cp's OEIS Frontend

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

If a(n) = 0, then n is a multiply-perfect number (A007691). - Alonso del Arte, Mar 30 2014

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)

Equivalently, Chowla function of k is congruent to 1 (mod k).

If p=2^i-3 is prime, then 2^(i-1)*p is a term of the sequence. 650 is in the sequence, but is not of this form.

Terms k from a(2) to a(14) satisfy sigma(k) = 2*k + 2, implying that sigma(k) == 0 (mod k+1). It is not known if this holds in general, for there might be solutions of sigma(k)=3k+2 or 4k+2 or ... (Comments from Jud McCranie and Dean Hickerson, updated by Jon E. Schoenfield, Sep 25 2021 and by Max Alekseyev, May 23 2025).

k | sigma(k) produces the multiperfect numbers (A007691). It is an open question whether k | sigma(k) - 1 iff k is a prime or 1. It is not known if there exist solutions to sigma(k) = 2k+1.

Sequence also gives the nonprime solutions to sigma(k) == 0 (mod k+1), k > 1. - Benoit Cloitre, Feb 05 2002

Sequence seems to give nonprime k such that the numerator of the sum of the reciprocals of the divisors of k equals k+1 (nonprime k such that A017665(k)=k+1). - Benoit Cloitre, Apr 04 2002

For k > 1, composite numbers k such that A108775(k) = floor(sigma(k)/k) = sigma(k) mod k = A054024(k). Complement of primes (A000040) with respect to A230606. There are no numbers k > 2 such that sigma(x) = k*(x+1) has a solution. - Jaroslav Krizek, Dec 05 2013

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

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 n such that a(n)=0 are: 1, 2, 24, 4320, 4680, ... (see A159907, conjecture by Jaroslav Krizek and further comments). - Michel Marcus, Sep 23 2013

Numbers n such that a(n)=n/2 are: 6, 28, 120, 496, 672, ... = A007691 \ {1}. - Michel Marcus, Sep 25 2013

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)

Intersection of A001599 and A003601.

The following are also in A046985: 1, 6, 672, 30240, 32760. Also contains multiply perfect (A007691) numbers. - Labos Elemer

The numbers whose average divisor is also a divisor. Ore's harmonic numbers A001599 without the numbers A046999. - Thomas Ordowski, Oct 26 2014, Apr 17 2022

Harmonic numbers k whose harmonic mean of divisors (A001600) is also a divisor of k. - Amiram Eldar, Apr 19 2022

From Jaroslav Krizek, Feb 18 2010: (Start)

Number of divisors d of number n such that d divides sigma(n).

a(n) = A000005(n) - A173438(n).

a(n) = A000005(n) for multiply-perfect numbers (A007691). (End)