A175805 -id:A175805 - cp's OEIS frontend

A024495 a(n) = C(n,2) + C(n,5) + ... + C(n, 3*floor(n/3)+2).

0, 0, 1, 3, 6, 11, 21, 42, 85, 171, 342, 683, 1365, 2730, 5461, 10923, 21846, 43691, 87381, 174762, 349525, 699051, 1398102, 2796203, 5592405, 11184810, 22369621, 44739243, 89478486, 178956971, 357913941, 715827882, 1431655765, 2863311531, 5726623062

Offset: 0

Clark Kimberling

Trisections give A082365, A132804, A132805. - Paul Curtz, Nov 18 2007
If the offset is changed to 1, this is the maximal number of closed regions bounded by straight lines after n straight line cuts in a plane: a(n) = a(n-1) + n - 3, a(1)=0; a(2)=0; a(3)=1; and so on. - Srikanth K S, Jan 23 2008
M^n * [1,0,0] = [A024493(n), a(n), A024494(n)]; where M = a 3x3 matrix [1,1,0; 0,1,1; 1,0,1]. Sum of terms = 2^n. Example: M^5 * [1,0,0] = [11, 11, 10], sum = 2^5 = 32. - Gary W. Adamson, Mar 13 2009
For n>=1, a(n-1) is the number of generalized compositions of n when there are i^2/2 - 3*i/2 + 1 different types of i, (i=1,2,...). - Milan Janjic, Sep 24 2010
Let M be any endomorphism on any vector space, such that M^3 = 1 (identity). Then (1+M)^n = A024493(n) + A024494(n)*M + a(n)*M^2. - Stanislav Sykora, Jun 10 2012
{A024493, A131708, A024495} is the difference analog of the hyperbolic functions {h_1(x), h_2(x), h_3(x)} of order 3. For the definitions of {h_i(x)} and the difference analog {H_i(n)} see [Erdelyi] and the Shevelev link respectively. - Vladimir Shevelev, Aug 01 2017
This is the p-INVERT of (1,1,1,1,1,...) for p(S) = 1 - S^3; see A291000.  - Clark Kimberling, Aug 24 2017

A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, 2nd. ed., Problem 38, p. 70.

Seiichi Manyama, Table of n, a(n) for n = 0..3000
Paul Barry, A note on Krawtchouk Polynomials and Riordan Arrays, JIS 11 (2008) 08.2.2, example 9.
Antoine-Augustin Cournot, Solution d'un problème d'analyse combinatoire, Bulletin des Sciences Mathématiques, Physiques et Chimiques, item 34, volume 11, 1829, pages 93-97. Also at Google Books. Page 97 case p=3 formula y^(2) = a(n).
Christian Ramus, Solution générale d'un problème d'analyse combinatoire, Journal für die Reine und Angewandte Mathematik (Crelle's journal), volume 11, 1834, pages 353-355. Page 353 case p=3 formula y^(2) = a(n).
Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
Eric Weisstein's World of Mathematics, Plane division by lines
Index entries for linear recurrences with constant coefficients, signature (3,-3,2).

Cf. A007283, A010892, A011655, A024493, A024494, A024495, A057079, A082365.
Cf. A083322, A139469, A111927, A113405, A131577, A131708, A132804, A132805.
Cf. A175805, A291000.
Sequences of the form 1/((1-x)^m - x^m): A000079 (m=1,2), this sequence (m=3), A000749 (m=4), A049016 (m=5), A192080 (m=6), A049017 (m=7), A290995 (m=8), A306939 (m=9).

R:=PowerSeriesRing(Integers(), 30); [0,0] cat Coefficients(R!( x^2/((1-x)^3-x^3) )); // G. C. Greubel, Apr 11 2023

a:= proc(n) option remember; `if`(n=0, 0, 2*a(n-1)+
      [-1, 0, 1, 1, 0, -1, -1][1+(n mod 6)])
    end:
seq(a(n), n=0..33); # Paul Weisenhorn, May 17 2020

