A112032 -id:A112032 - cp's OEIS frontend

A084964 Follow n+2 by n. Also solution of a(n+2)=a(n)+1, a(0)=2, a(1)=0.

2, 0, 3, 1, 4, 2, 5, 3, 6, 4, 7, 5, 8, 6, 9, 7, 10, 8, 11, 9, 12, 10, 13, 11, 14, 12, 15, 13, 16, 14, 17, 15, 18, 16, 19, 17, 20, 18, 21, 19, 22, 20, 23, 21, 24, 22, 25, 23, 26, 24, 27, 25, 28, 26, 29, 27, 30, 28, 31, 29, 32, 30, 33, 31, 34, 32, 35, 33, 36, 34, 37, 35, 38, 36, 39

Offset: 0

Michael Somos, Jun 15 2003

Stefano Spezia, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).
Index entries for two-way infinite sequences

Cf. A030451, A097065, A152832.

Cf. A217764(1,n) = a(n+2).

import Data.List (transpose)
a084964 n = a084964_list !! n
a084964_list = concat $ transpose [[2..], [0..]]
-- Reinhard Zumkeller, Apr 06 2015

&cat[ [n+2, n]: n in [0..37] ]; // Klaus Brockhaus, Nov 23 2009

A084964:=n->floor(n/2)+1+(-1)^n; seq(A084964(k), k=0..100); # Wesley Ivan Hurt, Nov 08 2013

lst={}; a=1; Do[a=n-a; AppendTo[lst, a], {n, 0, 100}]; lst (* Vladimir Joseph Stephan Orlovsky, Dec 14 2008 *)
Table[{n,n-2},{n,2,40}]//Flatten (* or *) LinearRecurrence[{1,1,-1},{2,0,3},80] (* Harvey P. Dale, Sep 12 2021 *)

PARI
```
a(n)=n\2-2*(n%2)+2
```

G.f.: (2-2x+x^2)/((1-x)(1-x^2)).
a(2n+1)=n. a(2n)=n+2. a(n+2)=a(n)+1. a(n)=-a(-3-n).
a(n) = floor(n/2) + 1 + (-1)^n. - Reinhard Zumkeller, Aug 27 2005
A112032(n)=2^a(n); A112033(n)=3*2^a(n); a(n)=A109613(n+2)-A052938(n). - Reinhard Zumkeller, Aug 27 2005
a(n) = n + 1 - a(n-1) (with a(0)=2). - Vincenzo Librandi, Aug 08 2010
a(n) = floor(n/2)*3 - floor((n-1)/2)*2. - Ross La Haye, Mar 27 2013
a(n) = 3*n - 3 - 5*floor((n-1)/2). - Wesley Ivan Hurt, Nov 08 2013
a(n) = (3 + 5*(-1)^n + 2*n)/4. - Wolfgang Hintze, Dec 13 2014
E.g.f.: ((4 + x)*cosh(x) - (1 - x)*sinh(x))/2. - Stefano Spezia, Jul 01 2023

First part of definition adjusted to match offset by Klaus Brockhaus, Nov 23 2009

A112030 a(n) = (2 + (-1)^n) * (-1)^floor(n/2).

Original entry on oeis.org

3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1, -3, -1, 3, 1

Offset: 0

Reinhard Zumkeller, Aug 27 2005

The fractions A112031(n)/A112032(n) give the partial sums of a(n)/floor((n+4)/2).

Sum of the two Cartesian coordinates from the image of the point (2,1) after n 90-degree counterclockwise rotations about the origin. - Wesley Ivan Hurt, Jul 06 2013

Index entries for linear recurrences with constant coefficients, signature (0,-1).

Cf. A010684, A016116, A112031, A112032, A112033.

A112030 := proc(n)
        (2 + (-1)^n) * (-1)^floor(n/2) ;
end proc: # R. J. Mathar, Jul 09 2013

LinearRecurrence[{0, -1}, {3, 1}, 100] (* Jean-François Alcover, Nov 24 2020 *)

a(n)=[3,1,-3,-1][n%4+1] \\ Charles R Greathouse IV, Aug 21 2011

def A112030(n): return (3, 1, -3, -1)[n&3] # Chai Wah Wu, Jan 31 2023

a(n) = A010684(n+1) * (-1)^floor(n/2).

O.g.f.: (3+x)/(1+x^2). - R. J. Mathar, Jan 09 2008

A164298 a(n) = ((1+4sqrt(2))(2+sqrt(2))^n + (1-4sqrt(2))(2-sqrt(2))^n)/2.

Original entry on oeis.org

1, 10, 38, 132, 452, 1544, 5272, 18000, 61456, 209824, 716384, 2445888, 8350784, 28511360, 97343872, 332352768, 1134723328, 3874187776, 13227304448, 45160842240, 154188760064, 526433355776, 1797355902976, 6136556900352, 20951515795456, 71532949381120

Offset: 0

Al Hakanson (hawkuu(AT)gmail.com), Aug 12 2009

Binomial transform of A048696. Second binomial transform of A164587. Inverse binomial transform of A164299.

