A105374 a(n) = 4*n^3 + 4*n.

0, 8, 40, 120, 272, 520, 888, 1400, 2080, 2952, 4040, 5368, 6960, 8840, 11032, 13560, 16448, 19720, 23400, 27512, 32080, 37128, 42680, 48760, 55392, 62600, 70408, 78840, 87920, 97672, 108120, 119288, 131200, 143880, 157352, 171640, 186768

Offset

0,2

Comments

For n > 1, the number of straight lines with n points in a 4-dimensional hypercube of with n points on each edge is 4n^3 + 12n^2 + 16n + 8, i.e., A105374(n+1).

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Formula

a(n) = A002522(n)*A008586(n).

G.f.: 8*x*(1 + x + x^2)/(1-x)^4. - Colin Barker, May 24 2012

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 26 2012

a(n) = 8* A006003(n). - Bruce J. Nicholson, Apr 18 2017

Example

a(5) = 4*5^3 + 4*5 = 500 + 20 = 520.

Mathematica

CoefficientList[Series[8*x*(1+x+x^2)/(1-x)^4, {x, 0, 40}], x] (* or *) LinearRecurrence[{4, -6, 4, -1}, {0, 8, 40, 120}, 50] (* Vincenzo Librandi, Jun 26 2012 *)

Prog

(Magma)  I:=[0, 8, 40, 120]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..45]]; // Vincenzo Librandi, Jun 26 2012
(PARI) a(n)=4*n^3+4*n \\ Charles R Greathouse IV, Oct 16 2015

Crossrefs

Essentially row or column of A102728 and A105374.

Cf. A006003.

Sequence in context: A279273 A143943 A135796 * A162668 A227733 A191903

Adjacent sequences: A105371 A105372 A105373 * A105375 A105376 A105377

Keyword

easy,nonn

Author

Henry Bottomley, Apr 02 2005

Status

approved