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A108946 a(2n) = A001570(n), a(2n+1) = -A007654(n+1). 4
1, -3, 13, -48, 181, -675, 2521, -9408, 35113, -131043, 489061, -1825200, 6811741, -25421763, 94875313, -354079488, 1321442641, -4931691075, 18405321661, -68689595568, 256353060613, -956722646883, 3570537526921, -13325427460800, 49731172316281 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
In reference to program code, 2baseiseq[X](n) = ((-1)^n)*A001353(n) (a(n)^2 + 1 is a perfect square.) 1tesseq[X](n) = (-1^(n+1))*A097948(n).
Floretion Algebra Multiplication Program, FAMP Code: 1ibaseiseq[X] with X = .5'i + .5i' + 'ii' - .5'jj' + 1.5'kk' - 1 (* Corrected by Creighton Dement, Dec 11 2009 *)
LINKS
FORMULA
G.f.: (x^2+x+1)/((1-x)*(x+1)*(x^2+4*x+1)).
Floor(((2 + sqrt(3))^n + (2 - sqrt(3))^n)/4) produces this sequence with a different offset and without signs. - James R. Buddenhagen, May 20 2010
Define c(n) = a(n) - 4*a(n+1) - a(n+2) and d(n) = -a(n) - 4*a(n+1) - a(n+2); Conjectures: I: c(2n) = 24*A076139(n); (Triangular numbers that are one-third of another triangular number) II: c(2n+1) = -A011943(n+1); (Numbers n such that any group of n consecutive integers has integral standard deviation) III: d(2n) = -2; IV: d(2n+1) = -1
MAPLE
seriestolist(series((x^2+x+1)/((1-x)*(x+1)*(x^2+4*x+1)), x=0, 25));
MATHEMATICA
LinearRecurrence[{-4, 0, 4, 1}, {1, -3, 13, -48}, 30] (* Harvey P. Dale, Jun 15 2018 *)
PROG
(Magma) /* By definition: */
m:=15; R<x>:=PowerSeriesRing(Integers(), m);
A001570:=Coefficients(R!((1-x)/(1-14*x+x^2)));
A007654:=Coefficients(R!(-3*x^2*(1+x)/(-1+x)/(1-14*x+x^2)));
&cat[[A001570[i], -A007654[i]]: i in [1..m-2]]; // Bruno Berselli, Feb 05 2013
CROSSREFS
Cf. A007654, A001570, A076139. See also A117808, A122571 (same except for signs).
Sequence in context: A193164 A122424 A027326 * A352028 A048482 A094978
KEYWORD
sign,easy
AUTHOR
Creighton Dement, Jul 21 2005
STATUS
approved

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Last modified April 23 08:33 EDT 2024. Contains 371905 sequences. (Running on oeis4.)