cp's OEIS Frontend

Restricted growth sequence transform of the ordered pair [A033630(n), A083206(n)].

Restricted growth sequence transform of the ordered pair [A046523(n), A033630(n)].

The odd bisection differs from A099774 (the odd bisection of A000005) apparently in the positions given by A005231 (odd abundant numbers): 945, 1575, 2205, 2835, 3465, ...

Equally, inverse Möbius transform of A378437, which is the Dirichlet inverse of A033630.

Restricted growth sequence transform of the ordered pair [A001065(n), A033630(n)].

A number k is abundant if sigma(k) > 2k (cf. A005101), perfect if sigma(k) = 2k (cf. A000396), or deficient if sigma(k) < 2k (this sequence), where sigma(k) is the sum of the divisors of k (A000203).

Also, numbers k such that A033630(k) = 1. - Reinhard Zumkeller, Mar 02 2007

According to Deléglise (1998), the abundant numbers have natural density 0.2474 < A(2) < 0.2480. Since the perfect numbers have density 0, the deficient numbers have density 0.7520 < 1 - A(2) < 0.7526. Thus the n-th deficient number is asymptotic to 1.3287*n < n/(1 - A(2)) < 1.3298*n. - Daniel Forgues, Oct 10 2015

The data begins with 3 runs of 5 consecutive terms, from 1 to 5, 7 to 11 and 13 to 17. The maximal length of a run of consecutive terms is 5 because 6 is a perfect number and its proper multiples are abundant numbers. - Bernard Schott, May 19 2019

If p and q are primes such that phi(p*q) > p+1, then p*q^n is a term in the sequence for all n >= 1 where phi is the Euler totient function. - Amrit Awasthi, Sep 10 2024

Equivalently, positive integers m such that every number k <= m is a sum of distinct divisors of m.

2^r is a member for all r as every number < = sigma(2^r) = 2^(r+1)-1 is a sum of a distinct subset of divisors {1, 2, 2^2, ..., 2^m}. - Amarnath Murthy, Apr 23 2004

Also, numbers m such that A030057(m) > m. This is a consequence of the following theorem (due to Stewart), found at the McLeman link: An integer m >= 2 with factorization Product_{i=1..k} p_i^e_i with the p_i in ascending order is practical if and only if p_1 = 2 and, for 1 < i <= k, p_i <= sigma(Product_{j < i} p_j^e_j) + 1. - Franklin T. Adams-Watters, Nov 09 2006

Practical numbers first appear in Srinivasan's short paper, which contains terms up to 200. Let m be a practical number. He states that (1) if m>2, m is a multiple of 4 or 6; (2) sigma(m) >= 2*m-1 (A103288); and (3) 2^t*m is practical. He also states that highly composite numbers (A002182), perfect numbers (A000396), and primorial numbers (A002110) are practical. - T. D. Noe, Apr 02 2010

Conjecture: The sequence a(n)^(1/n) (n=3,4,...) is strictly decreasing to the limit 1. - Zhi-Wei Sun, Jan 12 2013

Conjecture: For any positive rational number r, there are finitely many pairwise distinct practical numbers q(1)..q(k) such that r = Sum_{j=1..k} 1/q(j). For example, 2 = 1/1 + 1/2 + 1/4 + 1/6 + 1/12 with 1, 2, 4, 6 and 12 all practical, and 10/11 = 1/2 + 1/4 + 1/8 + 1/48 + 1/132 + 1/176 with 2, 4, 8, 48, 132 and 176 all practical. - Zhi-Wei Sun, Sep 12 2015

Analogous with the {1 union primes} (A008578), practical numbers form a complete sequence. This is because it contains all powers of 2 as a subsequence. - Frank M Jackson, Jun 21 2016

Sun's 2015 conjecture on the existence of Egyptian fractions with practical denominators for any positive rational number is true. See the link "Egyptian fractions with practical denominators". - David Eppstein, Nov 20 2016

Conjecture: if all divisors of m are 1 = d_1 < d_2 < ... < d_k = m, then m is practical if and only if d_(i+1)/d_i <= 2 for 1 <= i <= k-1. - Jianing Song, Jul 18 2018

The above conjecture is incorrect. The smallest counterexample is 78 (for which one of these quotients is 13/6; see A174973). m is practical if and only if the divisors of m form a complete subsequence. See Wikipedia links. - Frank M Jackson, Jul 25 2018

Reply to the comment above: Yes, and now I can show the opposite: The largest value of d_(i+1)/d_i is not bounded for practical numbers. Note that sigma(n)/n is not bounded for primorials, and primorials are practical numbers. For any constant c >= 2, let k be a practical number such that sigma(k)/k > 2c. By Bertrand's postulate there exists some prime p such that c*k < p < 2c*k < sigma(k), so k*p is a practical number with consecutive divisors k and p where p/k > c. For example, for k = 78 we have 13/6 > 2, and for 97380 we have 541/180 > 3. - Jianing Song, Jan 05 2019

Erdős (1950) and Erdős and Loxton (1979) proved that the asymptotic density of practical numbers is 0. - Amiram Eldar, Feb 13 2021

Let P(x) denote the number of practical numbers up to x. P(x) has order of magnitude x/log(x) (see Saias 1997). Moreover, we have P(x) = c*x/log(x) + O(x/(log(x))^2), where c = 1.33607... (see Weingartner 2015, 2020 and Remark 1 of Pomerance & Weingartner 2021). As a result, a(n) = k*n*log(n*log(n)) + O(n), where k = 1/c = 0.74846... - Andreas Weingartner, Jun 26 2021

From Hal M. Switkay, Dec 22 2022: (Start)

Every number of least prime signature (A025487) is practical, thereby including two classes of number mentioned in Noe's comment. This follows from Stewart's characterization of practical numbers, mentioned in Adams-Watters's comment, combined with Bertrand's postulate (there is a prime between every natural number and its double, inclusive).

Also, the first condition in Stewart's characterization (p_1 = 2) is equivalent to the second condition with index i = 1, given that an empty product is equal to 1. (End)

Conjecture: every odd number, beginning with 3, is the sum of a prime number and a practical number. Note that this conjecture occupies the space between the unproven Goldbach conjecture and the theorem that every even number, beginning with 2, is the sum of two practical numbers (Melfi's 1996 proof of Margenstern's conjecture). - Hal M. Switkay, Jan 28 2023

From Reinhard Zumkeller, Dec 11 2009: (Start)

For odd primes p: a(p^2) = p + 2; for n > 1: a(A001248(n)) = A052147(n);

For odd primes p > 3, a(3*p) = 2*p + 4; for n > 2: a(A001748(n)) = A100484(n) + 4. (End)

From Matthew Crawford, Jan 19 2021: (Start)

For a prime p, a(p^3) = (p^3 + p^2 + 2*p + 4)/2;

For distinct primes p and q, a(p*q) = (p+1)*(q+1)/2 + 2. (End)