A160444 -id:A160444 - cp's OEIS frontend

A160572 Elements of A160444, pairs of consecutive entries swapped.

1, 0, 1, 1, 4, 2, 10, 6, 28, 16, 76, 44, 208, 120, 568, 328, 1552, 896, 4240, 2448, 11584, 6688, 31648, 18272, 86464, 49920, 236224, 136384, 645376, 372608, 1763200, 1017984, 4817152, 2781184, 13160704, 7598336, 35955712, 20759040, 98232832

Offset: 1

Willibald Limbrunner (w.limbrunner(AT)gmx.de), May 20 2009

The case k=3 of a family of sequences defined by a(1)=1, a(2)=0, a(2n+1)=a(2n-1)+k*a(2n), a(2n+2)=a(2n)+a(2n-1), each congruent to one of the sequences mentioned in A160444 by pairwise interchanges. The case k=2 is covered by swapping pairs in A002965.
Each of the two subsequences b(n) obtained by bisection has a limiting ratio b(n+1)/b(n)=1+sqrt(k) by Binet's Formula. In a logarithmic plot of the sequence a(n) one therefore sees a staircase, the two edges at each step alternately marked by one of the two subsequences.
Matrix M = [[1 3] [1 1]] is iterated with starting vector [1 0]^T. Since M has eigenvectors [+-sqrt(3) 1]^T with eigenvalues 1 +- sqrt(3), we have lim xn/yn = 1+sqrt(3) for all nonzero integer starting vectors. - Hagen von Eitzen, May 22 2009

			k=2: 1,0,1,1,3,2,7,5,17,12,41,29,99,70,239,169,577,408,1393,985
k=3: 1,0,1,1,4,2,10,6,28,16,76,44,208,120,568,328,1552... (here)
k=4: 1,0,1,1,5,2,13,7,41,20,121,61,365,182,1093,547,3281,..
k=5: 1,0,1,1,6,2,16,8,56,24,176,80,576,256,1856,832,6016,2688,..
k=6: 1,0,1,1,7,2,19,9,73,28,241,101,847,342,2899,1189,..
k=7: 1,0,1,1,8,2,22,10,92,32,316,124,1184,440,4264,1624,..
k=8: 1,0,1,1,9,2,25,11,113,36,401,149,1593,550,5993,2143,..
k=9: 1,0,1,1,10,2,28,12,136,40,496,176,2080,672,8128,2752,..
k=10: 1,0,1,1,11,2,31,13,161,44,601,205,2651,806,10711,3457,..

W. Limbrunner, Das Quadrat, ein Wunder der Geometrie (in German)
Index entries for linear recurrences with constant coefficients, signature (0, 2, 0, 2).

Cf. A160444, A002605 (bisection), A026150 (bisection).

a(2*n+1)=A160444(2*n+2). a(2*n+2)=A160444(2*n+1).
G.f.: -x*(1-x^2+x^3)/(-1+2*x^2+(k-1)*x^4). a(n)=2*a(n-2)+(k-1)*a(n-4) at k=3. - R. J. Mathar, May 22 2009
a(1)=1, a(2)=0, and for n>=1: a(2*n+1) = a(2*n-1)+3*a(2*n), a(2*n+2) = a(2*n+1)+a(2*n). Or: Let c1 = 1+sqrt(3), c2 = 1-sqrt(3). Then a(2*n+1) = (c1^n + c2^n)/2, a(2*n+2) = (c1^n - c2^n)/(2*sqrt(3)) for n >= 0. - Hagen von Eitzen, May 22 2009

Edited by R. J. Mathar, May 22 2009

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A160572 Elements of A160444, pairs of consecutive entries swapped.

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