cp's OEIS Frontend

Successive derivatives of the zeta function evaluated at x=2 round to (-1)^n * n!, for the n-th derivative, and converge with increasing n. For example, in Mathematica, Derivative[5][Zeta][2] = -120.000824333. A direct formula for the n-th derivative of Zeta at x=2 is: (-1)^n*Sum_{k>=1} log(k)^n/k^2. See also A201994 and A201995. The values of successive derivatives of Zeta(x) as x->1 are given by A252898, and are also related to the factorials. - Richard R. Forberg, Dec 30 2014

Number of elements in the set {(x,y): x|n, y|n, x<=y, gcd(x,y)=1}. Number of divisors of n^2 less than or equal to n. - Vladeta Jovovic, May 03 2002

Equivalently, number of pairs (x,y) such that lcm(x,y)=n. - Benoit Cloitre, May 16 2002

Also, number of right triangles with an integer hypotenuse and height n. - Reinhard Zumkeller, Jul 10 2002

The triangles are to be considered as resting on their hypotenuse, with the height measured to the right angle. - Franklin T. Adams-Watters, Feb 19 2015

a(n) >= 2 for n>=2 because of the identities 1/n = 1/(2*n) + 1/(2*n) = 1/(n+1) + 1/(n*(n+1)). - Lekraj Beedassy, May 04 2004

a(n) is the number of divisors of n^2 that are <= n; e.g., a(12) counts these 8 divisors of 12: 1,2,3,4,6,8,9,12. - Clark Kimberling, Apr 21 2019

Also number of ways to write 1/n as sum of exactly two distinct unit fractions. - Thomas L. York, Jan 11 2014

Also number of positive integers m such that 1/n + 1/m is a unit fraction. - Jon E. Schoenfield, Apr 17 2018

If 1/n = 1/b - 1/c then n = bc/(c-b) and 1/n = 1/(2n-b) + 1/(c+2n) (though it is also the case that 1/n = 1/(2n) + 1/(2n) equivalent to b = c = 0).

Also number of divisors of n^2 less than n. - Vladeta Jovovic, Aug 13 2001

Number of elements in the set {(x,y): x|n, y|n, xVladeta Jovovic, May 03 2002

Also number of positive integers of the form k*n/(k+n). - Benoit Cloitre, Jan 04 2002

This is similar to A062799, having the same first 29 terms. But they are different sequences.

If A001221(n) = omega(n) <= 2, then a(n) = A062799(n); if A001221(n) > 2, then a(n) > A062799(n). - Matthew Vandermast, Aug 25 2004

Number of r X s integer-sided rectangles such that r + s = 4n, r < s and (s - r) | (s * r). - Wesley Ivan Hurt, Apr 24 2020

Also number of integer-sided right triangles with 2n as a leg. Equivalent to the even indices of A046079. - Nathaniel C Beckman, May 14 2020; Jun 26 2020

a(n) is the number of positive integers k such that k+n divides k*n. - Thomas Ordowski, Dec 02 2024

a(n) is odd iff n is odd square. - Vladeta Jovovic, Aug 27 2002

From Robert Israel, Dec 26 2015: (Start)

a(n) >= A000005(n), with equality iff n is squarefree (i.e., is in A005117).

a(n) = 2 iff n is prime. (End)

Seems to be equal to the number of unordered pairs of coprime divisors of n. (Checked up to 2*10^14.) - Charles R Greathouse IV, May 03 2013

Outline of a proof for this observation, R. J. Mathar, May 05 2013: (Start)

i) To construct the divisors of n, write n=product_i p_i^e_i as the standard prime power decomposition, take any subset of the primes p_i (including the empty set representing the 1) and run with the associated list exponents from 0 up to their individual e_i.

To construct the *nonsquare* divisors of n, ensure that one or more of the associated exponents is/are odd. (The empty set is interpreted as 1^0 with even exponent.) To construct the nonsquare divisors of n^2, the principle remains the same, although the exponents may individually range from 0 up to 2*e_i.

The nonsquare divisor is therefore a nonempty product of prime powers (at least one) with odd exponents times a (potentially empty) product of prime powers (of different primes) with even exponents.

The nonsquare divisors of n^2 have exponents from 0 up to 2*e_i, but the subset of exponents in the "even/square" factor has e_i candidates (range 2, 4, .., 2*e_i) and in the "odd/nonsquare" factor also only e_i candidates (range 1,3,5,2*e_i-1).

ii) To construct the pairs of coprime divisors of n, take any two non-intersecting subsets of the set of p_i (possibly the empty subset which represents the factor 1), and let the exponents run from 1 up to their individual e_i in each of the two products.

iii) The bijection between the sets constructed in i) and ii) is given by mapping the two non-intersection prime sets onto each other, and observing that the numbers of compositions of exponents have the same orders in both cases.

(End)

a(n) is the number of pairs of positive integers <= n with their LCM <= n. - Andrew Howroyd, Sep 01 2019

Inverse Moebius transform of A068068. Number of elements in the set {(x,y): x is odd, x|n, y|n, gcd(x,y)=1}.

