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%I #27 Mar 15 2024 15:20:15
%S 1,-3,13,-48,181,-675,2521,-9408,35113,-131043,489061,-1825200,
%T 6811741,-25421763,94875313,-354079488,1321442641,-4931691075,
%U 18405321661,-68689595568,256353060613,-956722646883,3570537526921,-13325427460800,49731172316281
%N a(2n) = A001570(n), a(2n+1) = -A007654(n+1).
%C 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).
%C 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 *)
%H Robert Munafo, <a href="http://www.mrob.com/pub/math/seq-floretion.html">Sequences Related to Floretions</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (-4,0,4,1).
%F G.f.: (x^2+x+1)/((1-x)*(x+1)*(x^2+4*x+1)).
%F 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
%F 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
%p seriestolist(series((x^2+x+1)/((1-x)*(x+1)*(x^2+4*x+1)), x=0,25));
%t LinearRecurrence[{-4,0,4,1},{1,-3,13,-48},30] (* _Harvey P. Dale_, Jun 15 2018 *)
%o (Magma) /* By definition: */
%o m:=15; R<x>:=PowerSeriesRing(Integers(), m);
%o A001570:=Coefficients(R!((1-x)/(1-14*x+x^2)));
%o A007654:=Coefficients(R!(-3*x^2*(1+x)/(-1+x)/(1-14*x+x^2)));
%o &cat[[A001570[i],-A007654[i]]: i in [1..m-2]]; // _Bruno Berselli_, Feb 05 2013
%Y Cf. A007654, A001570, A076139. See also A117808, A122571 (same except for signs).
%Y Cf. A001353, A097948.
%K sign,easy
%O 0,2
%A _Creighton Dement_, Jul 21 2005
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