A166920 -id:A166920 - cp's OEIS frontend

A000975 a(2n) = 2a(2n-1), a(2n+1) = 2a(2n)+1 (also a(n) is the n-th number without consecutive equal binary digits).

0, 1, 2, 5, 10, 21, 42, 85, 170, 341, 682, 1365, 2730, 5461, 10922, 21845, 43690, 87381, 174762, 349525, 699050, 1398101, 2796202, 5592405, 11184810, 22369621, 44739242, 89478485, 178956970, 357913941, 715827882, 1431655765, 2863311530, 5726623061, 11453246122

Offset: 0

Mira Bernstein, N. J. A. Sloane, Robert G. Wilson v, Sep 13 1996

Might be called the "Lichtenberg sequence" after Georg Christoph Lichtenberg, who discussed it in 1769 in connection with the Chinese Rings puzzle (baguenaudier). - Andreas M. Hinz, Feb 15 2017
Number of steps to change from a binary string of n 0's to n 1's using a Gray code. - Jon Stadler (jstadler(AT)coastal.edu)
Popular puzzles such as Spin-Out and The Brain Puzzler are based on the Gray binary system and require a(n) steps to complete for some number n.
Conjecture: {a(n)} also gives all j for which A048702(j) = A000217(j); i.e., if we take the a(n)-th triangular number (a(n)^2 + a(n))/2 and multiply it by 3, we get a(n)-th even-length binary palindrome A048701(a(n)) concatenated from a(n) and its reverse. E.g., a(4) = 10, which is 1010 in binary; the tenth triangular number is 55, and 55*3 = 165 = 10100101 in binary. - Antti Karttunen, circa 1999. (This has been now proved by Paul K. Stockmeyer in his arXiv:1608.08245 paper.) - Antti Karttunen, Aug 31 2016
Number of ways to tie a tie of n or fewer half turns, excluding mirror images. Also number of walks of length n or less on a triangular lattice with the following restrictions; given l, r and c as the lattice axes. 1. All steps are taken in the positive axis direction. 2. No two consecutive steps are taken on the same axis. 3. All walks begin with l. 4. All walks end with rlc or lrc. - Bill Blewett, Dec 21 2000
a(n) is the minimal number of vertices to be selected in a vertex-cover of the balanced tree B_n. - Sen-peng Eu, Jun 15 2002
A087117(a(n)) = A038374(a(n)) = 1 for n > 1; see also A090050. - Reinhard Zumkeller, Nov 20 2003
Intersection of A003754 and A003714; complement of A107907. - Reinhard Zumkeller, May 28 2005
Equivalently, numbers m whose binary representation is effectively, for some number k, both the lazy Fibonacci and the Zeckendorf representation of k (in which case k = A022290(m)). - Peter Munn, Oct 06 2022
a(n+1) gives row sums of Riordan array (1/(1-x), x(1+2x)). - Paul Barry, Jul 18 2005
Total number of initial 01's in all binary words of length n+1. Example: a(3) = 5 because the binary words of length 4 that start with 01 are (01)00, (01)(01), (01)10 and (01)11 and the total number of initial 01's is 5 (shown between parentheses). a(n) = Sum_{k >= 0} k*A119440(n+1, k). - Emeric Deutsch, May 19 2006
In Norway we call the 10-ring puzzle "strikketoy" or "knitwear" (see link). It takes 682 moves to free the two parts. - Hans Isdahl, Jan 07 2008
Equals A002450 and A020988 interlaced. - Zak Seidov, Feb 10 2008
For n > 1, let B_n = the complete binary tree with vertex set V where |V| = 2^n - 1. If VC is a minimum-size vertex cover of B_n, Sen-Peng Eu points out that a(n) = |VC|. It also follows that if IS = V\VC, a(n+1) = |IS|. - K.V.Iyer, Apr 13 2009
Starting with 1 and convolved with [1, 2, 2, 2, ...] = A000295. - Gary W. Adamson, Jun 02 2009
a(n) written in base 2 is sequence A056830(n). - Jaroslav Krizek, Aug 05 2009
This is the sequence A(0, 1; 1, 2; 1) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010
From Vladimir Shevelev, Jan 30 2012, Feb 13 2012: (Start)
1) Denote by {n, k} the number of permutations of 1, ..., n with the up-down index k (for definition, see comment in A203827). Then max_k{n, k} = {n, a(n)} = A000111(n).
2) a(n) is the minimal number > a(n-1) with the Hamming distance d_H(a(n-1), a(n)) = n. Thus this sequence is the Hamming analog of triangular numbers 0, 1, 3, 6, 10, ... (End)
From Hieronymus Fischer, Nov 22 2012: (Start)
Represented in binary form each term after the second one contains every previous term as a substring.
The terms a(2) = 2 and a(3) = 5 are the only primes. Proof: For even n we get a(n) = 2*(2^(2*n) - 1)/3, which shows that a(n) is even, too, and obviously a(n) > 2 for all even n > 2. For odd n we have a(n) = (2^(n+1) - 1)/3 = (2^((n+1)/2) - 1) * (2^((n+1)/2) + 1)/3. Evidently, at least one of these factors is divisible by 3, both are greater than 6, provided n > 3. Hence it follows that a(n) is composite for all odd n > 3.
Represented as a binary number, a(n+1) has exactly n prime substrings. Proof: Evidently, a(1) = 1_2 has zero and a(2) = 10_2 has 1 prime substring. Let n > 1. Written in binary, a(n+1) is 101010101...01 (if n + 1 is odd) and is 101010101...10 (if n + 1 is even) with n + 1 digits. Only the 2- and 3-digits substrings 10_2 (=2) and 101_2 (=5) are prime substrings. All the other substrings are nonprime since every substring is a previous term and all terms unequal to 2 and 5 are nonprime. For even n + 1, the number of prime substrings equal to 2 = 10_2 is (n+1)/2, and the number of prime substrings equal to 5 = 101_2 is (n-1)/2, makes a sum of n. For odd n + 1 we get n/2 for both, the number of 2's and 5's prime substrings, in any case, the sum is n. (End)
Number of different 3-colorings for the vertices of all triangulated planar polygons on a base with n+2 vertices if the colors of the two base vertices are fixed. - Patrick Labarque, Feb 09 2013
A090079(a(n)) = a(n) and A090079(m) <> a(n) for m < a(n). - Reinhard Zumkeller, Feb 16 2013
a(n) is the number of length n binary words containing at least one 1 and ending in an even number (possibly zero) of 0's. a(3) = 5 because we have: 001, 011, 100, 101, 111. - Geoffrey Critzer, Dec 15 2013
a(n) is the number of permutations of length n+1 having exactly one descent such that the first element of the permutation is an even number. - Ran Pan, Apr 18 2015
a(n) is the sequence of the last row of the Hadamard matrix H(2^n) obtained via Sylvester's construction: H(2) = [1,1;1,-1], H(2^n) = H(2^(n-1))*H(2), where * is the Kronecker product. - William P. Orrick, Jun 28 2015
Conjectured record values of A264784: a(n) = A264784(A155051(n-1)). - Reinhard Zumkeller, Dec 04 2015. (This is proved by Paul K. Stockmeyer in his arXiv:1608.08245 paper.) - Antti Karttunen, Aug 31 2016
Decimal representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 131", based on the 5-celled von Neumann neighborhood. See A279053 for references and links. - Robert Price, Dec 05 2016
For n > 4, a(n-2) is the second-largest number in row n of A127824. - Dmitry Kamenetsky, Feb 11 2017
Conjecture: a(n+1) is the number of compositions of n with two kinds of parts, n and n', where the order of the 1 and 1' does not matter. For n=2, a(3) = 5 compositions, enumerated as follows:  2; 2'; 1,1; 1',1 = 1',1; 1',1'. - Gregory L. Simay, Sep 02 2017
Conjecture proved by recognizing the appropriate g.f. is x/(1 - x)(1 - x)(1 - 2*x^2 - 2x^3 - ...) =  x/(1 - 2*x - x^2 + 2x^3). - Gregory L. Simay, Sep 10 2017
a(n) = 2^(n-1) + 2^(n-3) + 2^(n-5) + ... a(2*k -1) = A002450(k) is the sum of the powers of 4. a(2*k) = 2*A002450(k). - Gregory L. Simay, Sep 27 2017
a(2*n) = n times the string [10] in binary representation, a(2*n+1) = n times the string [10] followed with [1] in binary representation. Example: a(7) = 85 = (1010101) in binary, a(8) = 170 = (10101010) in binary. - Ctibor O. Zizka, Nov 06 2018
Except for 0, these are the positive integers whose binary expansion has cuts-resistance 1. For the operation of shortening all runs by 1, cuts-resistance is the number of applications required to reach an empty word. Cuts-resistance 2 is A329862. - Gus Wiseman, Nov 27 2019
From Markus Sigg, Sep 14 2020: (Start)
Let s(k) be the length of the Collatz orbit of k, e.g. s(1) = 1, s(2) = 2, s(3) = 5. Then s(a(n)) = n+3 for n >= 3. Proof by induction: s(a(3)) = s(5) = 6 = 3+3. For odd n >= 5 we have s(a(n)) = s(4*a(n-2)+1) = s(12*a(n-2)+4)+1 = s(3*a(n-2)+1)+3 = s(a(n-2))+2 = (n-2)+3+2 = n+3, and for even n >= 4 this gives s(a(n)) = s(2*a(n-1)) = s(a(n-1))+1 = (n-1)+3+1 = n+3.
Conjecture: For n >= 3, a(n) is the second largest natural number whose Collatz orbit has length n+3. (End)
From Gary W. Adamson, May 14 2021: (Start)
With offset 1 the sequence equals the numbers of 1's from n = 1 to 3, 3 to 7, 7 to 15, ...; of A035263; as shown below:
..1     3           7                      15...
..1  0  1  1  1  0  1  0  1  0  1  1  1  0  1...
..1.....2...........5......................10...; a(n) = Sum_(k=1..2n-1)A035263(k)
.....1...........2.......................5...; as to zeros.
..1's in the Tower of Hanoi game represent CW moves On disks in the pattern:
..0,  1,  2,  0,  1,  2, ... whereas even numbered disks move in the pattern:
..0,  2,  1,  0,  2,  1, ... (End)
Except for 0, numbers that are repunits in Gray-code representation (A014550). - Amiram Eldar, May 21 2021
From Gus Wiseman, Apr 20 2023: (Start)
Also the number of nonempty subsets of {1..n} with integer median, where the median of a multiset is the middle part in the odd-length case, and the average of the two middle parts in the even-length case. For example, the a(1) = 1 through a(4) = 10 subsets are:
  {1}  {1}  {1}      {1}
       {2}  {2}      {2}
            {3}      {3}
            {1,3}    {4}
            {1,2,3}  {1,3}
                     {2,4}
                     {1,2,3}
                     {1,2,4}
                     {1,3,4}
                     {2,3,4}
The complement is counted by A005578.
For mean instead of median we have A051293, counting empty sets A327475.
For normal multisets we have A056450, strongly normal A361202.
For partitions we have A325347, strict A359907, complement A307683.
(End)
In binary form, because the sequence 10101_2... keeps repeating, one can represent a(n) = floor(0.(10)2 * 2^n). Because 0.(10)_2 = 0.(10)_2 * 4 - 2, then 0.(10)_2 = 2/3. Substituting this value in the first expression, we obtain a(n) = floor(2^(n+1)/3). This also agrees with the formula of Paul Barry, Oct 08 2005. - _Giovanni Ciriani, Mar 31 2025