This sequence is part of a class of sequences defined by the recurrence a(n,m) = 2*(m+1)*a(n-1,m) - ((m+1)^2 - 2)*a(n-2,m) with a(0) = 1 and a(1) = m+9. The generating function is Sum_{n>=0} a(n,m)*x^n = (1 - (m-7)*x)/(1 - 2*(m+1)*x + ((m+1)^2 - 2)*x^2) and has a series expansion in terms of Pell-Lucas numbers defined by a(n, m) = (1/2)*Sum_{k=0..n} binomial(n,k)*m^(n-k)*(5*Q(k) + 4*Q(k-1)). - G. C. Greubel, Mar 12 2021

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-2).

Sequences in the class a(n, m): this sequence (m=1), A164299 (m=2), A164300 (m=3), A164301 (m=4), A164598 (m=5), A164599 (m=6), A081185 (m=7), A164600 (m=8).
Cf. A007070, A016921, A048696, A056236, A112032, A164587, A241204.
Cf. A016116(n+1).

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-2); S:=[ ((1+4*r)*(2+r)^n+(1-4*r)*(2-r)^n)/2: n in [0..27] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Aug 17 2009

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1+6*x)/(1-4*x+2*x^2) )); // G. C. Greubel, Dec 14 2018

a:=n->((1+4*sqrt(2))*(2+sqrt(2))^n+(1-4*sqrt(2))*(2-sqrt(2))^n)/2: seq(floor(a(n)),n=0..25); # Muniru A Asiru, Dec 15 2018

LinearRecurrence[{4,-2}, {1,10}, 50] (* or *) CoefficientList[Series[(1 + 6*x)/(1 - 4*x + 2*x^2), {x,0,50}], x] (* G. C. Greubel, Sep 12 2017 *)

my(x='x+O('x^50)); Vec((1+6*x)/(1-4*x+2*x^2)) \\ G. C. Greubel, Sep 12 2017

[( (1+6*x)/(1-4*x+2*x^2) ).series(x,n+1).list()[n] for n in (0..30)] # G. C. Greubel, Dec 14 2018; Mar 12 2021

a(n) = 4*a(n-1) - 2*a(n-2) for n > 1; a(0)=1, a(1)=10.
G.f.: (1+6*x)/(1-4*x+2*x^2).
E.g.f.: (cosh(sqrt(2)*x) + 4*sqrt(2)*sinh(sqrt(2)*x))*exp(2*x). - G. C. Greubel, Sep 12 2017
From G. C. Greubel, Mar 12 2021: (Start)
a(n) = A056236(n) + 8*A007070(n-1).
a(n) = (1/2)*Sum_{k=0..n} binomial(n,k)*(5*Q(k) + 4*Q(k-1)), where Q(n) = Pell-Lucas(n) = A002203(n). (End)

Edited and extended beyond a(5) by Klaus Brockhaus, Aug 17 2009

A164587 a(n) = 2*a(n - 2) for n > 2; a(1) = 1, a(2) = 8.

Original entry on oeis.org

1, 8, 2, 16, 4, 32, 8, 64, 16, 128, 32, 256, 64, 512, 128, 1024, 256, 2048, 512, 4096, 1024, 8192, 2048, 16384, 4096, 32768, 8192, 65536, 16384, 131072, 32768, 262144, 65536, 524288, 131072, 1048576, 262144, 2097152, 524288, 4194304, 1048576

Offset: 1

Klaus Brockhaus, Aug 17 2009

nonn

Interleaving of A000079 and A000079 without initial terms 1, 2, 4.

Binomial transform is A048696. Second binomial transform is A164298.

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

Equals A112032 without initial term 4.

Cf. A000079 (powers of 2), A048696, A164298.

[ n le 2 select 7*n-6 else 2*Self(n-2): n in [1..41] ];

CoefficientList[Series[(1 - x)/(1 - 10*x + 17*x^2), {x,0,50}], x] (* G. C. Greubel, Aug 12 2017 *)

x='x+O('x^50); Vec(x*(1+8*x)/(1-2*x^2)) \\ G. C. Greubel, Aug 12 2017

a(n) = (5 + 3*(-1)^n)*2^((2*n -5 +(-1)^n)/4).
G.f.: x*(1+8*x)/(1-2*x^2).
E.g.f.: 4*cosh(sqrt(2)*x) + (1/sqrt(2))*sinh(sqrt(2)*x) - 4. - G. C. Greubel, Aug 12 2017

A112033 a(n) = 3 * 2^(floor(n/2) + 1 + (-1)^n).

Original entry on oeis.org

12, 3, 24, 6, 48, 12, 96, 24, 192, 48, 384, 96, 768, 192, 1536, 384, 3072, 768, 6144, 1536, 12288, 3072, 24576, 6144, 49152, 12288, 98304, 24576, 196608, 49152, 393216, 98304, 786432, 196608, 1572864, 393216, 3145728, 786432, 6291456, 1572864

Offset: 0

Reinhard Zumkeller, Aug 27 2005

George Pólya and Gábor Szegő, Problems and Theorems in Analysis I (Springer 1924, reprinted 1972), Part One, Chapter 4, Sect. 1, Problem 148.

