A145979 a(n) = (2*n + 4)/gcd(n,4).

1, 6, 4, 10, 3, 14, 8, 18, 5, 22, 12, 26, 7, 30, 16, 34, 9, 38, 20, 42, 11, 46, 24, 50, 13, 54, 28, 58, 15, 62, 32, 66, 17, 70, 36, 74, 19, 78, 40, 82, 21, 86, 44, 90, 23, 94, 48, 98, 25, 102, 52, 106, 27, 110, 56, 114, 29, 118, 60, 122

Offset: 0

Author

Paul Curtz, Oct 26 2008

Keywords

• nonn
• easy

Comments

Previous name was: Square root of A061038(n+2).

a(n) = denominator(Sum_{k=1..n} 1/((k+1)*(k+2))), n > 0. This summation has a closed form of 1/2 - 1/(n+2) and numerator of A060819(n). - Gary Detlefs, Sep 16 2011

Prefixing this sequence with 2 makes it a shift of the involution b defined on positive integers by b(n) = n if 4|n, b(n) = 2n if n is odd, b(n) = n/2 if n mod 4 = 2. This sequence b, when n > 2, occurs as the number of congruent regular n-gons in various ways of making cycles of them by sticking them together along edges with constant rotation angle between the two stickings on any one n-gon. For example, it is well known that only the triangle, square and hexagon can make cycles going once around a common point. But allowing the n-gons to keep going any number of times around the common corner, they will eventually close up into a cycle for any n (since their corner interior angle is a rational multiple of Pi), and the number of n-gons in that cycle is b(n). - David Pasino, Nov 12 2017

Here is another example of b(n) in the behavior of regular polygons as said in the comment of Nov 12 2017. For integers n and k, both exceeding 2, consider congruent regular k-gon tiles arranged as a ring going once around a central region, each tile adjacent to two others by sharing an exact edge, such that, if possible for n and k, the centers of the k-gons are the vertices of a regular n-gon. Then for any given n, the numbers k for which this arrangement is possible are exactly the multiples of b(n). (In the cases where (n, k) is (3, 6) or (4, 4) or (6, 3), the central region is only a point.) - David Pasino, Feb 20 2018

The generating function of the rationals A060819(n)/a(n) = 1/2 - 1/(n+2), n >= 0, with A060819(0) = 0, mentioned in the comment on a sum by Gary Detlefs above is (1/2)*(1-hypergeom([1, 1], [3], -x/(1-x)))/(1-x) = (x*(2 - x) + 2*(1 - x)*log(1-x) )/(2*(1-x)*x^2). Thanks to him for leading me to Jolley's general remark (201) on p. 38 on such sums. - Wolfdieter Lang, Mar 08 2018

The above b(n) also relates rotoinversions (rotation + inversion through the origin) to rotoreflections (rotation + reflection in a plane normal to the rotation axis). An n-fold rotoinversion clockwise is the same as some number of b(n)-fold rotoreflections counterclockwise. The Schoenflies notation for point group symmetry common in chemistry describes improper rotations as rotoreflections, while the International (Hemann-Mauguin) notation favored in crystallography describes them as rotoinversions. - R. James Evans, Nov 06 2024

References

• L. B. W. Jolley, Summation of Series, Second revised ed., Dover, 1961, p. 38, (201). For the sum given in the comment by Gary Detlefs.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..5000
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,-1).

Crossrefs

Cf. A000027, A005408, A008586, A016825, A022998, A060819, A061038.

Programs

GAP

List([1..70],n->(2*n+4)/Gcd(n,4)); # Muniru A Asiru, Apr 08 2018

Magma

[(2*n+4)/GCD(n,4): n in [0..70]]; // Vincenzo Librandi, Jan 29 2016

Maple

seq(denom(1/2-1/(n+2)), n=0..25); # Gary Detlefs, Sep 16 2011

Mathematica

Table[(2*n + 4)/GCD[n, 4], {n, 0, 50}] (* G. C. Greubel, Jan 29 2016 *)
LinearRecurrence[{0, 0, 0, 2, 0, 0, 0, -1}, {1, 6, 4, 10, 3, 14, 8, 18}, 70] (* Vincenzo Librandi, Jan 29 2016 *)
CoefficientList[ Series[(-2x^7 + 2x^5 + x^4 + 10x^3 + 4x^2 + 6x + 1)/(x^4 - 1)^2, {x, 0, 60}], x] (* Robert G. Wilson v, Nov 25 2016 *)

PARI

a(n) = (2*n + 4)/gcd(n,4); \\ Michel Marcus, Jan 29 2016

Sage

a = lambda n: (2 + n) / (2 - (n % 2) / 2 - (n % 4 != 0))
[a(n) for n in range(60)] # Peter Luschny, Jan 17 2015

Formula

a(2n) = A022998(n+1). a(2n+1) = A016825(n+1) = 2*A005408(n+1).

a(4n) = 2n+1 = A005408(n). a(4n+2) = A008586(n+1) = 4*A000027(n+1).

From R. J. Mathar, Dec 08 2008: (Start)

a(n) = 2*a(n-4) - a(n-8).

G.f.: (1 + 6x + 4x^2 + 10x^3 + x^4 + 2x^5 - 2x^7) / ((x-1)^2*(1+x)^2*(x^2+1)^2). (End)

a(n) = (n+2)*(11 - 5*(-1)^n - i^n - (-i)^n)/8, where i is the imaginary unit. - Bruno Berselli, Feb 25 2011

a(n) = A060819(n) + A060819(n+4). - Paul Curtz, Mar 13 2011

a(n) = (2*n + 4)/gcd(n,4). - Joerg Arndt, Jan 17 2015

E.g.f.: (1/4)*(2*(4*x+3)*cosh(x) + (3*x+16)*sinh(x) + x*sin(x) - 2*cos(x)). - G. C. Greubel, Jan 29 2016

a(n) = b(n+2), for b as in comment of Nov 12 2017. Same b is b(n) = (2n)/gcd(2n, n+2). - David Pasino, Feb 20 2018

Sum_{k=0..n} a(k) ~ (11/16) * n^2. - Amiram Eldar, Oct 09 2023

Extensions

Edited by R. J. Mathar, Dec 08 2008

New name from Joerg Arndt, Jan 17 2015