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%I #37 Mar 15 2024 15:20:05
%S 0,1,6,28,136,656,3168,15296,73856,356608,1721856,8313856,40142848,
%T 193826816,935878656,4518821888,21818802176,105350496256,508677193728,
%U 2456110759936,11859151814656,57261050298368,276480808452096
%N a(n+1) = 4*(a(n)+a(n-1)) for n>1, a(1)=1, a(2)=6.
%C Let (a_n) be the sequence and (a_(n+1)) the sequence beginning at 1. Let B and iB be the binomial and inverse binomial transforms, respectively. Then B((a_n)) = A001108(n) (a(n)-th triangular number is a square); B((a_(n+1))) = A002315(n) (NSW Numbers); iB((a_(n+1))) = A096980(n). Note: a 2nd sequence generated by the same floretion is A057087 (Scaled Chebyshev U-polynomials evaluated at i. Generalized Fibonacci sequence.). As is often the case with two sequences corresponding to a single floretion, both satisfy the same recurrence relation.
%C Floretion Algebra Multiplication Program, FAMP Code: (a_n) = 2ibasekseq[A*B] (with initial term zero), (a_(n+1)) = 1tesseq[A*B], A = + .5'i - .5'j + .5'k + .5i' - .5j' + .5k' - .5'ij' - .5'ik' - .5'ji' - .5'ki'; B = - .5'i + .5'j + .5'k - .5i' + .5j' + .5k' - .5'ik' - .5'jk' - .5'ki' - .5'kj'
%H Vincenzo Librandi, <a href="/A108051/b108051.txt">Table of n, a(n) for n = 0..1000</a>
%H Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, <a href="http://www.emis.de/journals/JIS/VOL18/Szczyrba/sz3.pdf">Analytic Representations of the n-anacci Constants and Generalizations Thereof</a>, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
%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_02">Index entries for linear recurrences with constant coefficients</a>, signature (4,4).
%F a(n+1) = -(1/2)*(2-2*2^(1/2))^n*(-1+2^(1/2))-(1/2)*(2+2*2^(1/2))^n(-1-2^(1/2)); G.f.: x*(1+2*x)/(1-4*x-4*x^2).
%F a(n) = sum{k=0..n, (-1)^k*C(n-1, k)*(Pell(2n-2k)-Pell(2n-2k-1))}, n>0, where Pell(n) = A000129(n). - _Paul Barry_, Jun 07 2005
%F a(n+1) = ((3+sqrt18)(2+sqrt8)^n+(3-sqrt18)(2-sqrt8)^n)/6. - Al Hakanson (hawkuu(AT)gmail.com), Aug 15 2009, index corrected Jul 11 2012
%F a(n) = 2^(n-1) * A001333(n), n>0. - _Ralf Stephan_, Dec 02 2010
%F a(n) = A057087(n-1) + 2*A057087(n-2). - _R. J. Mathar_, Jul 11 2012
%t CoefficientList[Series[x*(1+2*x)/(1-4*x-4*x^2),{x,0,40}],x] (* _Vincenzo Librandi_, Jun 26 2012 *)
%o (Magma) I:=[0, 1, 6]; [n le 3 select I[n] else 4*Self(n-1)+4*Self(n-2): n in [1..30]]; // _Vincenzo Librandi_, Jun 26 2012
%Y Cf. A057087, A001108, A002315, A096980.
%K easy,nonn
%O 0,3
%A _Creighton Dement_, Jun 01 2005
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