LinearRecurrence[{3,-3,2},{0,0,1},40] (* Harvey P. Dale, Sep 20 2016 *)

a(n) = sum(k=0,n\3,binomial(n,3*k+2)) /* Michael Somos, Feb 14 2006 */

a(n)=if(n<0, 0, ([1,0,1;1,1,0;0,1,1]^n)[3,1]) /* Michael Somos, Feb 14 2006 */

def A024495(n): return (2^n - chebyshev_U(n, 1/2) - chebyshev_U(n-1, 1/2))/3
[A024495(n) for n in range(41)] # G. C. Greubel, Apr 11 2023

a(n) = ( 2^n + 2*cos((n-4)*Pi/3) )/3 = (2^n - A057079(n))/3.
a(n) = 2*a(n-1) + A010892(n-2) = a(n-1) + A024494(n-1). With initial zero, binomial transform of A011655 which is effectively A010892 unsigned. - Henry Bottomley, Jun 04 2001
a(2) = 1, a(3) = 3, a(n+2) = a(n+1) - a(n) + 2^n. - Benoit Cloitre, Sep 04 2002
a(n) = Sum_{k=0..n} 2^k*2*sin(Pi*(n-k)/3 + Pi/3)/sqrt(3) (offset 0). - Paul Barry, May 18 2004
G.f.: x^2/((1-x)^3 - x^3) =  x^2 / ( (1-2*x)*(1-x+x^2) ).
a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3). - Paul Curtz, Nov 18 2007
a(n) + A024493(n-1) = A131577(n). - Paul Curtz, Jan 24 2008
From Paul Curtz, May 29 2011: (Start)
a(n) + a(n+3) = 3*2^n = A007283(n).
a(n+6) - a(n) = 21*2^n = A175805(n).
a(n) + a(n+9) = 171*2^n.
a(n+12) - a(n) = 1365*2^n. (End)
a(n) = A113405(n) + A113405(n+1). - Paul Curtz, Jun 05 2011
Start with x(0)=1, y(0)=0, z(0)=0 and set x(n+1) = x(n) + z(n), y(n+1) = y(n) + x(n), z(n+1) = z(n) + y(n). Then a(n) = z(n). - Stanislav Sykora, Jun 10 2012
G.f.: -x^2/( x^3 - 1 + 3*x/Q(0) ) where Q(k) = 1 + k*(x+1) + 3*x - x*(k+1)*(k+4)/Q(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Mar 15 2013
a(n) = 1/18*(-4*(-1)^floor((n - 1)/3) - 6*(-1)^floor(n/3) - 3*(-1)^floor((n + 1)/3) + (-1)^(1 + floor((n + 2)/3)) + 3*2^(n + 1)). - John M. Campbell, Dec 23 2016
a(n) = (1/63)*(-40 + 21*2^n - 42*floor(n/6) + 32*floor((n+3)/6) + 16*floor((n+ 4)/6) - 24*floor((n+5)/6) - 22*floor((n+7)/6) + 21*floor((n+8)/6) + 10*floor((n+9)/6) + 5*floor((n+10)/6) + 3*floor((n+11)/6) + floor((n+ 13)/6)). - John M. Campbell, Dec 24 2016
a(n+m) = a(n)*A024493(m) + A131708(n)*A131708(m) + A024493(n)*a(m). - Vladimir Shevelev, Aug 01 2017
From Kevin Ryde, Sep 24 2020: (Start)
a(n) = (1/3)*2^n - (1/3)*cos((1/3)*Pi*n) - (1/sqrt(3))*sin((1/3)*Pi*n). [Cournot]
a(n) + A111927(n) + A131708(n) = 2^n - 1. [Cournot, page 96 last formula, but misprint should be 2^x - 1 rather than 2^p - 1] (End)
E.g.f.: (exp(2*x) - exp(x/2)*(cos(sqrt(3)*x/2) + sqrt(3)*sin(sqrt(3)*x/2)))/3. - Stefano Spezia, Feb 06 2025

A005009 a(n) = 7*2^n.

Original entry on oeis.org

7, 14, 28, 56, 112, 224, 448, 896, 1792, 3584, 7168, 14336, 28672, 57344, 114688, 229376, 458752, 917504, 1835008, 3670016, 7340032, 14680064, 29360128, 58720256, 117440512, 234881024, 469762048, 939524096, 1879048192, 3758096384

Offset: 0

N. J. A. Sloane

The first differences are the sequence itself. - Alexandre Wajnberg & Eric Angelini, Sep 07 2005

