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If the canonical factorization of n into prime powers is Product p^e(p) then d(n) = Product (e(p) + 1). More generally, for k > 0, sigma_k(n) = Product_p ((p^((e(p)+1)*k))-1)/(p^k-1) is the sum of the k-th powers of the divisors of n.

Number of ways to write n as n = x*y, 1 <= x <= n, 1 <= y <= n. For number of unordered solutions to x*y=n, see A038548.

Note that d(n) is not the number of Pythagorean triangles with radius of the inscribed circle equal to n (that is A078644). For number of primitive Pythagorean triangles having inradius n, see A068068(n).

Number of factors in the factorization of the polynomial x^n-1 over the integers. - T. D. Noe, Apr 16 2003

Also equal to the number of partitions p of n such that all the parts have the same cardinality, i.e., max(p)=min(p). - Giovanni Resta, Feb 06 2006

Equals A127093 as an infinite lower triangular matrix * the harmonic series, [1/1, 1/2, 1/3, ...]. - Gary W. Adamson, May 10 2007

For odd n, this is the number of partitions of n into consecutive integers. Proof: For n = 1, clearly true. For n = 2k + 1, k >= 1, map each (necessarily odd) divisor to such a partition as follows: For 1 and n, map k + (k+1) and n, respectively. For any remaining divisor d <= sqrt(n), map (n/d - (d-1)/2) + ... + (n/d - 1) + (n/d) + (n/d + 1) + ... + (n/d + (d-1)/2) {i.e., n/d plus (d-1)/2 pairs each summing to 2n/d}. For any remaining divisor d > sqrt(n), map ((d-1)/2 - (n/d - 1)) + ... + ((d-1)/2 - 1) + (d-1)/2 + (d+1)/2 + ((d+1)/2 + 1) + ... + ((d+1)/2 + (n/d - 1)) {i.e., n/d pairs each summing to d}. As all such partitions must be of one of the above forms, the 1-to-1 correspondence and proof is complete. - Rick L. Shepherd, Apr 20 2008

Number of subgroups of the cyclic group of order n. - Benoit Jubin, Apr 29 2008

Equals row sums of triangle A143319. - Gary W. Adamson, Aug 07 2008

Equals row sums of triangle A159934, equivalent to generating a(n) by convolving A000005 prefaced with a 1; (1, 1, 2, 2, 3, 2, ...) with the INVERTi transform of A000005, (A159933): (1, 1,-1, 0, -1, 2, ...). Example: a(6) = 4 = (1, 1, 2, 2, 3, 2) dot (2, -1, 0, -1, 1, 1) = (2, -1, 0, -2, 3, 2) = 4. - Gary W. Adamson, Apr 26 2009

Number of times n appears in an n X n multiplication table. - Dominick Cancilla, Aug 02 2010

Number of k >= 0 such that (k^2 + k*n + k)/(k + 1) is an integer. - Juri-Stepan Gerasimov, Oct 25 2015

The only numbers k such that tau(k) >= k/2 are 1,2,3,4,6,8,12. - Michael De Vlieger, Dec 14 2016

a(n) is also the number of partitions of 2*n into equal parts, minus the number of partitions of 2*n into consecutive parts. - Omar E. Pol, May 03 2017

From Tomohiro Yamada, Oct 27 2020: (Start)

Let k(n) = log d(n)*log log n/(log 2 * log n), then lim sup k(n) = 1 (Hardy and Wright, Chapter 18, Theorem 317) and k(n) <= k(6983776800) = 1.537939... (the constant A280235) for every n (Nicolas and Robin, 1983).

There exist infinitely many n such that d(n) = d(n+1) (Heath-Brown, 1984). The number of such integers n <= x is at least c*x/(log log x)^3 (Hildebrand, 1987) but at most O(x/sqrt(log log x)) (Erdős, Carl Pomerance and Sárközy, 1987). (End)

Number of 2D grids of n congruent rectangles with two different side lengths, in a rectangle, modulo rotation (cf. A038548 for squares instead of rectangles). Also number of ways to arrange n identical objects in a rectangle (NOT modulo rotation, cf. A038548 for modulo rotation); cf. A007425 and A140773 for the 3D case. - Manfred Boergens, Jun 08 2021

