) Apart from 1, there are no common terms with A349753.

Note: If any of the last four conjectures could be proved, it would refute the existence of odd perfect numbers at once. Note that it seems that gcd(sigma(k), A003961(k)) < k, for all k except these four: 1, 2, 20, 160.

Questions:

(1) For any term x here, can 2*x be in A349745? (Partial answer: at least x should be in A191218 and should not be a multiple of 3). Would this then imply that x is an odd perfect number? (Which could explain the points (1) and (4) in above, assuming the nonexistence of opn's).

Numerator of ratio A003961(n) / A000203(n). Sequence A349162 gives the denominators.

Numerator of ratio A003961(n) / A161942(n). Sequence A348992 gives the denominators.

Both ratios are multiplicative because the constituent sequences are.

No 1's occur as terms after a(2), because for n > 2, sigma(n) < A003961(n). (See A286385).

Denominator of ratio A003961(n) / A000203(n).

Small values are rare, but are not limited to the beginning. For example in range 1 .. 2^25, a(n) = 4 at n = 3, 6, 24, 792, 2720, 122944, 31307472.

Question: Would it be possible to prove that a(n) > 1 for all n > 2?

Obviously, 1's may occur only on squares & twice squares (A028982). See also comments in A350072. - Antti Karttunen, Feb 16 2022