			a(4)=10 since 0001, 0011, 0010, 0110, 0111, 0101, 0100, 1100, 1101, 1111 are the 10 binary strings switching 0000 to 1111.
a(3) = 1 because "lrc" is the only way to tie a tie with 3 half turns, namely, pass the business end of the tie behind the standing part to the left, bring across the front to the right, then behind to the center. The final motion of tucking the loose end down the front behind the "lr" piece is not considered a "step".
a(4) = 2 because "lrlc" is the only way to tie a tie with 4 half turns. Note that since the number of moves is even, the first step is to go to the left in front of the tie, not behind it. This knot is the standard "four in hand", the most commonly known men's tie knot. By contrast, the second most well known tie knot, the Windsor, is represented by "lcrlcrlc".
a(n) = (2^0 - 1) XOR (2^1 - 1) XOR (2^2 - 1) XOR (2^3 - 1) XOR ... XOR (2^n - 1). - _Paul D. Hanna_, Nov 05 2011
G.f. = x + 2*x^2 + 5*x^3 + 10*x^4 + 21*x^5 + 42*x^6 + 85*x^7 + 170*x^8 + ...
a(9) = 341 = 2^8 + 2^6 + 2^4 + 2^2 + 2^0 = 4^4 + 4^3 + 4^2 + 4^1 + 4^0 = A002450(5). a(10) = 682 = 2*a(9) = 2*A002450(5). - _Gregory L. Simay_, Sep 27 2017

Thomas Fink and Yong Mao, The 85 Ways to Tie a Tie, Broadway Books, New York (1999), p. 138.
Clifford A. Pickover, The Math Book, From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics, Sterling Publ., NY, 2009.