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

Sequences of the form (2*m+1)*2^n: A000079 (m=0), A007283 (m=1), A020714 (m=2), this sequence (m=3), A005010 (m=4), A005015 (m=5), A005029 (m=6), A110286 (m=7), A110287 (m=8), A110288 (m=9), A175805 (m=10), A248646 (m=11), A164161 (m=12), A175806 (m=13), A257548 (m=15).
Row sums of (6, 1)-Pascal triangle A093563 and of (1, 6)-Pascal triangle A096956, n>=1.
Cf. A000120, A014311, A118416, A173787, A212191.

a005009 = (* 7) . (2 ^) -- Reinhard Zumkeller, May 03 2012

[7*2^n:n in [0..50]]; // Vincenzo Librandi, Sep 20 2011

7*2^Range[0,50] (* Vladimir Joseph Stephan Orlovsky, Mar 14 2011 *)
NestList[2#&,7,30] (* Harvey P. Dale, Aug 10 2024 *)

a(n)=7<Charles R Greathouse IV, Dec 22 2011

[7*2^n for n in range(51)] # G. C. Greubel, Jan 05 2023

G.f.: 7/(1-2*x).
a(n) = A118416(n+1,4) for n > 3. - Reinhard Zumkeller, Apr 27 2006
a(n) = 2*a(n-1), for n > 0, with a(0)=7 . - Philippe Deléham, Nov 23 2008
a(n) = 7 * A000079(n). - Omar E. Pol, Dec 16 2008
a(n) = A173787(n+3,n). - Reinhard Zumkeller, Feb 28 2010
Intersection of A014311 and A212191: all terms and their squares are the sum of exactly three distinct powers of 2, A000120(a(n)) = A000120(a(n)^2) = 3. - Reinhard Zumkeller, May 03 2012
G.f.: 2/x/G(0) - 1/x + 9, where G(k)= 1 + 1/(1 - x*(7*k+2)/(x*(7*k+9) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 03 2013
E.g.f.: 7*exp(2*x). - Stefano Spezia, May 15 2021

A091629 Product of digits associated with A091628(n). Essentially the same as A007283.

Original entry on oeis.org

6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072, 6144, 12288, 24576, 49152, 98304, 196608, 393216, 786432, 1572864, 3145728, 6291456, 12582912, 25165824, 50331648, 100663296, 201326592, 402653184, 805306368, 1610612736, 3221225472

Offset: 1

Enoch Haga, Jan 24 2004

Sequence arising in Farideh Firoozbakht's solution to Prime Puzzle 251 - 23 is the only pointer prime (A089823) not containing digit "1".

The monotonic increasing value of successive product of digits strongly suggests that in successive n the digit 1 must be present.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Tanya Khovanova, Recursive Sequences
Carlos Rivera, Puzzle 251, Pointer primes, The Prime Puzzles and Problems Connection.
Index entries for linear recurrences with constant coefficients, signature (2).

Sequences of the form (2*m+1)*2^n: A000079 (m=0), A007283 (m=1), A020714 (m=2), A005009 (m=3), A005010 (m=4), A005015 (m=5), A005029 (m=6), A110286 (m=7), A110287 (m=8), A110288 (m=9), A175805 (m=10), A248646 (m=11), A164161 (m=12), A175806 (m=13), A257548 (m=15).
Cf. A089823, A091628, A091630, A091631, A091632.
Similar to A003945, A042950, A058764, A087009, A081808.

[3*2^n : n in [1..40]]; // Wesley Ivan Hurt, Jul 17 2020

3*2^Range[1, 60] (* Vladimir Joseph Stephan Orlovsky, Jun 09 2011 *)

[3*2^n for n in range(1,51)] # G. C. Greubel, Jan 05 2023

a(n) = 3 * 2^n = product of digits of A091628(n).
From Philippe Deléham, Nov 23 2008: (Start)
a(n) = 6*2^(n-1).
a(n) = 2*a(n-1), with a(1) = 6.
G.f.: 6*x/(1-2*x). (End)
E.g.f.: 3*(exp(2*x) - 1). - G. C. Greubel, Jan 05 2023

Edited and extended by Ray Chandler, Feb 07 2004

A110287 a(n) = 17*2^n.

Original entry on oeis.org

17, 34, 68, 136, 272, 544, 1088, 2176, 4352, 8704, 17408, 34816, 69632, 139264, 278528, 557056, 1114112, 2228224, 4456448, 8912896, 17825792, 35651584, 71303168, 142606336, 285212672, 570425344, 1140850688, 2281701376, 4563402752, 9126805504, 18253611008

Offset: 0

Alexandre Wajnberg, Sep 07 2005

The first differences are the sequence itself. Doubling the terms gives the same sequence (beginning one step further).

