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A145979 a(n) = (2*n + 4)/gcd(n,4). 23

%I #75 Oct 09 2023 02:20:21

%S 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,

%T 54,28,58,15,62,32,66,17,70,36,74,19,78,40,82,21,86,44,90,23,94,48,98,

%U 25,102,52,106,27,110,56,114,29,118,60,122

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

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

%C 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

%C 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

%C 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

%C 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

%D 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_.

%H G. C. Greubel, <a href="/A145979/b145979.txt">Table of n, a(n) for n = 0..5000</a>

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,2,0,0,0,-1).

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

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

%F From _R. J. Mathar_, Dec 08 2008: (Start)

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

%F 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)

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

%F a(n) = A060819(n) + A060819(n+4). - _Paul Curtz_, Mar 13 2011

%F a(n) = (2*n + 4)/gcd(n,4). - _Joerg Arndt_, Jan 17 2015

%F 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

%F 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

%F Sum_{k=0..n} a(k) ~ (11/16) * n^2. - _Amiram Eldar_, Oct 09 2023

%p seq(denom(1/2-1/(n+2)), n=0..25); # _Gary Detlefs_, Sep 16 2011

%t Table[(2*n + 4)/GCD[n, 4], {n, 0, 50}] (* _G. C. Greubel_, Jan 29 2016 *)

%t LinearRecurrence[{0, 0, 0, 2, 0, 0, 0, -1}, {1, 6, 4, 10, 3, 14, 8, 18}, 70] (* _Vincenzo Librandi_, Jan 29 2016 *)

%t 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 *)

%o (Sage)

%o a = lambda n: (2 + n) / (2 - (n % 2) / 2 - (n % 4 != 0))

%o [a(n) for n in range(60)] # _Peter Luschny_, Jan 17 2015

%o (Magma) [(2*n+4)/GCD(n,4): n in [0..70]]; // _Vincenzo Librandi_, Jan 29 2016

%o (PARI) a(n) = (2*n + 4)/gcd(n,4); \\ _Michel Marcus_, Jan 29 2016

%o (GAP) List([1..70],n->(2*n+4)/Gcd(n,4)); # _Muniru A Asiru_, Apr 08 2018

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

%K nonn,easy

%O 0,2

%A _Paul Curtz_, Oct 26 2008

%E Edited by _R. J. Mathar_, Dec 08 2008

%E New name from _Joerg Arndt_, Jan 17 2015

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