A177206 - OEIS

%I #32 Apr 09 2026 17:28:11

%S 6,15,36,77,148,261,430,671,1002,1443,2016,2745,3656,4777,6138,7771,

%T 9710,11991,14652,17733,21276,25325,29926,35127,40978,47531,54840,

%U 62961,71952,81873,92786,104755,117846,132127,147668,164541,182820,202581

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

%C We consider n points in the plane, {A1, A2, ..., An}, n > = 4, lying on a line; a(n-4) is the number of points of intersections of the circles with diameters AiAj (i<>j).

%C The above comment does not hold for all possible sets of collinear points; for example, if the points are equidistant, some intersections coincide when n >= 9. _Giovanni Resta_, Apr 20 2025

%D J. M. Monier. Algèbre & Géometrie, Dunod 1996. Exercise p. 62.

%H Michel Lagneau, <a href="/A177206/a177206.pdf">Illustration for n=4</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F a(n) = (n^4 + 10n^3 + 35n^2 + 62n + 12)/12. - _Gary Detlefs_, Jun 06 2010

%F G.f.: -(3*x^4-13*x^3+21*x^2-15*x+6) / (x-1)^5. - _Colin Barker_, Dec 20 2012

%e For n = 5, we obtain 2*5 + 5 = 15 intersections.

%e From _Michel Lagneau_, May 04 2010: (Start)

%e For n = 4, we obtain 2*1 + 4 = 6 intersections (including tangential circles);

%e For n = 5, we obtain 2*5 + 5 = 15 intersections (including tangential circles). (End)

%p n0:=75: T:=array(1..n0-3):for n from 4 to n0 do: T[n-3]:= 2*binomial(n,4)+n:od:print(T):

%p # Alternative:

%p seq(2*binomial(n+4,4)+n+4,n=0..39); # _Gary Detlefs_, Jun 06 2010

%t 2Binomial[#+4,4]+#+4&/@Range[0,40]  (* _Harvey P. Dale_, Feb 08 2011 *)

%Y Cf. A290447.

%K nonn,easy

%O 0,1

%A _Michel Lagneau_, May 04 2010

%E Definition corrected by _Gary Detlefs_, Jun 06 2010