17 times powers of 2. - Omar E. Pol, Dec 17 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..235
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2).

Sequences of the form (2*m+1)*2^n: A000079 (m=0), A003945 (m=1), A020714 (m=2), A005009 (m=3), A005010 (m=4), A005015 (m=5), A005029 (m=6), A110286 (m=7), this sequence (m=8), A110288 (m=9), A175805 (m=10), A248646 (m=11), A164161 (m=12), A175806 (m=13), A257548 (m=15).

Cf. A007283.

[17*2^n: n in [0..40]]; // Vincenzo Librandi, Apr 28 2011

17*2^Range[0, 60] (* Vladimir Joseph Stephan Orlovsky, Jun 09 2011 *)
NestList[2#&,17,30] (* Harvey P. Dale, Jul 21 2014 *)

a(n)=17*2^n \\ Charles R Greathouse IV, Oct 07 2015

[17*2^n for n in range(51)] # G. C. Greubel, Jan 05 2023

G.f.: 17/(1-2*x). - Philippe Deléham, Nov 23 2008
a(n) = 17*A000079(n). - Omar E. Pol, Dec 17 2008
a(n) = 2*a(n-1) (with a(0)=17). - Vincenzo Librandi, Dec 26 2010
a(n) = A173786(n+4, n) for n>3. - Reinhard Zumkeller, Feb 28 2010
E.g.f.: 17*exp(2*x). - G. C. Greubel, Jan 05 2023

Edited by Omar E. Pol, Dec 16 2008

A110288 a(n) = 19*2^n.

Original entry on oeis.org

19, 38, 76, 152, 304, 608, 1216, 2432, 4864, 9728, 19456, 38912, 77824, 155648, 311296, 622592, 1245184, 2490368, 4980736, 9961472, 19922944, 39845888, 79691776, 159383552, 318767104, 637534208, 1275068416, 2550136832, 5100273664, 10200547328, 20401094656

Offset: 0

Alexandre Wajnberg, Sep 07 2005

The first differences are the sequence itself. Doubling the terms gives the same sequence (beginning one step further).

19 times powers of 2. - Omar E. Pol, Dec 17 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..235
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2).

Sequences of the form (2*m+1)*2^n: A000079 (m=0), A007283 (m=1), A020714 (m=2), A005009 (m=3), A005010 (m=4), A005015 (m=5), A005029 (m=6), A110286 (m=7), A110287 (m=8), this sequence (m=9), A175805 (m=10), A248646 (m=11), A164161 (m=12), A175806 (m=13), A257548 (m=15).

[19*2^n: n in [0..40]]; // Vincenzo Librandi, Apr 28 2011

19*2^Range[0, 60] (* Vladimir Joseph Stephan Orlovsky, Jun 09 2011 *)
NestList[2#&,19,30] (* Harvey P. Dale, May 11 2018 *)

[19*2^n for n in range(41)] # G. C. Greubel, Jan 04 2023

G.f.: 19/(1-2*x). - Philippe Deléham, Nov 23 2008
a(n) = A000079(n)*19. - Omar E. Pol, Dec 17 2008
E.g.f.: 19*exp(2*x). - G. C. Greubel, Jan 04 2023

Edited by Omar E. Pol, Dec 16 2008

A081374 Size of "uniform" Hamming covers of distance 1, that is, Hamming covers in which all vectors of equal weight are treated the same, included or excluded from the cover together.

Original entry on oeis.org

1, 2, 2, 5, 10, 22, 43, 86, 170, 341, 682, 1366, 2731, 5462, 10922, 21845, 43690, 87382, 174763, 349526, 699050, 1398101, 2796202, 5592406, 11184811, 22369622, 44739242, 89478485, 178956970, 357913942, 715827883, 1431655766, 2863311530, 5726623061

Offset: 1

David Applegate, Aug 22 2003

nonn

Motivation: consideration of the "hats" problem (which boils down to normal hamming covering codes) in the case when the people are indistinguishable or unlabeled.
From Paul Curtz, May 26 2011: (Start)
If we add a(0)=1 in front and build the table of a(n) and iterated differences in further rows we get:
1,    1,  2,  2,  5, 10,
0,    1,  0,  3,  5, 12,
1,   -1,  3,  2,  7,  9,
-2,   4, -1,  5,  2, 13,
6,   -5,  6, -3, 11,  6
-11, 11, -9, 14, -5, 21.
The first column is the inverse binomial transform, which is 1,0 followed by (-1)^n*A083322(n-1), n>=2.
The main diagonal in the table above is A001045, the adjacent upper diagonals are A078008, A048573 and A062092. (End)

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

Cf. A083322.

