A276918 a(2n) = A060867(n+1), a(2n+1) = A092440(n+1).

1, 5, 9, 25, 49, 113, 225, 481, 961, 1985, 3969, 8065, 16129, 32513, 65025, 130561, 261121, 523265, 1046529, 2095105, 4190209, 8384513, 16769025, 33546241, 67092481, 134201345, 268402689, 536838145, 1073676289, 2147418113, 4294836225, 8589803521, 17179607041

Offset

0,2

Comments

In binary there is a pattern in how the zeros and ones appear:

a(0) = 01

a(1) = 101

a(2) = 1001

a(3) = 11001

a(4) = 110001

a(5) = 1110001

a(6) = 11100001

a(7) = 111100001

a(8) = 1111000001

a(9) = 11111000001

a(10) = 111110000001

a(11) = 1111110000001

a(12) = 11111100000001

a(13) = 111111100000001

a(14) = 1111111000000001

a(15) = 11111111000000001

Graphically, each term can be obtained by successively and alternately forming squares and centered squares as shown in the illustration.

Links

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..1000

Daniel Poveda Parrilla, Illustration of initial terms

Index entries for linear recurrences with constant coefficients, signature (3,0,-6,4).

Formula

a(n) = 1 + 2^(n+2) - 2^(1 + n/2) + (-1)^(n+1)*2^(1 + n/2) - 2^((n+1)/2) + (-1)^(n+2)*2^((n+1)/2).

a(n) = 3*a(n-1) - 6*a(n-3) + 4*a(n-4) for n>3.

G.f.: (-1-2*x+6*x^2-4*x^3)/(-1+3*x-6*x^3+4*x^4).

Mathematica

Table[1+2^(n+2)-2^(1+n/2)+(-1)^(n+1) 2^(1+n/2)-2^((n+1)/2)+(-1)^(n+2) 2^((n+1)/2), {n, 0, 28}] (*or*)
CoefficientList[Series[(-1 - 2 x + 6 x^2 - 4 x^3)/(-1 + 3 x - 6 x^3 + 4 x^4), {x, 0, 28}], x] (*or*)
LinearRecurrence[{3, 0, -6, 4}, {1, 5, 9, 25}, 29]

Prog

(PARI) Vec((-1-2*x+6*x^2-4*x^3) / (-1+3*x-6*x^3+4*x^4) + O(x^29))

Crossrefs

Cf. A000225, A060867, A092440, A224195.

Sequence in context: A025624 A147496 A147383 * A127976 A147286 A137199

Adjacent sequences: A276915 A276916 A276917 * A276919 A276920 A276921

Keyword

nonn

Author

Daniel Poveda Parrilla, Jan 26 2017

Status

approved