OFFSET
3,1
COMMENTS
There are (nC3)^2 ways of choosing two triples out of n points with repetition.
There are 3! = 6 ways of mapping the points of one triple to the other.
However, given each triple pair, there is one case where each of the initial three points is mapped to itself, resulting in the identity Moebius transformation.
There are nC3 cases of this, all but one redundant.
REFERENCES
Michael P. Hitchman, Geometry With an Introduction to Cosmic Topology, Jones and Bartlett Publishers, 2009, pages 59-60.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 3..1000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).
FORMULA
M(n) = 6*C(n,3)^2 - C(n,3) + 1.
M(n) = 1/6*(n^6-6*n^5+13*n^4-13*n^3+7*n^2-2*n+6).
G.f.: x^3*(6+51*x+66*x^2-13*x^3+15*x^4-6*x^5+x^6)/(1-x)^7. - Colin Barker, May 02 2012
EXAMPLE
For n=3, M(3) = 3! = 6, since there aren't any redundancies.
For n=4, M(4) = (6*4^2) - 3 = 93, since there are 3 redundant mappings.
MATHEMATICA
CoefficientList[Series[(6 + 51 x + 66 x^2 - 13 x^3 + 15 x^4 - 6 x^5 + x^6) / (1 - x)^7, {x, 0, 30}], x] (* Vincenzo Librandi, Aug 14 2013 *)
LinearRecurrence[{7, -21, 35, -35, 21, -7, 1}, {6, 93, 591, 2381, 7316, 18761, 42253}, 30] (* Harvey P. Dale, Mar 07 2020 *)
PROG
(PARI) a(n) = 6* binomial(n, 3)^2 - binomial(n, 3) + 1; \\ Michel Marcus, Aug 13 2013
(Magma) I:=[6, 93, 591, 2381, 7316, 18761, 42253]; [n le 7 select I[n] else 7*Self(n-1)-21*Self(n-2)+35*Self(n-3)-35*Self(n-4)+21*Self(n-5)-7*Self(n-6)+Self(n-7): n in [1..30]]; // Vincenzo Librandi, Aug 14 2013
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Matthew Lehman, Mar 12 2009
EXTENSIONS
More terms from Michel Marcus, Aug 13 2013
STATUS
approved