A288732 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.

2, 4, 6, 8, 10, 14, 18, 22, 26, 34, 42, 50, 58, 74, 90, 106, 122, 154, 186, 218, 250, 314, 378, 442, 506, 634, 762, 890, 1018, 1274, 1530, 1786, 2042, 2554, 3066, 3578, 4090, 5114, 6138, 7162, 8186, 10234, 12282, 14330, 16378, 20474, 24570, 28666, 32762

Offset: 0

Author

Clark Kimberling, Jun 16 2017

Keywords

• nonn
• easy

Comments

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.

From Michel Dekking, Mar 25 2018: (Start)

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

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

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

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

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

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

(End)

Links

• Clark Kimberling, Table of n, a(n) for n = 0..10000
• Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, 2, -2).

Crossrefs

Cf. A288729.

Programs

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

Maple

f:= gfun:-rectoproc({a(n) = a(n-1) + 2*a(n-4) - 2*a(n-5),
a(0) = 2, a(1) = 4, a(2) = 6, a(3) = 8, a(4) = 10},a(n),remember):
map(f, [$0..50]); # Robert Israel, Mar 25 2018

Mathematica

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

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

Formula

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.

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

Extensions

a(41)-a(49) from Muniru A Asiru, Mar 22 2018