Matthew Lehman - cp's OEIS frontend

User: Matthew Lehman

Matthew Lehman has authored 3 sequences.

A158121 Given n points in the complex plane, let M(n) the number of distinct Moebius transformations that take 3 distinct points to 3 distinct points. Note that the triples may have some or all of the points in common.

6, 93, 591, 2381, 7316, 18761, 42253, 86281, 163186, 290181, 490491, 794613, 1241696, 1881041, 2773721, 3994321, 5632798, 7796461, 10612071, 14228061, 18816876, 24577433, 31737701, 40557401, 51330826, 64389781, 80106643, 98897541

Offset: 3

Matthew Lehman, Mar 12 2009

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.

			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.

Michael P. Hitchman, Geometry With an Introduction to Cosmic Topology, Jones and Bartlett Publishers, 2009, pages 59-60.

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).

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

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 *)

a(n) = 6* binomial(n, 3)^2 - binomial(n, 3) + 1; \\ Michel Marcus, Aug 13 2013

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

More terms from Michel Marcus, Aug 13 2013

A141214 Defining A to be the interior angle of a regular polygon, the number of constructible regular polygons such that A is in a field extension <= degree 2^n, starting with n=0. This is also the number of values of x such that phi(x)/2 is a power of 2 <= 2^n (where phi is Euler's phi function), also starting with n=0.

Original entry on oeis.org

3, 7, 12, 18, 25, 33, 42, 52, 63, 75, 88, 102, 117, 133, 150, 168, 187, 207, 228, 250, 273, 297, 322, 348, 375, 403, 432, 462, 493, 525, 558, 592, 626, 660, 694

Offset: 0

Matthew Lehman, Jun 14 2008

nonn

			For degree 2^0, there are 3 polygons of sides 3, 4 & 6.
For degree 2^1, there are 4 polygons of sides 5, 8, 10 & 12.
For degree 2^2 there are 5 (15, 16, 20, 24 & 30).
For n<=31, for degree 2^n, there are n+3 polygons.
For n>= 31 there are 34 polygons.
Assuming there are only 5 Fermat primes, this is the value of the sum 3+4+5+... up to 31 (and 32) terms, after which each term is 34.

The first 32 terms are identical to A055998 and A027379.

For 0=31 34n-462 The formulas are identical when n=31 f(31)=592

A141213 Defining A to be the interior angle of a regular polygon, the number of constructible regular polygons such that A is in a field extension = degree 2^n, starting with n=0. This is also the number of values of x such that phi(x)/2 = 2^n (where phi is Euler's phi function), also starting with n=0.

Original entry on oeis.org

3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 34, 34

Offset: 0

Matthew Lehman, Jun 14 2008

nonn

			For degree 2^0, there are 3 polygons with 3, 4 & 6 sides.
For degree 2^1, there are 4 polygons with 5, 8, 10 & 12 sides.
For degree 2^2 there are 5 polygons with 15, 16, 20, 24 & 30 sides.
For n<=31, for degree 2^n, there are n+3 polygons.
For n>= 31 there are 34 polygons.
Assuming there are only five Fermat primes, the sequence will continue repeating 34 forever.

Cf. A141214.

For n<=31, f(n)=n+3; for n>=31, f(n)=34.

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User: Matthew Lehman

A158121 Given n points in the complex plane, let M(n) the number of distinct Moebius transformations that take 3 distinct points to 3 distinct points. Note that the triples may have some or all of the points in common.

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