A108051 a(n+1) = 4*(a(n)+a(n-1)) for n>1, a(1)=1, a(2)=6.

0, 1, 6, 28, 136, 656, 3168, 15296, 73856, 356608, 1721856, 8313856, 40142848, 193826816, 935878656, 4518821888, 21818802176, 105350496256, 508677193728, 2456110759936, 11859151814656, 57261050298368, 276480808452096

Offset: 0

Author

Creighton Dement, Jun 01 2005

Keywords

• easy
• nonn

Comments

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.

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'

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
• Robert Munafo, Sequences Related to Floretions
• Index entries for linear recurrences with constant coefficients, signature (4,4).

Crossrefs

Cf. A057087, A001108, A002315, A096980.

Programs

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

Mathematica

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

Formula

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).

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

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

a(n) = 2^(n-1) * A001333(n), n>0. - Ralf Stephan, Dec 02 2010

a(n) = A057087(n-1) + 2*A057087(n-2). - R. J. Mathar, Jul 11 2012