The constant quoted above from Nicolas and Robin, 6983776800 = 2^5 * 3^3 * 5^2 * 7 * 11 * 13 * 17 * 19, appears arbitrary, but interestingly equals 2 * A095849(36). That second factor is highly composite and deeply composite. - Hal M. Switkay, Aug 08 2025

The sequence contains the smaller member of every pair of twin primes (A001359) and all squarefree semiprimes m such that m+4 is also a squarefree semiprime (A255746). Can one prove that this is an infinite sequence? - Vladimir Shevelev, Jul 11 2015

The sequence does not contain perfect squares. Indeed, let a(m)=k^2. Then d(k^2+d(k^2)) = d(k^2). Note that d(k^2) is odd. On the other hand, it is known (A046522) that d(k^2)<2*k. Hence, (k+1)^2 - k^2 > d(k^2). Thus k^2Vladimir Shevelev, Feb 10 2017

If p is prime and t+1 is odd prime, then p^t is not in the sequence. Indeed, if d(p^t+t+1)=t+1, then p^t+t+1=q^t, where q is prime > p (if p^t+t+1= say q^l*r^m, then (l+1)*(m+1)=t+1 which is impossible by the condition). But q>=p+2 and p^t+t+1>=p^t+2*t*p^(t-1) or t+1>=2*t*p^(t-1) which trivially has only solution t=1; however, by the condition t>=2. - Vladimir Shevelev, Feb 18 2017

If an odd integer k is in this sequence, so is 2k. - Charlie Neder, Jan 14 2019

This sequences uses the same rules as Recamán's sequence A005132 except that, instead of adding or subtracting n each term, the number of divisors of n is used. See A000005.

For the first 10 million terms the smallest value not appearing is 28. The data indicate that a(n)/n approaches 1 as n goes to infinity. As tau(n) <= 2*sqrt(n) (see A046522), it implies that 28 and other small unvisited values will never be visited.

In the same range the maximum value is a(9998226) = 10987569, and 2202001 terms repeat a previously visited value, the first time this occurs is a(21) = a(16) = 16. The longest run of consecutive increasing terms is 30, starting at a(1115610) = 1217112, while the longest run of consecutive decreasing terms is 534, starting at a(9960335) = 10946233.

The sequence contains the smaller member of every pair of cousin primes (A023200).

The sequence contains no perfect squares. Indeed, let a(m) = k^2 for some m. Then, by the definition, d(k^2 + 2*d(k^2)) = d(k^2). Note that d(k^2) is odd. On the other hand, it is known (cf. A046522) that d(k^2) < 2*k. Hence (k+2)^2 - k^2 = 4*k + 4 > 2*d(k^2). Thus k^2 < k^2 + 2*d(k^2) < (k+2)^2. Since, evidently, k^2 + 2*d(k^2) cannot be (k+1)^2, then k^2 + 2*d(k^2) cannot be a square. Therefore, d(k^2 + 2*d(k^2)) is even, which is a contradiction.

The sequence contains the smaller member of every pair of sexy primes (A023201).

The sequence contains no perfect squares. Indeed, let a(m) = k^2 for some m. Then, by the definition, d(k^2 + 3*d(k^2)) = d(k^2). Note that d(k^2) is odd. On the other hand, it is known (cf. A046522) that d(k^2) < 2*k. Hence (k+3)^2 - k^2 = 6*k + 1 > 3*d(k^2). Thus k^2 < k^2 + 3*d(k^2) < (k+3)^2. Note that, evidently, k^2 + 3*d(k^2) cannot be (k+2)^2. Let us also show that k^2 + 3*d(k^2) cannot be (k+1)^2, or, equivalently, 3*d(k^2) cannot be equal to 2*k + 1. Indeed, let 3*d(k^2) = 2*k + 1. For some prime p, let p^a || k (that is, p^a | k, but p^(a+1) !| k), a > 0, so 2*k + 1 == 1 (mod p). But now we have 3*p^(a+1) | 3*d(k^2) and thus 3*p^(a+1)|2*k + 1, so 2*k + 1 == 0 (mod p). Contradiction. Therefore, we conclude that k^2 + 3*d(k^2) cannot be a square. Hence, d(k^2 + 3*d(k^2)) is even, which is a contradiction.