Reinhard Zumkeller, Table of n, a(n) for n = 0..100
Index entries for linear recurrences with constant coefficients, signature (0,2).

Cf. A084964, A112030, A112031, A112032.

A112033:=n->3*2^(floor(n/2) + 1 + (-1)^n); seq(A112033(k), k=0..50); # Wesley Ivan Hurt, Nov 01 2013

Table[3*2^(Floor[n/2] + 1 + (-1)^n), {n,0,50}] (* Wesley Ivan Hurt, Nov 01 2013 *)

a(n) = 3 * 2^(n\2 + 1 + (-1)^n); \\ Michel Marcus, Nov 02 2013

def A112033(n): return 3*(1<<(n>>1)+(int(not n&1)<<1)) # Chai Wah Wu, Jan 17 2023

a(n) = 1 / abs(A112031(n)/A112032(n) - 2/3). (previous name)
a(n) = 3*2^A084964(n) = 3*A112032(n).
From Ralf Stephan, Jul 16 2013: (Start)
Recurrence: a(n) = 2a(n-2), a(0)=12, a(1)=3.
G.f.: (6*x+24)/(1-2*x^2). (End)
From Amiram Eldar, May 11 2025: (Start)
Sum_{n>=0} 1/a(n) = 5/6.
Sum_{n>=0} (-1)^n/a(n) = -1/2. (End)

A112031 Numerator of 3/4 + 1/4 - 3/8 - 1/8 + 3/16 + 1/16 - 3/32 - 1/32 + 3/64 + ....

Original entry on oeis.org

3, 1, 5, 1, 11, 3, 21, 5, 43, 11, 85, 21, 171, 43, 341, 85, 683, 171, 1365, 341, 2731, 683, 5461, 1365, 10923, 2731, 21845, 5461, 43691, 10923, 87381, 21845, 174763, 43691, 349525, 87381, 699051, 174763, 1398101, 349525, 2796203, 699051, 5592405

Offset: 0

Reinhard Zumkeller, Aug 27 2005

Numerator of partial sums of A112030(n)/A016116(n+4), denominators = A112032;
a(n)/A112032(n) - 2/3 = (-1)^floor(n/2) / A112033(n);
lim_{n->infinity} a(n)/A112032(n) = 2/3.

G. Pólya and G. Szegő, Problems and Theorems in Analysis I (Springer 1924, reprinted 1972), Part One, Chap. 4, Sect. 1, Problem 148.

Vincenzo Librandi, Table of n, a(n) for n = 0..3000
Index entries for linear recurrences with constant coefficients, signature (0,1,0,2).

Cf. A016116, A112030, A112032, A112033, A001045 (bisections).

[(2^(Floor(n/2) + 2 + (-1)^n) + (-1)^Floor(n/2)) / 3: n in [0..50]]; // Vincenzo Librandi, Aug 17 2011

LinearRecurrence[{0,1,0,2},{3,1,5,1},50] (* Harvey P. Dale, Dec 31 2017 *)

m=50; v=concat([3,1,5,1], vector(m-4)); for(n=5, m, v[n]=v[n-2] +2*v[n-4]); v \\ G. C. Greubel, Nov 08 2018

a(n) = (2^(floor(n/2) + 2 + (-1)^n) + (-1)^floor(n/2)) / 3.
From Colin Barker, Apr 05 2013: (Start)
a(n) = a(n-2) + 2*a(n-4);
g.f.: (2*x^2+x+3) / ((1+x^2)*(1-2*x^2)). (End)

a(22) corrected by Vincenzo Librandi, Aug 17 2011

A112034 1 / (A010684(n)/A016116(n+5) - 1/A112033(n)).

Original entry on oeis.org

6, 24, 12, 48, 24, 96, 48, 192, 96, 384, 192, 768, 384, 1536, 768, 3072, 1536, 6144, 3072, 12288, 6144, 24576, 12288, 49152, 24576, 98304, 49152, 196608, 98304, 393216, 19660, 8, 786432, 393216, 1572864, 786432, 3145728, 1572864, 6291456

Offset: 0

Reinhard Zumkeller, Aug 27 2005

nonn

a(n) = 3*2^A052938(n).

Cf. A112030, A112031, A112032.

cp's OEIS Frontend

A084964 Follow n+2 by n. Also solution of a(n+2)=a(n)+1, a(0)=2, a(1)=0.

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A112030 a(n) = (2 + (-1)^n) * (-1)^floor(n/2).

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A164298 a(n) = ((1+4*sqrt(2))*(2+sqrt(2))^n + (1-4*sqrt(2))*(2-sqrt(2))^n)/2.

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A164587 a(n) = 2*a(n - 2) for n > 2; a(1) = 1, a(2) = 8.

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A112033 a(n) = 3 * 2^(floor(n/2) + 1 + (-1)^n).

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A112031 Numerator of 3/4 + 1/4 - 3/8 - 1/8 + 3/16 + 1/16 - 3/32 - 1/32 + 3/64 + ....

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A164298 a(n) = ((1+4sqrt(2))(2+sqrt(2))^n + (1-4sqrt(2))(2-sqrt(2))^n)/2.