G. C. Greubel, Table of n, a(n) for n = 0..3300 (terms 0..300 from T. D. Noe)
Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
Thomas Baruchel, Properties of the cumulated deficient binary digit sum, arXiv:1908.02250 [math.NT], 2019.
Sergei L. Bezrukov et al., The congestion of n-cube layout on a rectangular grid, Discrete Mathematics 213.1-3 (2000): 13-19. See Theorem 1.
F. Chapoton and S. Giraudo, Enveloping operads and bicoloured noncrossing configurations, arXiv:1310.4521 [math.CO], 2013. Is the sequence in Table 2 this sequence? - _N. J. A. Sloane_, Jan 04 2014. (Yes, it is. See Stockmeyer's arXiv:1608.08245 2016 paper for the proof.)
Ji Young Choi, Ternary Modified Collatz Sequences And Jacobsthal Numbers, Journal of Integer Sequences, Vol. 19 (2016), #16.7.5.
Ji Young Choi, A Generalization of Collatz Functions and Jacobsthal Numbers, J. Int. Seq., Vol. 21 (2018), Article 18.5.4.
Madeleine Goertz and Aaron Williams, The Quaternary Gray Code and How It Can Be Used to Solve Ziggurat and Other Ziggu Puzzles, arXiv:2411.19291 [math.CO], 2024. See pp. 1, 17, 42.
David Hayes, Kaveh Khodjasteh, Lorenza Viola and Michael J. Biercuk, Reducing sequencing complexity in dynamical quantum error suppression by Walsh modulation, arXiv:1109.6002 [quant-ph], 2011.
Clemens Heuberger and Daniel Krenn, Esthetic Numbers and Lifting Restrictions on the Analysis of Summatory Functions of Regular Sequences, arXiv:1808.00842 [math.CO], 2018. See p. 10.
Clemens Heuberger and Daniel Krenn, Asymptotic Analysis of Regular Sequences, arXiv:1810.13178 [math.CO], 2018. See p. 29.
Andreas M. Hinz, The Lichtenberg sequence, Fib. Quart., 55 (2017), 2-12.
Andreas M. Hinz, Sandi Klavžar, Uroš Milutinović, and Ciril Petr, The Tower of Hanoi - Myths and Maths, Birkhäuser 2013. See page 56. Book's website
Andreas M. Hinz and Paul K. Stockmeyer, Precious Metal Sequences and Sierpinski-Type Graphs, J. Integer Seq., Vol 25 (2022), Article 22.4.8.
Jia Huang, Nonassociativity of the Norton Algebras of some distance regular graphs, arXiv:2001.05547 [math.CO], 2020.
Jia Huang, Norton algebras of the Hamming Graphs via linear characters, arXiv:2101.05711 [math.CO], 2021.
Jia Huang and Erkko Lehtonen, Associative-commutative spectra for some varieties of groupoids, arXiv:2401.15786 [math.CO], 2024. See p. 17.
Jia Huang, Madison Mickey, and Jianbai Xu, The Nonassociativity of the Double Minus Operation, Journal of Integer Sequences, Vol. 20 (2017), #17.10.3.
Ryota Inagaki, Tanya Khovanova, and Austin Luo, On Chip-Firing on Undirected Binary Trees, Ann. Comb. (2025). See p. 10.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 394
Hans Isdahl, "Knitwear" puzzle
D. E. Knuth and O. P. Lossers, Partitions of a circular set, Problem 11151 in Amer. Math. Monthly 114 (3) (2007) p 265, E_3.
S. Lafortune, A. Ramani, B. Grammaticos, Y. Ohta and K.M. Tamizhmani, Blending two discrete integrability criteria: singularity confinement and algebraic entropy, arXiv:nlin/0104020 [nlin.SI], 2001.
Robert L. Lamphere, A Recurrence Relation in the Spinout Puzzle, The College Mathematics Journal, Vol. 27, Nbr. 4, Page 289, Sep. 96.
Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences.
Georg Christoph Lichtenberg, Vermischte Schriften, Band 6 (1805). See chapter 6, p. 257.
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.
Richard Moot, Partial Orders, Residuation, and First-Order Linear Logic, arXiv:2008.06351 [cs.LO], 2020.
Gregg Musiker and Son Nguyen, Labeled Chip-firing on Binary Trees, arXiv:2206.02007 [math.CO], 2022.
Kival Ngaokrajang, Illustration of initial terms
Ahmet Öteleş, On the sum of Pell and Jacobsthal numbers by the determinants of Hessenberg matrices, AIP Conference Proceedings 1863, 310003 (2017).
Vladimir Shevelev, On the Basis Polynomials in the Theory of Permutations with Prescribed Up-Down Structure, arXiv:0801.0072 [math.CO], 2007-2010. See Example 3.
Chloe E. Shiff and Noah A. Rosenberg, Enumeration of rooted binary perfect phylogenies, arXiv:2410.14915 [q-bio.PE], 2024. See pp. 15, 17.
A. V. Sills and H. Wang, On the maximal Wiener index and related questions, Discrete Applied Mathematics, Volume 160, Issues 10-11, July 2012, Pages 1615-1623 - _N. J. A. Sloane_, Sep 21 2012
N. J. A. Sloane, Transforms
Elen Viviani Pereira Spreafico, Francisco Regis Vieira Alves, Paula Maria Machado Cruz Catarino, and Eudes Antonio Costa, A brief study of the one parameter Lichtenberg numbers, VI Int'l Conf. Math. Appl. Sci. Eng. (ICMASE 2025).
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Index entries for sequences related to binary expansion of n
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).

Partial sums of A001045.
Row sums of triangle A013580.
Equals A026644/2.
Union of the bijections A002450 and A020988. - Robert G. Wilson v, Jun 09 2014
Column k=3 of A261139.
Cf. A000295, A005578, A015441, A043291, A053404, A059260, A077854, A119440, A127824, A130125, A135228, A155051, A179970, A264784.
Complement of A107907.
Row 3 of A300653.
Other sequences that relate to the binary representation of the terms: A003714, A003754, A007088, A022290, A056830, A104161, A107909.
Cf. A000120, A001511, A003242, A027383, A070939, A164707, A295235, A319416.
Cf. A005186, A033491, A153772.
Cf. A014550, A035263
Cf. A051293, A067659, A079309, A231147, A325347, A359893, A359907, A361801.

List([0..35],n->(2^(n+1)-2+(n mod 2))/3); # Muniru A Asiru, Nov 01 2018

a000975 n = a000975_list !! n
a000975_list = 0 : 1 : map (+ 1)
   (zipWith (+) (tail a000975_list) (map (* 2) a000975_list))
-- Reinhard Zumkeller, Mar 07 2012

[(2^(n+1) - 2 + (n mod 2))/3: n in [0..40]]; // Vincenzo Librandi, Mar 18 2015

A000975 := proc(n) option remember; if n <= 1 then n else if n mod 2 = 0 then 2*A000975(n-1) else 2*A000975(n-1)+1 fi; fi; end;
seq(iquo(2^n,3),n=1..33); # Zerinvary Lajos, Apr 20 2008
f:=n-> if n mod 2 = 0 then (2^n-1)/3 else (2^n-2)/3; fi; [seq(f(n),n=0..40)]; # N. J. A. Sloane, Mar 21 2017

Array[Ceiling[2(2^# - 1)/3] &, 41, 0]
RecurrenceTable[{a[0] == 0, a[1] == 1, a[n] == a[n - 1] + 2a[n - 2] + 1}, a, {n, 40}] (* or *)
LinearRecurrence[{2, 1, -2}, {0, 1, 2}, 40] (* Harvey P. Dale, Aug 10 2013 *)
f[n_] := Block[{exp = n - 2}, Sum[2^i, {i, exp, 0, -2}]]; Array[f, 33] (* Robert G. Wilson v, Oct 30 2015 *)
f[s_List] := Block[{a = s[[-1]]}, Append[s, If[OddQ@ Length@ s, 2a + 1, 2a]]]; Nest[f, {0}, 32] (* Robert G. Wilson v, Jul 20 2017 *)
NestList[2# + Boole[EvenQ[#]] &, 0, 39] (* Alonso del Arte, Sep 21 2018 *)