I:=[1,2,2,5]; [n le 4 select I[n] else 2*Self(n-1)-Self(n-3)+2*Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jul 08 2016

hatwork := proc(n,i,covered) local val, val2; options remember;
# computes the minimum cover of the i-bit through n-bit words.
# if covered is true the i-bit words are already covered (by the (i-1)-bit words)
if (i>n or (i = n and covered)) then 0; elif (i = n and not covered) then 1; else
# one choice is to include the i-bit words in the cover
val := hatwork(n, i+1, true) + binomial(n,i);
# the other choice is not to include the i-bit words in the cover
if (covered) then val2 := hatwork (n, i+1, false); if (val2 < val) then val := val2; fi; else
# if the i-bit words were not covered by (i-1), they must be covered by the (i+1)-bit words
if (i <= n) then val2 := hatwork (n, i+2, true) + binomial(n,i+1); if (val2 < val) then val := val2; fi; fi; fi; val; fi; end proc;
A081374 := proc (n) hatwork(n, 0, false); end proc;

LinearRecurrence[{2,0,-1,2},{1,2,2,5},40] (* Harvey P. Dale, Feb 11 2015 *)

If (n mod 6 = 5) then sum(binomial(n, 3*i+1), i=0..n/3); elif (n mod 6 = 2) then sum(binomial(n, 3*i), i=0..n/3)+1; else sum(binomial(n, 3*i), i=0..n/3); fi;
G.f.: x*(2*x^3-2*x^2+1)/( (1-2*x)*(1+x)*(1-x+x^2) ).
a(n)=2*a(n-1)-a(n-3)+2*a(n-4).
From Paul Curtz, May 26 2011: (Start)
a(n+1) - 2*a(n) has period length 6: repeat 0, -2, 1, 0, 2, -1 (see A080425).
a(n) - A083322(n-1) = A010892(n-1) has period length 6.
a(n) + a(n+3) = 3*2^n = A007283(n).
a(n+6)-a(n) = 21*2^n = A175805(n).
a(n) - A131708(n) = -A131531(n).  (End)

A083322 a(n) = 2^n - A081374(n).

Original entry on oeis.org

1, 2, 6, 11, 22, 42, 85, 170, 342, 683, 1366, 2730, 5461, 10922, 21846, 43691, 87382, 174762, 349525, 699050, 1398102, 2796203, 5592406, 11184810, 22369621, 44739242, 89478486, 178956971, 357913942, 715827882, 1431655765, 2863311530, 5726623062

Offset: 1

David Applegate, Aug 22 2003

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

Cf. A081374.

Trisections: A082365, A007613, A132804.

I:=[1,2,6,11]; [n le 4 select I[n] else 2*Self(n-1)-Self(n-3)+2*Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jul 08 2016

CoefficientList[Series[(1 + 2 x^2) / ((1 - 2 x) (1 + x) (1 - x + x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Jul 08 2016 *)
LinearRecurrence[{2,0,-1,2},{1,2,6,11},40] (* Harvey P. Dale, Jan 30 2024 *)

G.f.: x*(1+2*x^2) / ( (1-2*x)*(1+x)*(1-x+x^2) ). - R. J. Mathar, May 27 2011
From Paul Curtz, May 27 2011: (Start)
a(n) = 2*a(n-1) - a(n-3) + 2*a(n-4).
a(n)+a(n+3) = 3*2^(n+1) = A007283(n+1).
a(n+6)-a(n) = 21*2^(n+1) = A175805(n+1).
(End)

A171389 a(n) = 21*2^n - 1.

Original entry on oeis.org

20, 41, 83, 167, 335, 671, 1343, 2687, 5375, 10751, 21503, 43007, 86015, 172031, 344063, 688127, 1376255, 2752511, 5505023, 11010047, 22020095, 44040191, 88080383, 176160767, 352321535, 704643071, 1409286143, 2818572287

Offset: 0

Vincenzo Librandi, Dec 07 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Gennady Eremin, Partitioning the set of natural numbers into Mersenne trees and into arithmetic progressions; Natural Matrix and Linnik's constant, arXiv:2405.16143 [math.CO], 2024. See pp. 3, 14.
Index entries for linear recurrences with constant coefficients, signature (3,-2).

