A051032 -id:A051032 - cp's OEIS frontend

A163403 a(n) = 2*a(n-2) for n > 2; a(1) = 1, a(2) = 2.

1, 2, 2, 4, 4, 8, 8, 16, 16, 32, 32, 64, 64, 128, 128, 256, 256, 512, 512, 1024, 1024, 2048, 2048, 4096, 4096, 8192, 8192, 16384, 16384, 32768, 32768, 65536, 65536, 131072, 131072, 262144, 262144, 524288, 524288, 1048576, 1048576, 2097152, 2097152

Offset: 1

Klaus Brockhaus, Jul 26 2009

a(n+1) is the number of palindromic words of length n using a two-letter alphabet. - Michael Somos, Mar 20 2011

			x + 2*x^2 + 2*x^3 + 4*x^4 + 4*x^5 + 8*x^6 + 8*x^7 + 16*x^8 + 16*x^9 + 32*x^10 + ...

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,2).

Equals A016116 without initial 1. Unsigned version of A152166.
Partial sums are in A136252.
Binomial transform is A078057, second binomial transform is A007070, third binomial transform is A102285, fourth binomial transform is A163350, fifth binomial transform is A163346.
Cf. A000079 (powers of 2), A009116, A009545, A051032.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A029744 = {s(n), n>=1}, the numbers 2^k and 3*2^k, as the parent: A029744 (s(n)); A052955 (s(n)-1), A027383 (s(n)-2), A354788 (s(n)-3), A347789 (s(n)-4), A209721 (s(n)+1), A209722 (s(n)+2), A343177 (s(n)+3), A209723 (s(n)+4); A060482, A136252 (minor differences from A354788 at the start); A354785 (3*s(n)), A354789 (3*s(n)-7). The first differences of A029744 are 1,1,1,2,2,4,4,8,8,... which essentially matches eight sequences: A016116, A060546, A117575, A131572, A152166, A158780, A163403, A320770. The bisections of A029744 are A000079 and A007283. - N. J. A. Sloane, Jul 14 2022

[ n le 2 select n else 2*Self(n-2): n in [1..43] ];

LinearRecurrence[{0, 2}, {1, 2}, 50] (* Paolo Xausa, Feb 02 2024 *)

{a(n) = if( n<1, 0, 2^(n\2))} /* Michael Somos, Mar 20 2011 */

def A163403():
    x, y = 1, 1
    while True:
        yield x
        x, y = x + y, x - y
a = A163403(); [next(a) for i in range(40)]  # Peter Luschny, Jul 11 2013

a(n) = 2^((1/4)*(2*n - 1 + (-1)^n)).
G.f.: x*(1 + 2*x)/(1 - 2*x^2).
a(n) = A051032(n) - 1.
G.f.: x / (1 - 2*x / (1 + x / (1 + x))) = x * (1 + 2*x / (1 - x / (1 - x / (1 + 2*x)))). - Michael Somos, Jan 03 2013
From R. J. Mathar, Aug 06 2009: (Start)
a(n) = A131572(n).
a(n) = A060546(n-1), n > 1. (End)
a(n+3) = a(n+2)*a(n+1)/a(n). - Reinhard Zumkeller, Mar 04 2011
a(n) = |A009116(n-1)| + |A009545(n-1)|. - Bruno Berselli, May 30 2011
E.g.f.: cosh(sqrt(2)*x) + sinh(sqrt(2)*x)/sqrt(2) - 1. - Stefano Spezia, Feb 05 2023

A183978 1/4 the number of (n+1) X 2 binary arrays with all 2 X 2 subblock sums the same.

Original entry on oeis.org

4, 6, 9, 15, 25, 45, 81, 153, 289, 561, 1089, 2145, 4225, 8385, 16641, 33153, 66049, 131841, 263169, 525825, 1050625, 2100225, 4198401, 8394753, 16785409, 33566721, 67125249, 134242305, 268468225, 536920065, 1073807361, 2147581953, 4295098369

Offset: 1

R. H. Hardin, Jan 08 2011

Column 1 of A183986

Based on the conjectured recursion formula, it is also the number of notches in a sheet of paper when you fold it n times and cut off the four corners (see A274230). - Philippe Gibone, Jul 06 2016

			Some solutions for 5X2
..0..1....1..0....1..0....1..1....0..1....1..0....1..0....0..1....0..1....0..1
..0..0....1..0....1..0....1..0....1..0....1..0....1..0....0..1....1..0....0..1
..1..0....1..0....0..1....1..1....0..1....0..1....0..1....1..0....0..1....1..0
..0..0....1..0....1..0....0..1....1..0....1..0....0..1....1..0....0..1....0..1
..1..0....1..0....1..0....1..1....1..0....0..1....0..1....1..0....1..0....0..1

R. H. Hardin, Table of n, a(n) for n = 1..46
Robert Israel, Proof of empirical formulas for A183978
Index entries for linear recurrences with constant coefficients, signature (3, 0, -6, 4).

Cf. A274230.

Conjectured to be the main diagonal of A274636.

seq((1+2^floor((n-1)/2))*(1+2^ceil((n-1)/2)), n=1..20); # Robert Israel, May 21 2019

Empirical: a(n) = 3*a(n-1) - 6*a(n-3) + 4*a(n-4)
Based on the conjectured recursion formula, we may prove (by a tedious induction) that a(n) = (2^ceiling(n/2) + 1) * (2^floor(n/2) + 1) = A051032(n) * A051032(n-1) for n >= 1. - Philippe Gibone, Jul 06 2016, corrected by Robert Israel, May 21 2019
Empirical: G.f.: -x*(4-6*x-9*x^2+12*x^3) / ( (x-1)*(2*x-1)*(2*x^2-1) ). - R. J. Mathar, Jul 15 2016
Empirical formulas verified (see link): Robert Israel, May 21 2019.
2*a(n) = 2+2^n+A029744(n+3). - R. J. Mathar, Jul 19 2024

cp's OEIS Frontend

A163403 a(n) = 2*a(n-2) for n > 2; a(1) = 1, a(2) = 2.

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