{a(n) = if( n<0, 0, 2 * 2^n \ 3)}; /* Michael Somos, Sep 04 2006 */

a(n)=if(n<=0,0,bitxor(a(n-1),2^n-1)) \\ Paul D. Hanna, Nov 05 2011

concat(0, Vec(x/(1-2*x-x^2+2*x^3) + O(x^100))) \\ Altug Alkan, Oct 30 2015

{a(n) = (4*2^n - 3 - (-1)^n) / 6}; /* Michael Somos, Jul 23 2017 */

def a(n): return (2**(n+1) - 2 + (n%2))//3
print([a(n) for n in range(35)]) # Michael S. Branicky, Dec 19 2021

a(n) = ceiling(2*(2^n-1)/3).
Alternating sum transform (PSumSIGN) of {2^n - 1} (A000225).
a(n) = a(n-1) + 2*a(n-2) + 1.
a(n) = 2*2^n/3 - 1/2 - (-1)^n/6.
a(n) = Sum_{i = 0..n} A001045(i), partial sums of A001045. - Bill Blewett
a(n) = n + 2*Sum_{k = 1..n-2} a(k).
G.f.: x/((1+x)*(1-x)*(1-2*x)) = x/(1-2*x-x^2+2*x^3). - Paul Barry, Feb 11 2003
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3). - Paul Barry, Feb 11 2003
a(n) = Sum_{k = 0..floor((n-1)/2)} 2^(n-2*k-1). - Paul Barry, Nov 11 2003
a(n+1) = Sum_{k=0..floor(n/2)} 2^(n-2*k); a(n+1) = Sum_{k = 0..n} Sum_{j = 0..k} (-1)^(j+k)*2^j. - Paul Barry, Nov 12 2003
(-1)^(n+1)*a(n) = Sum_{i = 0..n} Sum_{k = 1..i} k!*k* Stirling2(i, k)*(-1)^(k-1) = (1/3)*(-2)^(n+1)-(1/6)(3*(-1)^(n+1)-1). - Mario Catalani (mario.catalani(AT)unito.it), Dec 22 2003
a(n+1) = (n!/3)*Sum_{i - (-1)^i + j = n, i = 0..n, j = 0..n} 1/(i - (-1)^i)!/j!. - Benoit Cloitre, May 24 2004
a(n) = A001045(n+1) - A059841(n). - Paul Barry, Jul 22 2004
a(n) = Sum_{k = 0..n} 2^(n-k-1)*(1-(-1)^k), row sums of A130125. - Paul Barry, Jul 28 2004
a(n) = Sum_{k = 0..n} binomial(k, n-k+1)2^(n-k); a(n) = Sum_{k = 0..floor(n/2)} binomial(n-k, k+1)2^k. - Paul Barry, Oct 07 2004
a(n) = A107909(A104161(n)); A007088(a(n)) = A056830(n). - Reinhard Zumkeller, May 28 2005
a(n) = floor(2^(n+1)/3) = ceiling(2^(n+1)/3) - 1 = A005578(n+1) - 1. - Paul Barry, Oct 08 2005
Convolution of "Number of fixed points in all 231-avoiding involutions in S_n." (A059570) with "1-n" (A024000), treating the result as if offset was 0. - Graeme McRae, Jul 12 2006
a(n) = A081254(n) - 2^n. - Philippe Deléham, Oct 15 2006
Starting (1, 2, 5, 10, 21, 42, ...), these are the row sums of triangle A135228. - Gary W. Adamson, Nov 23 2007
Let T = the 3 X 3 matrix [1,1,0; 1,0,1; 0,1,1]. Then T^n * [1,0,0] = [A005578(n), A001045(n), a(n-1)]. - Gary W. Adamson, Dec 25 2007
2^n = 2*A005578(n-1) + 2*A001045(n) + 2*a(n-2). - Gary W. Adamson, Dec 25 2007
If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*stirling2(k,j)*x^(m-k) then a(n-3) = (-1)^(n-1)*f(n,3,-2), (n >= 3). - Milan Janjic, Apr 26 2009
a(n) + A001045(n) = A166920(n). a(n) + A001045(n+2) = A051049(n+1). - Paul Curtz, Oct 29 2009
a(n) = floor(A051049(n+1)/3). - Gary Detlefs, Dec 19 2010
a(n) = round((2^(n+2)-3)/6) = floor((2^(n+1)-1)/3) = round((2^(n+1)-2)/3); a(n) = a(n-2) + 2^(n-1), n > 1. - Mircea Merca, Dec 27 2010
a(n) = binary XOR of 2^k-1 for k=0..n. - Paul D. Hanna, Nov 05 2011
E.g.f.: 2/3*exp(2*x) - 1/2*exp(x) - 1/6*exp(-x) = 2/3*U(0); U(k) = 1 - 3/(4*(2^k) - 4*(2^k)/(1+3*(-1)^k - 24*x*(2^k)/(8*x*(2^k)*(-1)^k - (k+1)/U(k+1)))); (continued fraction). - Sergei N. Gladkovskii, Nov 21 2011
Starting with "1" = triangle A059260 * [1, 2, 2, 2, ...] as a vector. - Gary W. Adamson, Mar 06 2012
a(n) = 2*(2^n - 1)/3, for even n; a(n) = (2^(n+1) - 1)/3 = (1/3)*(2^((n+1)/2) - 1)*(2^((n+1)/2) + 1), for odd n. - Hieronymus Fischer, Nov 22 2012
a(n) + a(n+1) = 2^(n+1) - 1. - Arie Bos, Apr 03 2013
G.f.: Q(0)/(3*(1-x)), where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k - 1/(1 + 1/(2*4^k - 8*x*16^k/(4*x*4^k + 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 21 2013
floor(a(n+2)*3/5) = A077854(n), for n >= 0. - Armands Strazds, Sep 21 2014
a(n) = (2^(n+1) - 2 + (n mod 2))/3. - Paul Toms, Mar 18 2015
a(0) = 0, a(n) = 2*(a(n-1)) + (n mod 2). - Paul Toms, Mar 18 2015
Binary: a(n) = (a(n-1) shift left 1) + (a(n-1)) NOR (...11110). - Paul Toms, Mar 18 2015
Binary: for n > 1, a(n) = 2*a(n-1) OR a(n-2). - Stanislav Sykora, Nov 12 2015
a(n) = A266613(n) - 20*2^(n-5), for n > 2. - Andres Cicuttin, Mar 31 2016
From Michael Somos, Jul 23 2017: (Start)
a(n) = -(2^n)*a(-n) for even n; a(n) = -(2^(n+1))*a(-n) + 1 for odd n.
0 = +a(n)*(+2 +4*a(n) -4*a(n+1)) + a(n+1)*(-1 +a(n+1)) for all n in Z. (End)
G.f.: (x^1+x^3+x^5+x^7+...)/(1-2*x). - Gregory L. Simay, Sep 27 2017
a(n+1) = A051049(n) + A001045(n). - Yuchun Ji, Jul 12 2018
a(n) = A153772(n+3)/4. - Markus Sigg, Sep 14 2020
a(4*k+d) = 2^(d+1)*a(4*k-1) + a(d), a(n+4) = a(n) + 2^n*10, a(0..3) = [0,1,2,5]. So the last digit is always 0,1,2,5 repeated. - Yuchun Ji, May 22 2023

Additional comments from Barry E. Williams, Jan 10 2000

A014551 Jacobsthal-Lucas numbers.