Cf. A086224, A175805.

for j = 0 to 30 : print str$((21*2^j)-1)+", "; : next j [Jeremy Gardiner, Oct 23 2011]

I:=[20, 41]; [n le 2 select I[n] else 3*Self(n-1)-2*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 06 2012

A171389:=n->21*2^n-1; seq(A171389(n), n=0..40); # Wesley Ivan Hurt, Jun 11 2014

CoefficientList[Series[(20-19*x)/((1-x)*(1-2*x)),{x,0,40}],x] (* Vincenzo Librandi, Jul 06 2012 *)

a(n+1) = 2*a(n) + 1.
G.f.: (20-19*x)/((1-x)*(1-2*x)). - Vincenzo Librandi, Jul 06 2012
a(n) = 3*a(n-1) - 2*a(n-2). - Vincenzo Librandi, Jul 06 2012
a(n) + a(n-1)^2 = (a(n-1) + 1)^2. - Vincenzo Librandi, Jun 11 2014

Edited by Jon E. Schoenfield, Jun 23 2010

Offset changed to 0 and first formula corrected by Jeremy Gardiner, Oct 23 2011

A171160 a(n) = a(n-1) + 2*a(n-2) with a(0)=3, a(1)=4.

Original entry on oeis.org

3, 4, 10, 18, 38, 74, 150, 298, 598, 1194, 2390, 4778, 9558, 19114, 38230, 76458, 152918, 305834, 611670, 1223338, 2446678, 4893354, 9786710, 19573418, 39146838, 78293674, 156587350, 313174698, 626349398, 1252698794, 2505397590, 5010795178, 10021590358

Offset: 0

Paul Curtz, Dec 04 2009

J. Mulder, Table of n, a(n) for n = 0..2999
Index entries for linear recurrences with constant coefficients, signature (1,2).

Cf. A001045, A078008, A175805.
Cf. A062510, A083582, (-1)^n*A140966.
Cf. A092297, A154879.
Cf. A062092, A083595.

f[n_]:=2/(n+1);x=5;Table[x=f[x];Numerator[x],{n,0,5!}] (* Vladimir Joseph Stephan Orlovsky, Mar 12 2010 *)
LinearRecurrence[{1,2},{3,4},40] (* Harvey P. Dale, Sep 04 2013 *)

Vec(-(x+3)/((x+1)*(2*x-1)) + O(x^100)) \\ Colin Barker, Feb 10 2015

a(n) = (1/3)*(2*(-1)^n + 7*2^n), with n>=0. - Paolo P. Lava, Dec 14 2009
G.f.: -(x+3) / ((x+1)*(2*x-1)). - Colin Barker, Feb 10 2015
From Paul Curtz, Jun 03 2022: (Start)
a(n) = A078008(n) + A078008(n+1) + A078008(n+2).
a(n) = 2^(n+1) + A078008(n).
a(n) = A001045(n+3) - A001045(n).
(a(n) + a(n+1) = a(n+2) - a(n) = A005009(n).)
a(n) + a(n+3) = A175805(n).
a(n) = A062510(n) + A083582(n-1) with A083582(-1) = 3.
a(n) = A092297(n) + A154879(n). (End)
a(n) = 2*A062092(n-1), for n>0; 2*a(n) = A083595(n+1). - Paul Curtz, Jun 08 2022

Edited by N. J. A. Sloane, Dec 05 2009
More terms from J. Mulder (jasper.mulder(AT)planet.nl), Jan 28 2010
More terms from Max Alekseyev, Apr 24 2010

A242563 a(n) = 2a(n-1) - a(n-3) + 2a(n-4), a(0)=a(1)=0, a(2)=2, a(3)=3.

