A177206 a(n) = 2*binomial(n+4, 4) + n + 4.

6, 15, 36, 77, 148, 261, 430, 671, 1002, 1443, 2016, 2745, 3656, 4777, 6138, 7771, 9710, 11991, 14652, 17733, 21276, 25325, 29926, 35127, 40978, 47531, 54840, 62961, 71952, 81873, 92786, 104755, 117846, 132127, 147668, 164541, 182820, 202581

Offset

0,1

Comments

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

We consider n aligned points of the plane {A1, A2, ..., An}, n > = 4. The sequence a(n) is the number of points of intersections of the circles with diameters AiAj (i<>j). - Michel Lagneau, May 04 2010

References

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

Links

Table of n, a(n) for n=0..37.

Michel Lagneau, Illustration for n=4

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

a(n) = (n^4 + 10n^3 + 35n^2 + 62n + 12)/12. - Gary Detlefs, Jun 06 2010

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

Example

For n = 5, we obtain 2*5 + 5 = 15 intersections.
From Michel Lagneau, May 04 2010: (Start)
For n = 4, we obtain 2*1 + 4 = 6 intersections (including tangential circles);
For n = 5, we obtain 2*5 + 5 = 15 intersections (including tangential circles). (End)

Maple

with(numtheory): n0:=75: T:=array(1..n0-4):for n from 5 to n0 do: T[n-4]:= 2*binomial(n, 4)+n:od:print(T):
with(numtheory): 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): # Michel Lagneau, May 04 2010
seq(2*binomial(n+4, 4)+n+4, n=0..39); # Gary Detlefs, Jun 06 2010

Mathematica

2Binomial[#+4, 4]+#+4&/@Range[0, 40]  (* Harvey P. Dale, Feb 08 2011 *)

Crossrefs

Sequence in context: A273411 A273490 A324221 * A128443 A245470 A135854

Adjacent sequences: A177203 A177204 A177205 * A177207 A177208 A177209

Keyword

nonn,easy

Author

Michel Lagneau, May 04 2010

Extensions

Definition corrected by Gary Detlefs, Jun 06 2010

Status

approved