A083322 -id:A083322 - 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

A048573 a(n) = a(n-1) + 2*a(n-2), a(0)=2, a(1)=3.

Original entry on oeis.org

2, 3, 7, 13, 27, 53, 107, 213, 427, 853, 1707, 3413, 6827, 13653, 27307, 54613, 109227, 218453, 436907, 873813, 1747627, 3495253, 6990507, 13981013, 27962027, 55924053, 111848107, 223696213, 447392427, 894784853, 1789569707, 3579139413, 7158278827, 14316557653

Offset: 0

Michael Somos, Jun 17 1999

Number of positive integers requiring exactly n signed bits in the modified non-adjacent form representation. - Ralf Stephan, Aug 02 2003
The n-th entry (n>1) of the sequence is equal to the 1,1-entry of the n-th power of the unnormalized 4 X 4 Haar matrix: [1 1 1 0 / 1 1 -1 0 / 1 1 0 1 / 1 1 0 -1]. - Simone Severini, Oct 27 2004
Pisano period lengths:  1, 1, 6, 2, 2, 6, 6, 2, 18, 2, 10, 6, 12, 6, 6, 2, 8, 18, 18, 2, ... - R. J. Mathar, Aug 10 2012
For n >= 1, a(n) is the number of ways to tile a strip of length n+2 with blue squares and blue and red dominos, with the restriction that the first two tiles must be the same color. - Guanji Chen and Greg Dresden, Jul 15 2024

			G.f. = 2 + 3*x + 7*x^2 + 13*x^3 + 27*x^4 + 53*x^5 + 107*x^6 + 213*x^7 + 427*x^8 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wieb Bosma, Signed bits and fast exponentiation, Journal de théorie des nombres de Bordeaux, Vol. 13, No. 1 (2001), pp. 27-41.
Karl Dilcher and Larry Ericksen, Continued fractions and Stern polynomials, Ramanujan Journal 45.3 (2018): 659-681. See Table 2.
Karl Dilcher and Hayley Tomkins, Square classes and divisibility properties of Stern polynomials, Integers, Vol. 18 (2018), Article #A29.
Petro Kosobutskyy, The Collatz problem as a reverse n->0 problem on a graph tree formed from theta*2^n Jacobsthal-type numbers, arXiv:2306.14635 [math.GM], 2023.
Petro Kosobutskyy and Dariia Rebot, Collatz conjecture 3n+/-1 as a Newton binomial problem, Comp. Des. Sys. Theor. Prac., Lviv Nat'l Polytech. Univ. (Ukraine 2023) Vol. 5, No. 1, 137-145. See p. 140.
Saad Mneimneh, Simple Variations on the Tower of Hanoi to Guide the Study of Recurrences and Proofs by Induction, Department of Computer Science, Hunter College, CUNY, 2019.
Sam Northshield, Stern's diatomic sequence 0, 1, 1, 2, 1, 3, 2, 3, 1, 4, ..., Amer. Math. Monthly, Vol. 117, No. 7 (2010), pp. 581-598.
Index entries for linear recurrences with constant coefficients, signature (1,2).

Cf. A084214 (first differences).
Cf. A001045, A014551, A024493, A062510.
Cf. A007283, A007679, A078008, A081254.
Cf. A083322, A083944, A083581, A084170.
Cf. A102713, A127978, A130755, A139763.
Cf. A140360, A140966, A166249, A168642.
Cf. A280560, A290604, A340627.
Cf. A112387, A135318.

[(5*2^n+(-1)^n)/3: n in [0..35]]; // Vincenzo Librandi, Jul 05 2011

LinearRecurrence[{1,2},{2,3},40] (* Harvey P. Dale, Dec 11 2017 *)

{a(n) = if( n<0, 0, (5*2^n + (-1)^n) / 3)};

{a(n) = if (n<0 ,0, if( n<2, n+2, a(n-1) + 2*a(n-2)))};

[(5*2^n+(-1)^n)/3 for n in range(35)] # G. C. Greubel, Apr 10 2019

G.f.: (2 + x) / (1 - x - 2*x^2).
a(n) = (5*2^n + (-1)^n) / 3.
a(n) = 2^(n+1) - A001045(n).
a(n) = A084170(n)+1 = abs(A083581(n)-3) = A081254(n+1) - A081254(n) = A084214(n+2)/2.
a(n) = 2*A001045(n+1) + A001045(n) (note that 2 is the limit of A001045(n+1)/A001045(n)). - Paul Barry, Sep 14 2009
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-3, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=-charpoly(A,-1). - Milan Janjic, Jan 27 2010
Equivalently, with different offset, a(n) = b(n+1) with b(0)=1 and b(n) = Sum_{i=0..n-1} (-1)^i (1 + (-1)^i b(i)). - Olivier Gérard, Jul 30 2012
a(n) = A000975(n-2)*10 + 5 + 2*(-1)^(n-2), a(0)=2, a(1)=3. - Yuchun Ji, Mar 18 2019
a(n+1) = Sum_{i=0..n} a(i) + 1 + (1-(-1)^n)/2, a(0)=2. - Yuchun Ji, Apr 10 2019
a(n) = 2^n + J(n+1) = J(n+2) + J(n+1) - J(n), where J is A001045. - Yuchun Ji, Apr 10 2019
a(n) = A001045(n+2) + A078008(n) = A062510(n+1) - A078008(n+1) = (A001045(n+2) + A062510(n+1))/2 = A014551(n) + 2*A001045(n). - Paul Curtz, Jul 14 2021
From Thomas Scheuerle, Jul 14 2021: (Start)
a(n) = A083322(n) + A024493(n).
a(n) = A127978(n) - A102713(n).
a(n) = A130755(n) - A166249(n).
a(n) = A007679(n) + A139763(n).
a(n) = A168642(n) XOR A007283(n).
a(n) = A290604(n) + A083944(n). (End)
From Paul Curtz, Jul 21 2021: (Start)
a(n) = 5*A001045(n) - A280560(n+1) = abs(A140360(n+1)) - A280560(n+1).
a(n) = 2^n + A001045(n+1) = A001045(n+3) - A000079(n).
a(n) = A001045(n+4) - A340627(n). (End)
a(n) = A001045(n+5) - A005010(n).
a(n+1) + a(n) = a(n+2) - a(n) = 5*2^n. - Michael Somos, Feb 22 2023
a(n) = A135318(2*n) + A135318(2*n+1) = A112387(2*n) + A112387(2*n+1). - Paul Curtz, Jun 26 2024
E.g.f.: (cosh(x) + 5*cosh(2*x) - sinh(x) + 5*sinh(2*x))/3. - Stefano Spezia, May 18 2025

Formula of Milan Janjic moved here from wrong sequence by Paul D. Hanna, May 29 2010

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)

cp's OEIS Frontend

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

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A048573 a(n) = a(n-1) + 2*a(n-2), a(0)=2, a(1)=3.

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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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Magma

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