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a(n) = a(n-1) + 2*a(n-4) - 2*a(n-5) for n >= 5, where a(0) = 2, a(1) = 4, a(2) = 6, a(3) = 8, a(4) = 10.
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%I #28 Apr 06 2020 20:24:59

%S 2,4,6,8,10,14,18,22,26,34,42,50,58,74,90,106,122,154,186,218,250,314,

%T 378,442,506,634,762,890,1018,1274,1530,1786,2042,2554,3066,3578,4090,

%U 5114,6138,7162,8186,10234,12282,14330,16378,20474,24570,28666,32762

%N a(n) = a(n-1) + 2*a(n-4) - 2*a(n-5) for n >= 5, where a(0) = 2, a(1) = 4, a(2) = 6, a(3) = 8, a(4) = 10.

%C Conjecture: a(n) is the number of letters (0's and 1's) in the n-th iterate of the mapping 00->1000, 10->01, starting with 00; see A288729.

%C From _Michel Dekking_, Mar 25 2018: (Start)

%C Note that a(n) - a(n-1) = 2*[a(n-4) - a(n-5)] for n>4.

%C It follows that this sequence is a union of four simple sequences:

%C a(4k-4) = 4*2^k - 6 = A131130(k) for k = 1,2,3,...

%C a(4k-3) = 5*2^k - 6 = A020714(k) - 6 for k = 1,2,3...

%C a(4k-2) = 6*2^k - 6 = A007283(k+1) - 6 for k = 1,2,3, ...

%C a(4k-1) = 7*2^k - 6 = A048489(k) for k = 1,2,3...

%C (End)

%H Clark Kimberling, <a href="/A288732/b288732.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1, 0, 0, 2, -2).

%F a(n) = a(n-1) + 2*a(n-4) - 2*a(n-5) for n >= 5, where a(0) = 2, a(1) = 4, a(2) = 6, a(3) = 8, a(4) = 10.

%F G.f.: -((2*(-1 - x - x^2 - x^3 + x^4))/(1 - x - 2*x^4 + 2*x^5)).

%p f:= gfun:-rectoproc({a(n) = a(n-1) + 2*a(n-4) - 2*a(n-5),

%p a(0) = 2, a(1) = 4, a(2) = 6, a(3) = 8, a(4) = 10},a(n),remember):

%p map(f, [$0..50]); # _Robert Israel_, Mar 25 2018

%t LinearRecurrence[{1, 0, 0, 2, -2}, {2, 4, 8, 8, 10}, 40]

%o (GAP) a:=[2,4,6,8,10];; for n in [6..45] do a[n]:=a[n-1]+2*a[n-4]-2*a[n-5]; od; a; # _Muniru A Asiru_, Mar 22 2018

%o (PARI) x='x+O('x^99); Vec(2*(1+x+x^2+x^3-x^4)/(1-x-2*x^4+2*x^5)) \\ _Altug Alkan_, Mar 22 2018

%Y Cf. A288729.

%K nonn,easy

%O 0,1

%A _Clark Kimberling_, Jun 16 2017

%E a(41)-a(49) from _Muniru A Asiru_, Mar 22 2018