Original entry on oeis.org

2, 1, 5, 7, 17, 31, 65, 127, 257, 511, 1025, 2047, 4097, 8191, 16385, 32767, 65537, 131071, 262145, 524287, 1048577, 2097151, 4194305, 8388607, 16777217, 33554431, 67108865, 134217727, 268435457, 536870911, 1073741825, 2147483647, 4294967297, 8589934591

Offset: 0

Eric W. Weisstein

Also gives the number of points of period n in the subshift of finite type corresponding to the square matrix A=[1,2;1,0] (this is then given by trace(A^n)). - Thomas Ward, Mar 07 2001
Sequence is identical to its signed inverse binomial transform (autosequence of the second kind). - Paul Curtz, Jul 11 2008
a(n) can be expressed in terms of values of the Fibonacci polynomials F_n(x), computed at x=1/sqrt(2). - Tewodros Amdeberhan (tewodros(AT)math.mit.edu), Dec 15 2008
Pisano period lengths: 1, 1, 2, 2, 4, 2, 6, 2, 6, 4, 10, 2, 12, 6, 4, 2, 8, 6, 18, 4, ... - R. J. Mathar, Aug 10 2012
Let F(x) = Product_{n >= 0} (1 - x^(3*n+1))/(1 - x^(3*n+2)). This sequence is the simple continued fraction expansion of the real number 1 + F(-1/2) = 2.83717 78068 73232 99799 ... = 2 + 1/(1 + 1/(5 + 1/(7 + 1/(17 + ...)))). See A111317. - Peter Bala, Dec 26 2012
With different signs,  2, -1, 5, -7, 17, -31, 65, -127, 257, -511, 1025, -2047, ... is the Lucas V(-1,-2) sequence. - R. J. Mathar, Jan 08 2013
The identity 2 = 2/2 + 2^2/(2*1) - 2^3/(2*1*5) - 2^4/(2*1*5*7) + 2^5/(2*1*5*7*17) + 2^6/(2*1*5*7*17*31) - - + + can be viewed as a generalized Engel-type expansion of the number 2 to the base 2. Compare with A062510. - Peter Bala, Nov 13 2013
For n >= 2, a(n) is the number of ways to tile a 2 X n strip, where the first two columns have an extra cell at the top, with 1 X 2 dominoes and 2 X 2 squares. Shown here is one of the a(7)=127 ways for the n=7 case:
  ._.
  |_|_________.
  | |   | |_| |
  ||__|_|_|_|. - Greg Dresden, Sep 26 2021
Named by Horadam (1988) after the German mathematician Ernst Jacobsthal (1882-1965) and the French mathematician Édouard Lucas (1842-1891). - Amiram Eldar, Oct 02 2023

G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. pp. 180, 255.
Lind and Marcus, An Introduction to Symbolic Dynamics and Coding, Cambridge University Press, 1995. (General material on subshifts of finite type)
Kritkhajohn Onphaeng and Prapanpong Pongsriiam. Jacobsthal and Jacobsthal-Lucas Numbers and Sums Introduced by Jacobsthal and Tverberg. Journal of Integer Sequences, Vol. 20 (2017), Article 17.3.6.
Abdelmoumène Zekiri, Farid Bencherif, Rachid Boumahdi, Generalization of an Identity of Apostol, J. Int. Seq., Vol. 21 (2018), Article 18.5.1.

T. D. Noe, Table of n, a(n) for n = 0..200
Kunle Adegoke, Robert Frontczak, and Taras Goy, Partial sum of the products of the Horadam numbers with subscripts in arithmetic progression, Notes on Num. Theor. and Disc. Math. (2021) Vol. 27, No. 2, 54-63.
Tewodros Amdeberhan, A note on Fibonacci-type polynomials, arXiv:0811.4652 [math.NT], 2008.
Hacène Belbachir, Amine Belkhir, and Ihab-Eddin Djellas, Permanent of Toeplitz-Hessenberg Matrices with Generalized Fibonacci and Lucas entries, Applications and Applied Mathematics: An International Journal (AAM 2022), Vol. 17, Iss. 2, Art. 15, 558-570.
Paula Catarino, Helena Campos, and Paulo Vasco. On the Mersenne sequence. Annales Mathematicae et Informaticae, 46 (2016), pp. 37-53.
Charles K. Cook and Michael R. Bacon, Some identities for Jacobsthal and Jacobsthal-Lucas numbers satisfying higher order recurrence relations, Annales Mathematicae et Informaticae, 41 (2013), pp. 27-39.
Fatih Erduvan and Refik Keskin, Fibonacci And Lucas Numbers Which Are Product Of Two Jacobsal-Lucas Numbers [sic], Appl. Math. E-Notes (2023) Vol. 23, 60-70.
M. C. Firengiz and A. Dil, Generalized Euler-Seidel method for second order recurrence relations, Notes on Number Theory and Discrete Mathematics, Vol. 20, 2014, No. 4, 21-32.
Élis Gardel da Costa Mesquita, Eudes Antonio Costa, Paula M. M. C. Catarino, and Francisco R. V. Alves, Jacobsthal-Mulatu Numbers, Latin Amer. J. Math. (2025) Vol. 4, No. 1, 23-45. See p. 24.
A. F. Horadam, Jacobsthal and Pell Curves, Fib. Quart. 26, 79-83, 1988.
A. F. Horadam, Jacobsthal Representation Numbers, Fib Quart. 34, 40-54, 1996.
D. Jhala, G. P. S. Rathore, and K. Sisodiya, Some Properties of k-Jacobsthal Numbers with Arithmetic Indexes, Turkish Journal of Analysis and Number Theory, 2014, Vol. 2, No. 4, 119-124.
Thomas Koshy and Ralph P. Grimaldi, Ternary words and Jacobsthal numbers, Fib. Quart., 55 (No. 2, 2017), 129-136.
Kritkhajohn Onphaeng, Tammatada Khemaratchatakumthorn, and Prapanpong Pongsriiam, Inequalities for Inclusion-Exclusion-Like Sums Involving the Ceiling and the Nearest Integer Functions, Integers (2025) Vol. 25, Art. No. A45. See p. 3.
Mihai Prunescu and Lorenzo Sauras-Altuzarra, On the representation of C-recursive integer sequences by arithmetic terms, arXiv:2405.04083 [math.LO], 2024. See p. 16.
Yash Puri and Thomas Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), Article 01.2.1.
M. Rahmani, The Akiyama-Tanigawa matrix and related combinatorial identities, Linear Algebra and its Applications 438 (2013) 219-230. - From _N. J. A. Sloane_, Dec 26 2012
Mario Raso, Integer Sequences in Cryptography: A New Generalized Family and its Application, Ph. D. Thesis, Sapienza University of Rome (Italy 2025). See p. 41.
Yüksel Soykan, On Summing Formulas For Generalized Fibonacci and Gaussian Generalized Fibonacci Numbers, Advances in Research (2019) Vol. 20, No. 2, 1-15, Article AIR.51824.
Yüksal Soykan, On Summing Formulas for Horadam Numbers, Asian Journal of Advanced Research and Reports (2020) Vol. 8, Issue 1, 45-61.
Yüksel Soykan, Generalized Fibonacci Numbers: Sum Formulas, Journal of Advances in Mathematics and Computer Science (2020) Vol. 35, No. 1, 89-104.
Yüksel Soykan, Closed Formulas for the Sums of Squares of Generalized Fibonacci Numbers, Asian Journal of Advanced Research and Reports (2020) Vol. 9, No. 1, 23-39, Article no. AJARR.55441.
Yüksel Soykan, Closed Formulas for the Sums of Cubes of Generalized Fibonacci Numbers: Closed Formulas of Sum_{k=0..n} W_k^3 and Sum_{k=1..n} W_(-k)^3, Archives of Current Research International (2020) Vol. 20, Issue 2, 58-69.
Yüksel Soykan, A Study on Generalized Fibonacci Numbers: Sum Formulas Sum_{k=0..n} k * x^k * W_k^3 and Sum_{k=1..n} k * x^k W_-k^3 for the Cubes of Terms, Earthline Journal of Mathematical Sciences (2020) Vol. 4, No. 2, 297-331.
Yüksel Soykan, On Generalized (r, s)-numbers, Int. J. Adv. Appl. Math. and Mech. (2020) Vol. 8, No. 1, 1-14.
Yüksel Soykan, Erkan Taşdemir, and İnci Okumuş, On Dual Hyperbolic Numbers With Generalized Jacobsthal Numbers Components, Zonguldak Bülent Ecevit University, (Zonguldak, Turkey, 2019).
Anetta Szynal-Liana, Iwona Włoch, and Mirosław Liana, Generalized commutative quaternion polynomials of the Fibonacci type, Annales Math. Sect. A, Univ. Mariae Curie-Skłodowska (Poland 2022) Vol. 76, No. 2, 33-44.
Elif Tan, Luka Podrug, and Vesna Iršič Chenoweth, Horadam-Lucas Cubes, Axioms (2024) Vol. 13, No. 12, 837.
Kai Wang, On Horadam Sequences and Related Infinite Series, (2020).
Kai Wang, General Identities for Horadam Sequences, (2020).
Eric Weisstein's World of Mathematics, Jacobsthal Number.
Wikipedia, Lucas sequence.
OEIS Wiki, Autosequence.
Volkan Yildiz, Some divisibility properties of Jacobsthal numbers, arXiv:2212.08814 [math.CO], 2022.
Index entries for linear recurrences with constant coefficients, signature (1,2).
Index entries for Lucas sequences.
Index entries for sequences related to Chebyshev polynomials.