The number of Pythagorean points (x,y), 0 < x < y, located on the hyperbola y = 2n(x-n)/(x-2n) and having "excess" x+y-z = 2n. - Seppo Mustonen, Jun 07 2005

a(n) is the number of Pythagorean triangles with radius of the inscribed circle equal to n. For number of primitive Pythagorean triangles having inradius n, see A068068(n). - Ant King, Mar 06 2006

Dirichlet convolution of A048691 and A154269. - R. J. Mathar, Jun 01 2011

Number of distinct L-shapes of thickness n where the L area equals the rectangular area that it "contains". Visually can be thought as those areas of A156688 (surrounded by equal border of thickness n: 2xy = (x+2n)(y+2n), x and y positive integers) where both x and y are even, so they can be split into L-shapes. So L-shapes have formula: 2xy = (x+n)(y+n). - Juhani Heino, Jul 23 2012

Equivalently, a(n) is the number of nonisomorphic unitary rings whose rank is 2 when viewed as a free module over the ring (Z_n, +, *). - Jianing Song, Feb 23 2021

Every unitary ring with additive group (Z/nZ)^2 must be commutative, and is of the form Z_n[x]/(x^2 + b*x + c) for some b, c in Z_n, where (x^2 + b*x + c) stands for the ideal of Z_n[x] generated by x^2 + b*x + c. Proof: Let R be a unitary commutative ring with additive group (Z/nZ)^2. Suppose e is the identity element of R, x is an element such that {e, x} is a basis for R as a free module over Z_n (such a basis must exist, see my note in the link section), then every element can be written as the form u*x + v*e for 0 <= u, v <= n-1. If x^2 = -p*x - q*e, it turns out that R is isomorphic to Z_n[x]/(x^2 + [p]*x + [q]). - Jianing Song, Apr 23 2021

Equivalently, a(n) is the number of nonisomorphic rings of the form Z[x]/(n, x^2 + p*x + q), where (n, x^2 + p*x + q) is the ideal of Z[x] generated by n and x^2 + p*x + q. - Jianing Song, Feb 15 2021

Theorem. R_1 = Z_n[x]/(x^2 + b*x + c) and R_2 = Z_n[y]/(y^2 + b'*y + c') are isomorphic if and only if there exists some k in Z, t in Z_n such that gcd(k, n) = 1 and that b' == b*k + 2*t (mod n), c' == t^2 + b*k*t + c*k^2 (mod n).

Proof: "<=": Note that y^2 + (b*k + 2*t)*y + (t^2 + b*k*t + c*k^2) = (y + t)^2 + b*k*(y + t) + c*k^2, so a mapping from R_1 to R_2 is given by f(x) = (y + t)/k and f(r*x + s) = r*f(x) + s. Since gcd(k, n) = 1, f is an isomorphic mapping.

"=>": If R_1 and R_2 are isomorphic, there exists some isomorphic mapping from R_2 to R_1 such that f(y) = k*x - t. If gcd(k, n) > 1, since f(r*y + s) = r*f(y) + s = r*(k*x - t), there is no element in R_2 such that f(y) = x, a contradiction. So this isomorphic mapping sends x in R_1 to (y + t)/k, then (y + t)^2 + b*k*(y + t) + c*k^2 = 0. The corresponding coefficients must be equal modulo n, so b' == b*k + 2*t (mod n), c' == t^2 + b*k*t + c*k^2 (mod n).

Now note that without loss of generality we can suppose that b = 0 or -1, because we can always find some t such that b*k + 2*t == 0 or -1 (mod n). Furthermore, if n is an odd number, we can suppose that b = 0.

Case (i): n is an odd number, then a unitary ring with additive group (Z/nZ)^2 is of the form Z_n[x]/(x^2 - c). From the theorem above we can see that R_1 = Z_n[x]/(x^2 - c) and R_2 = Z_n[y]/(y^2 - c') are isomorphic if and only if there exists some k such that gcd(k, n) = 1 and that c*k^2 == c' (mod n). So the number of such rings is A092089(n).

Case (ii): n is an even number, then a unitary ring with additive group (Z/nZ)^2 is of the form Z_n[x]/(x^2 - c) or Z_n[x]/(x^2 - x - (c - 1)/4), c in Z_{4n}, c == 1 (mod 4). From the theorem above we can see that R_1 = Z_n[x]/(x^2 - c) and R_2 = Z_n[y]/(y^2 - c') are isomorphic if and only if there exists some k such that gcd(k, n) = 1 and that c*k^2 == c' (mod n) or c*k^2 + n^2/4 == c' (mod n) (with t = 0 and t = n/2 respectively); R_3 = Z_n[x]/(x^2 - x - (c - 1)/4)) and R_4 = Z_n[y]/(y^2 - y - (c' - 1)/4)) are isomorphic if and only if there exists some k such that gcd(k, 4*n) = 1 and that c*k^2 == c' (mod 4*n) or c*k^2 - n^2 + 2*n == c' (mod 4*n) (with t = (k - 1)/2 and t = (n + k - 1)/2 respectively).

(a) if n == 2 (mod 4), then the number of rings of the form is Z_n[x]/(x^2 - c) is A092089(n/2), and the number of rings of the form Z_n[x]/(x^2 - x - (c - 1)/4) is equal to the number of inequivalent residue classes modulo 4*n that are congruent to 1 modulo 4 where the equivalence relation is defined as [a] ~ [b] (mod 4*n) if and only if there exists some k such that gcd(k, 4*n) = 1 and that a*k^2 == b (mod 4*n). The number of the even inequivalent residue classes modulo 4*n is equal to the number of inequivalent residue classes modulo 2*n, and the number of inequivalent residue classes modulo 4*n that are congruent to 1 modulo 4 is equal to the number of those that are congruent to 3 modulo 4. So the total number if A092089(n/2) + (A092089(4*n) - A092089(2*n))/2.

(b) if n == 0 (mod 4). Similarly, the number of rings of the form is Z_n[x]/(x^2 - c) is A092089(n), and the number of rings of the form Z_n[x]/(x^2 - x - (c - 1)/4) is (A092089(2*n) - A092089(n))/2.