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%I #57 Sep 28 2023 04:15:43
%S 1,1,1,2,3,4,5,8,11,16,21,32,43,64,85,128,171,256,341,512,683,1024,
%T 1365,2048,2731,4096,5461,8192,10923,16384,21845,32768,43691,65536,
%U 87381,131072,174763,262144,349525,524288,699051,1048576,1398101,2097152,2796203
%N The Kentucky-2 sequence: a(n) = a(n-2) + 2*a(n-4), with a[0..3] = [1, 1, 1, 2].
%C Shifted Jacobsthal recurrence.
%C From _L. Edson Jeffery_, Apr 21 2011: (Start)
%C Let U be the unit-primitive matrix (see [Jeffery])
%C U=U_(6,2)=
%C (0 0 1)
%C (0 2 0)
%C (2 0 1),
%C let i in {0,1}, m>=0 an integer and n=2*m+i. Then a(n)=a(2*m+i)=Sum_{j=0..2} (U^m)_(i,j). (End)
%C a(n) is also the pebbling number of the cycle graph C_{n+1} for n > 1. - _Eric W. Weisstein_, Jan 07 2021
%C From _Greg Dresden_ and Ziyi Xie, Aug 25 2023: (Start)
%C a(n) is the number of ways to tile a zig-zag strip of n cells using squares (of 1 cell) and triangles (of 3 cells). Here is the zig-zag strip corresponding to n=11, with 11 cells:
%C ___ ___
%C ___| |___| |___
%C | |___| |___| |___
%C |___| |___| |___| |
%C | |___| |___| |___|
%C |___| |___| |___|,
%C and here are the two types of triangles (where one is just a reflection of the other):
%C ___ ___
%C | |___ ___| |
%C | | | |
%C | ___| and |___ |
%C |___| |___|.
%C As an example, here is one of the a(11) = 32 ways to tile the zig-zag strip of 11 cells:
%C ___ ___
%C ___| |___| |___
%C | |___| | |___
%C | |___ | |
%C | ___| |___| ___|
%C |___| |___| |___|. (End)
%H Vincenzo Librandi, <a href="/A135318/b135318.txt">Table of n, a(n) for n = 0..5000</a>
%H Minerva Catral et al., <a href="https://arxiv.org/abs/1409.0488">Generalizing Zeckendorf's Theorem: The Kentucky Sequence</a>, arXiv:1409.0488 [math.NT], 2014. See 1.3 p. 2, same sequence without the first 2 terms.
%H L. E. Jeffery, <a href="/wiki/User:L._Edson_Jeffery/Unit-Primitive_Matrices">Unit-primitive matrices</a>.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CycleGraph.html">Cycle Graph</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PebblingNumber.html">Pebbling Number</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,1,0,2).
%F From _R. J. Mathar_, Feb 19 2008: (Start)
%F O.g.f.: [1/(1+x^2)+(-2-3*x)/(2*x^2-1)]/3.
%F a(2n) = A001045(n+1).
%F a(2n+1) = A000079(n). (End)
%F From _L. Edson Jeffery_, Apr 21 2011: (Start)
%F G.f.: (1+x+x^3)/((1+x^2)*(1-2*x^2)).
%F a(n) = (((-i)^(n+1)-i^(n+1))*2*i*sqrt(2)+3*(1+(-1)^(n+1))*2^((n+2)/2)+(1-(-1)^(n+1))*2^((n+5)/2))/(12*sqrt(2)), where i=sqrt(-1). (End)
%F a(n) = (2^floor(n/2)*(5-(-1)^n)+(-1)^floor(n/2)*(1+(-1)^n))/6 = (A016116(n)*A010711(n)+2*A056594(n))/6. - _Bruno Berselli_, Apr 21 2011
%F a(2n) = 2*a(2n-1) - a(2n-2); a(2n+1) = a(2n) + a(2n-2). - _Richard R. Forberg_, Aug 19 2013
%F a(n) = A112387(n + (-1)^n). - _Alois P. Heinz_, Sep 28 2023
%e Let i=0 and m=3. Then U^3 = (2,0,3;0,8,0;6,0,5), and the first-row sum (corresponding to i=0) is 2 + 0 + 3 = 5. Hence a(n) = a(2*m+i) = a(2*3+0) = a(6) = 2 + 3 = 5.
%p a:= n-> (<<0|1>, <2|1>>^(iquo(n, 2, 'm')). <<1, 1+m>>)[1,1]:
%p seq(a(n), n=0..50); # _Alois P. Heinz_, May 30 2022
%t LinearRecurrence[{0,1,0,2},{1,1,1,2},40] (* _Harvey P. Dale_, Oct 14 2015 *)
%o (Magma) [(2^Floor(n/2)*(5-(-1)^n)+(-1)^Floor(n/2)*(1+(-1)^n))/6: n in [0..50]]; // _Vincenzo Librandi_, Aug 10 2011
%Y Cf. A000079, A001045, A112387.
%K nonn,easy
%O 0,4
%A _Paul Curtz_, Feb 16 2008
%E More terms from _R. J. Mathar_, Feb 19 2008
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