Cf. A001045 (companion "autosequence"), A019322, A066845, A111317.
Cf. A135440 (first differences), A166920 (partial sums).
Cf. A000079, A033999, A102345, A105723.
Cf. A006995.

a014551 n = a000079 n + a033999 n
a014551_list = map fst $ iterate (\(x,s) -> (2 * x - 3 * s, -s)) (2, 1)
-- Reinhard Zumkeller, Jan 02 2013

[2^n + (-1)^n: n in [0..30]]; // G. C. Greubel, Dec 17 2017

f[n_]:=2/(n+1);x=4;Table[x=f[x];Denominator[x],{n,0,5!}] (* Vladimir Joseph Stephan Orlovsky, Mar 12 2010 *)
nxt[{n_,a_}]:={n+1,2a-3(-1)^(n+1)}; Transpose[NestList[nxt,{1,2},40]] [[2]] (* Harvey P. Dale, May 27 2013 *)
LinearRecurrence[{1, 2}, {2, 1}, 40] (* Jean-François Alcover, Jan 07 2019 *)

a(n)=2^n+(-1)^n \\ Charles R Greathouse IV, Nov 20 2012

[lucas_number2(n,1,-2) for n in range(0, 32)] # Zerinvary Lajos, Apr 30 2009

a(n+1) = 2 * a(n) - (-1)^n * 3.
From Len Smiley, Dec 07 2001: (Start)
a(n) = 2^n + (-1)^n.
G.f.: (2-x)/(1-x-2*x^2). (End)
E.g.f.: exp(x) + exp(-2*x) produces a signed version. - Paul Barry, Apr 27 2003
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-1, 2*k)*3^(2*k)/2^(n-2). - Paul Barry, Feb 21 2003
0, 1, 5, 7 ... is 2^n - 2*0^n + (-1)^n, the 2nd inverse binomial transform of (2^n-1)^2 (A060867). - Paul Barry, Sep 05 2003
a(n) = 2*T(n, i/(2*sqrt(2))) * (-i*sqrt(2))^n with i^2=-1. - Paul Barry, Nov 17 2003
a(n) = A078008(n) + A001045(n+1). - Paul Barry, Feb 12 2004
a(n) = 2*A001045(n+1) - A001045(n). - Paul Barry, Mar 22 2004
a(0)=2, a(1)=1, a(n) = a(n-1) + 2*a(n-2) for n > 1. - Philippe Deléham, Nov 07 2006
a(2*n+1) = Product_{d|(2*n+1)} cyclotomic(d,2). a(2^k*(2*n+1)) = Product_{d|(2*n+1)} cyclotomic(2*d,2^(2^k)). - Miklos Kristof, Mar 12 2007
a(n) = 2^{(n-1)/2}F_{n-1}(1/sqrt(2)) + 2^{(n+2)/2}F_{n-2}(1/sqrt(2)). - Tewodros Amdeberhan (tewodros(AT)math.mit.edu), Dec 15 2008
E.g.f.: U(0) where U(k) = 1 + (-1)^k/(2^k - 4^k*x*2/(2*x*2^k + (-1)^k*(k+1)/U(k+1))) ; (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Nov 02 2012
G.f.: U(0) where U(k) = 1 + (-1)^k/(2^k - 4^k*x*2/(2*x*2^k + (-1)^k/U(k+1))) ; (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Nov 02 2012
a(n) = sqrt(9*(A001045)^2 + (-1)^n*2^(n+2)). - Vladimir Shevelev, Mar 13 2013
G.f.: 2 + G(0)*x*(1+4*x)/(2-x), where G(k) = 1 + 1/(1 - x*(9*k-1)/( x*(9*k+8) - 2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 13 2013
a(n) = [x^n] ( (1 + x + sqrt(1 + 2*x + 9*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
For n >= 1: a(n) = A006995(2^((n+2)/2)) when n is even, a(n) = A006995(3*2^((n-1)/2) - 1) when n is odd. - Bob Selcoe, Sep 04 2017
a(n) = J(n) + 4*J(n-1), a(0)=2, where J is A001045. - Yuchun Ji, Apr 23 2019
For n >= 0, 1/(2*a(n+1)) = Sum_{m>=n} a(m)/(a(m+1)*a(m+2)). - Kai Wang, Mar 03 2020
For 4 > h >= 0, k >= 0, a(4*k+h) mod 5 = a(h) mod 5. - Kai Wang, May 06 2020
From Kai Wang, May 30 2020: (Start)
(2 - a(n+1)/a(n))/9 = Sum_{m>=n} (-2)^m/(a(m)*a(m+1)).
a(n) = 2*A001045(n+1) - A001045(n).
a(n)^2 = a(2*n) + 2*(-2)^n.
a(n)^2 = 9*A001045(n)^2 + 4*(-2)^n.
a(2*n) = 9*A001045(n)^2 + 2*(-2)^n.
2*A001045(m+n) = A001045(m)*a(n) + a(m)*A001045(n).
2*(-2)^n*A001045(m-n) = A001045(m)*a(n) - a(m)*A001045(n).
A001045(m+n) + (-2)^n*A001045(m-n) = A001045(m)*a(n).
A001045(m+n) - (-2)^n*A001045(m-n) = a(m)*A001045(n).
2*a(m+n) = 9*A001045(m)*A001045(n) + a(m)*a(n).
2*(-2)^n*a(m-n) = a(m)*a(n) - 9*A001045(m)*A001045(n).
a(m+n) - (-2)^n*a(m-n) = 9*A001045(m)*A001045(n).
a(m+n) + (-2)^n*a(m-n) = a(m)*a(n).
a(m+n)*a(m-n) - a(m)*a(m) = 9*(-2)^(m-n)*A001045(n)^2.
a(m+1)*a(n) - a(m)*a(n+1) = 9*(-2)^n*A001045(m-n). (End)
a(n) = F(n+1) + F(n-1) + Sum_{k=0..(n-2)} a(k)*F(n-1-k) for F(n) the Fibonacci numbers and for n > 1. - Greg Dresden, Jun 03 2020

