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A145979
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a(n) = (2*n + 4)/gcd(n,4).
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23
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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
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OFFSET
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0,2
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COMMENTS
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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
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REFERENCES
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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.
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LINKS
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FORMULA
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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
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
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MAPLE
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seq(denom(1/2-1/(n+2)), n=0..25); # Gary Detlefs, Sep 16 2011
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MATHEMATICA
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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 *)
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PROG
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(Sage)
a = lambda n: (2 + n) / (2 - (n % 2) / 2 - (n % 4 != 0))
(GAP) List([1..70], n->(2*n+4)/Gcd(n, 4)); # Muniru A Asiru, Apr 08 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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