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A100215 Expansion of (4-7*x+2*x^2)/((1-2*x)*(1-2*x+2*x^2)). 3
4, 9, 14, 18, 24, 44, 104, 248, 544, 1104, 2144, 4128, 8064, 16064, 32384, 65408, 131584, 263424, 525824, 1049088, 2095104, 4189184, 8382464, 16775168, 33562624, 67129344, 134242304, 268443648, 536838144 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

COMMENTS

a(n) = (-1)^n*A009116(n+3) + A100216(n) + A038503(n+1), where A009116, A100216 and A038503 can be generated by the operators jes, les and tes of the Floretion algebra, which is a product factor space Q x Q /{(1,1), (-1,-1)}.

Binomial transform of the sequence 4,5,0,-1 (repeated with period length 4). [From R. J. Mathar, Apr 18 2009]

LINKS

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

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

C. K. Dement, Floretion. [From R. J. Mathar, Apr 18 2009]

FORMULA

a(n) = 4a(n-1) - 6a(n-2) + 4a(n-3).

a(n) = vesseq(.5 'j + .5 'k + .5 j' + .5 k' + 1 'ii' + 1 e), where ves sums over all floretion basis vector coefficients.

a(n) = 2^(n+1)+2*A099087(n)+A099087(n-1). [From R. J. Mathar, Apr 18 2009]

EXAMPLE

a(2) = 14 because (.5 'j + .5 'k + .5 j' + .5 k' + 1 'ii' + 1 e)^3 = 1'j + 1'k + 1j' + 1k' + 3'ii' + 2'jj' + 2'kk' + 1'jk' + 1'kj' + 1e

and the sum of these coefficients is: 1+1+1+1+3+2+2+1+1+1 = 14 (see comment).

MATHEMATICA

LinearRecurrence[{4, -6, 4}, {4, 9, 14}, 40] (* Vincenzo Librandi, Jun 25 2012

PROG

(MAGMA) I:=[4, 9, 14]; [n le 3 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3): n in [1..35]]; // Vincenzo Librandi, Jun 25 2012

CROSSREFS

Cf. A100213 - A100216, A009116, A038503.

Sequence in context: A228177 A172329 A182778 * A100213 A043365 A023738

Adjacent sequences:  A100212 A100213 A100214 * A100216 A100217 A100218

KEYWORD

nonn,easy

AUTHOR

Creighton Dement, Nov 11 2004

EXTENSIONS

Replaced the definition by the more precise g.f. - R. J. Mathar, Nov 17 2010

STATUS

approved

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Last modified March 29 13:14 EDT 2017. Contains 284270 sequences.