A166956 a(n) = 2^n +(-1)^n - 2.

Original entry on oeis.org

0, -1, 3, 5, 15, 29, 63, 125, 255, 509, 1023, 2045, 4095, 8189, 16383, 32765, 65535, 131069, 262143, 524285, 1048575, 2097149, 4194303, 8388605, 16777215, 33554429, 67108863, 134217725, 268435455, 536870909, 1073741823, 2147483645, 4294967295, 8589934589

Offset: 0

Paul Curtz, Oct 25 2009

The inverse binomial transform yields 0,-1,5,-7,17,-31,..., a sign alternating variant of A014551.
In a table of a(n) and higher-order differences in successive rows, the main diagonal contains 0, 4, 8, 16, ... (zero followed by A020707).
Similar to the decimal representation of the diagonal from the corner to the origin of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 899", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero, which begins with 1,3,5,15,29,63,125. - Robert Price, Aug 08 2017

S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.

Vincenzo Librandi, Table of n, a(n) for n = 0..240
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton
S. Wolfram, A New Kind of Science
Wolfram Research, Wolfram Atlas of Simple Programs
Index entries for sequences related to cellular automata
Index to 2D 5-Neighbor Cellular Automata
Index to Elementary Cellular Automata
Index entries for linear recurrences with constant coefficients, signature (2,1,-2)

Cf. A166920, A166956, A290660, A290661, A290662.

[2^n-2+(-1)^n: n in [0..40]]; // Vincenzo Librandi, Apr 28 2011

LinearRecurrence[{2,1,-2},{0,-1,3},20] (* G. C. Greubel, May 29 2016 *)

a(n) = A000079(n) - A010684(n).
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3).
G.f.: x*(5*x -1)/((1-x)*(1-2*x)*(1+x)).
E.g.f.: exp(2*x) - 2*exp(x) + exp(-x). - G. C. Greubel, May 29 2016

Edited and extended by R. J. Mathar, Mar 02 2010

A320933 a(n) = 2^n - floor((n+3)/2).

Original entry on oeis.org

0, 0, 2, 5, 13, 28, 60, 123, 251, 506, 1018, 2041, 4089, 8184, 16376, 32759, 65527, 131062, 262134, 524277, 1048565, 2097140, 4194292, 8388595, 16777203, 33554418, 67108850, 134217713, 268435441, 536870896

Offset: 0

Paul Curtz, Oct 28 2018

The sequence 0, 0, a(n) is an autosequence of the second kind. The difference table is:
   0,   0,   0,   0,   2,   5,  13, ...
   0,   0,   0,   2,   3,   8,  15, ...
   0,   0,   2,   1,   5,   7,  17, ...
   0,   2,  -1,   4,   2,  10,  14, ...
   2,  -3,   5,  -2,   8,   4,  20, ...
  -5,   8,  -7,  10,  -4,  16,   8, ...
  13, -15,  17, -14,  20,  -8,  32, ...
etc.

Colin Barker, Table of n, a(n) for n = 0..1000
OEIS Wiki, Autosequence
Index entries for linear recurrences with constant coefficients, signature (3,-1,-3,2).

Cf. A000079, A004526, A011377, A014551, A060182, A166920.

List([0..40],n->2^n-Int((n+3)/2)); # Muniru A Asiru, Oct 28 2018

[((-1)^n+2^(n+2)-2*n-5)/4: n in [0..40]]; // G. C. Greubel, Jun 04 2019

seq(2^n-floor((n+3)/2),n=0..40); # Muniru A Asiru, Oct 28 2018

a[n_]:=2^n - Floor[(n+3)/2]; Array[a, 40, 0] (* or *) CoefficientList[ Series[x^2*(2-x)/((1-x)^2*(1-x-2*x^2)), {x, 0, 40}], x] (* Stefano Spezia, Oct 28 2018 *)

concat([0,0], Vec(x^2*(2-x)/((1-x)^2*(1+x)*(1-2*x)) + O(x^40))) \\ Colin Barker, Oct 28 2018

[((-1)^n+2^(n+2)-2*n-5)/4 for n in (0..40)] # G. C. Greubel, Jun 04 2019

a(n) = 3*a(n-1) - a(n-2) - 3*a(n-3) + a(n-4).
a(n+1) = a(n) + A166920(n).
a(n+4) - a(n) = 13, 28, 58, 118, ... = 15*2^n - 2 = A060182(n+2).
With b(n) = 0, 0, 0, A011377(n) = 0, 0, 0, 1, 3, 8, 18, ..., then a(n) = 2*b(n+1) - b(n).
a(n+2) - 2*a(n+1) + a(n) = A014551(n).
G.f.: x^2*(2 - x)/((1-x)^2*(1 - x - 2*x^2)). - Stefano Spezia, Oct 28 2018
a(n) = ((-1)^n + 2^(n+2) - 2*n - 5) / 4. - Colin Barker, Oct 28 2018

Three terms corrected by Colin Barker, Oct 28 2018

A128932 Define the Fibonacci polynomials by F[1] = 1, F[2] = x; for n > 2, F[n] = xF[n-1] + F[n-2] (cf. A049310, A053119). Swamy's inequality implies that F[n] <= F[n]^2 <= G[n] = (x^2 + 1)^2(x^2 + 2)^(n-3) for n >= 3 and x >= 1. The sequence gives a triangle of coefficients of G[n] - F[n] read by rows.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, -2, 5, -1, 4, 0, 1, 3, 0, 9, 0, 12, 0, 6, 0, 1, 8, -3, 28, -4, 38, -1, 25, 0, 8, 0, 1, 15, 0, 58, 0, 99, 0, 87, 0, 41, 0, 10, 0, 1, 32, -4, 144, -10, 272, -6, 280, -1, 170, 0, 61, 0, 12, 0, 1, 63, 0, 310, 0, 673, 0, 825, 0, 619, 0, 292, 0, 85, 0, 14, 0, 1