Original entry on oeis.org

0, 0, 2, 3, 6, 10, 21, 42, 86, 171, 342, 682, 1365, 2730, 5462, 10923, 21846, 43690, 87381, 174762, 349526, 699051, 1398102, 2796202, 5592405, 11184810, 22369622, 44739243, 89478486, 178956970, 357913941, 715827882, 1431655766, 2863311531, 5726623062, 11453246122

Offset: 0

Paul Curtz, May 17 2014

Generally, a(n) is an autosequence if its inverse binomial transform is (-1)^n*a(n). It is of the first kind if the main diagonal is 0's and the first two upper diagonals (just above the main one) are the same. It is of the second kind if the main diagonal is equal to the first upper diagonal multiplied by 2. If the first upper diagonal is an autosequence, the sequence is a super autosequence. Example: A113405. The first upper diagonal is A001045(n). Another super autosequence: 0, 0, 0 followed by A059633(n). The first upper diagonal is A000045(n).
Difference table of a(n):
   0,  0,  2, 3, 6, 10, 21, 42, ...
   0,  2,  1, 3, 4, 11, 21, 44, ...
   2, -1,  2, 1, 7, 10, 23, 41, ...
  -3,  3, -1, 6, 3, 13, 18, 45, ... .
This is an autosequence of the second kind. The main diagonal is 2*A001045(n) = A078008(n). More precisely it is a super autosequence, companion of A113405(n).
a(n+1) mod 10 = period 12: repeat 0, 2, 3, 6, 0, 1, 2, 6, 1, 2, 2, 5.
It is shifted A081374(n+1) mod 10 =
    period 12: repeat 1, 2, 2, 5, 0, 2, 3, 6, 0, 1, 2, 6.
a(n) mod 9 = period 18:
repeat 0, 0, 2, 3, 6, 1, 3, 6, 5, 0, 0, 7, 6, 3, 8, 6, 3, 4 = c(n).
c(n) + c(n+9) = 0, 0, 9, 9, 9, 9, 9, 9, 9.

			G.f. = 2*x^2 + 3*x^3 + 6*x^4 + 10*x^5 + 21*x^6 + 42*x^7 + 86*x^8 + ...

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

Cf. A000032, 1/(n+1), A164555/A027642 (all autosequences of 2nd kind). A007283, A175805.

a[n_] := (m = Mod[n, 6]; 1/3*(2^n + (-1)^n + 1/120*(m-6)*(m+1)*(m^3-29*m+40))); Table[a[n], {n, 0, 35}] (* Jean-François Alcover, May 19 2014, a non-recursive formula, after Mathematica's RSolve *)
LinearRecurrence[{2, 0, -1, 2}, {0, 0, 2, 3},50] (* G. C. Greubel, Feb 21 2017 *)

concat([0,0], Vec(x^2*(x-2)/((x+1)*(2*x-1)*(x^2-x+1)) + O(x^100))) \\ Colin Barker, May 18 2014

a(n+3) = 3*2^n - a(n), a(0)=a(1)=0, a(2)=2.
a(n) = 2*A113405(n+1) - A113405(n).
a(n+1) = 2*a(n) + period 6: repeat 0, 2, -1, 0, -2, 1. a(0)=0.
a(n) = 2^n - A081374(n+1).
a(n+3) = a(n+1) + A130755(n).
G.f.: x^2*(x-2) / ((x+1)*(2*x-1)*(x^2-x+1)). - Colin Barker, May 18 2014
a(n) = A024495(n) + A131531(n).
a(n+6) = a(n) + 21*2^n, a(0)=a(1)=0, a(2)=2, a(3)=3, a(4)=6, a(5)=10.
a(n) = A001045(n) - A092220(n).
a(n+12) = a(n) + 1365*2^n. First 12 values in the Data. (A024495(n+12) = A024495(n) + 1365*2^n).
a(3n)   = A132805(n) = 3*A015565(n).
a(3n+1) = A132804(n) = 6*A015565(n).
a(3n+2) = A132397(n) = 2*A082311(n).
a(n) = 1/3*((-1)^n - 2*cos((n*Pi)/3) + 2^n). - Alexander R. Povolotsky,  Jun 02 2014

More terms from Colin Barker, May 18 2014

cp's OEIS Frontend

A024495 a(n) = C(n,2) + C(n,5) + ... + C(n, 3*floor(n/3)+2).

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A005009 a(n) = 7*2^n.

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A091629 Product of digits associated with A091628(n). Essentially the same as A007283.

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A110287 a(n) = 17*2^n.

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A110288 a(n) = 19*2^n.

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A081374 Size of "uniform" Hamming covers of distance 1, that is, Hamming covers in which all vectors of equal weight are treated the same, included or excluded from the cover together.

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A242563 a(n) = 2a(n-1) - a(n-3) + 2a(n-4), a(0)=a(1)=0, a(2)=2, a(3)=3.