Offset: 3

N. J. A. Sloane, Apr 28 2007

From Petros Hadjicostas, Jun 10 2020: (Start)
Swamy's (1966) inequality states that F[n]^2 <= G[n] for all real x and all integers n >= 3. Because F[n] >= 1 for all real x >= 1, we get F[n] <= G[n] for all integers n >= 3 and all real x >= 1.
Row n >= 3 of this irregular table gives the coefficients of the polynomial G[n] - F[n] (with exponents in increasing order). The degree of G[n] - F[n] is 2*n - 2, so row n >= 3 contains 2*n - 1 terms.
Guilfoyle (1967) notes that F[n] = det(A_n), where A_n is the (n-1) X (n-1) matrix [[x, -1, 0, 0, ..., 0, 0, 0], [1, x, -1, 0, ..., 0, 0, 0], [0, 1, x, -1, ..., 0, 0, 0], ..., [0, 0, 0, 0, ..., 1, x, -1], [0, 0, 0, 0, ..., 0, 1, x]], and Swamy's original inequality follows from Hadamard's inequality.
Koshy (2019) writes Swamy's original inequality in the form x^(n-3)*F[n]^2 <= F[3]^2*F[4]^(n-3) for x >= 1, and gives a counterpart inequality for Lucas polynomials. Notice, however, that the original form of Swamy's inequality is true for all real x. (End)

			Triangle T(n,k) (with rows n >= 3 and columns k = 0..2*n-2) begins:
   0,  0,  1,  0,  1;
   2, -2,  5, -1,  4,  0,  1;
   3,  0,  9,  0, 12,  0,  6, 0,  1;
   8, -3, 28, -4, 38, -1, 25, 0,  8, 0,  1;
  15,  0, 58,  0, 99,  0, 87, 0, 41, 0, 10, 0, 1;
  ...

Thomas Koshy, Fibonacci and Lucas numbers with Applications, Vol. 2, Wiley, 2019; see p. 33. [Vol. 1 was published in 2001.]
D. S. Mitrinovic, Analytic Inequalities, Springer-Verlag, 1970; p. 232, Sect. 3.3.38.

Richard Guilfoyle, Comment to the solution of Problem E1846, Amer. Math. Monthly, 74(5), 1967, 593. [It is pointed out that the inequality is a special case of Hadamard's inequality.]
M. N. S. Swamy, Problem E1846 proposed for solution, Amer. Math. Monthly, 73(1) (1966), 81.
M. N. S. Swamy and R. E. Giudici, Solution to Problem E1846, Amer. Math. Monthly, 74(5), 1967, 592-593.
M. N. S. Swamy and Joel Pitcain, Comment to Problem E1846, Amer. Math. Monthly, 75(3) (1968), 295. [It is pointed out that I^{n-1}*F[n](x) = U_{n-1}(I*x/2), where U_{n-1}(cos(t)) = sin(n*t)/sin(t) and I = sqrt(-1); Cf. A049310 and A053119, but with different notation.]
Wikipedia, Fibonacci polynomials.
Wikipedia, Hadamard's inequality.

Cf. A000045, A003946, A049310, A053119, A166920.

lista(nn) = {my(f=vector(nn)); my(g=vector(nn)); my(h=vector(nn)); f[1]=1; f[2]=x; g[1]=0; g[2]=0; for(n=3, nn, g[n] = (x^2+1)^2*(x^2+2)^(n-3)); for(n=3, nn, f[n] = x*f[n-1]+f[n-2]); for(n=1, nn, h[n] = g[n]-f[n]); for(n=3, nn, for(k=0, 2*n-2, print1(polcoef(h[n], k, x), ",")); print(););} \\ Petros Hadjicostas, Jun 10 2020

From Petros Hadjicostas, Jun 10 2020: (Start)
T(n,0) = 2^(n-3) - (1 - (-1)^n)/2 = A166920(n-3) for n >= 3.
Sum_{k=0}^{2*n-2} T(n,k) = 4*3^(n-3) - Fib(n) = A003946(n-2) - A000045(n) for n >= 3. (End)

Name edited by Petros Hadjicostas, Jun 10 2020

A166918 Triangle T(n,k) read by rows: T(n,0) = n mod 2. T(n,k) = 2^(k-1), 0

Original entry on oeis.org

0, 1, 1, 0, 1, 2, 1, 1, 2, 4, 0, 1, 2, 4, 8, 1, 1, 2, 4, 8, 16, 0, 1, 2, 4, 8, 16, 32, 1, 1, 2, 4, 8, 16, 32, 64, 0, 1, 2, 4, 8, 16, 32, 64, 128, 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 0, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024

Offset: 0

Paul Curtz, Oct 23 2009

			0;
1,1;
0,1,2;
1,1,2,4;
0,1,2,4,8;
1,1,2,4,8,16;
0,1,2,4,8,16,32;
1,1,2,4,8,16,32,64;

Cf. A166692, A059268.

 t[n_, 0] := Mod[n, 2]; t[n_, k_] := 2^(k - 1); Table[ t[n, k], {n, 0, 11}, {k, 0, n}] // Flatten

T(2n,k) = A131577(k). T(2n+1,k) = A011782(k).

sum_{k=0..n} T(n,k) = A166920(n).

A277078 Triangular array similar to A255935 but with 0's and 2's swapped in the trailing diagonal. The columns alternate in signs.

Original entry on oeis.org

2, 1, 0, 1, -2, 2, 1, -3, 3, 0, 1, -4, 6, -4, 2, 1, -5, 10, -10, 5, 0, 1, -6, 15, -20, 15, -6, 2, 1, -7, 21, -35, 35, -21, 7, 0, 1, -8, 28, -56, 70, -56, 28, -8, 2, 1, -9, 36, -84, 126, -126, 84, -36, 9, 0, 1, -10, 45, -120, 210, -252, 210, -120, 45, -10, 2

Offset: 0

Paul Curtz, Oct 23 2016

a(n)=
2,
1,  0,
1, -2, 2,
1, -3, 3,  0,
1, -4, 6, -4, 2,
etc.
transforms every sequence s(n) in an autosequence of the second kind via the multiplication by the triangle
s0,           T2
s0, s1,
s0, s1, s2,
s0, s1, s2, s3,
etc.
which is the reluctant form of s(n).
Example.
s(n) = A131577(n) = 0, 1, 2, 4, ... .
The multiplication gives 0, 0, 2, 3, 8, 15, 32, 63, ... = 0 followed by A166920.
a(n) comes from alternate sum and difference of s(n) and t(n), its inverse binomial transform. In the example (t(n) = periodic 2: repeat 0, 1) the first terms are: 0+0, 1-1, 2+0, 4-1, 8+0, 16-1, 32+0, 64-1, ... .

Cf. A000035, A007318, A054977, A131577, A166920, A197870, A255935.

a[n_, k_] := If[k == n, 2 - 2*Mod[n, 2], (-1)^k*Binomial[n, k]]; Table[a[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 16 2016 *)

a(n) = A007318(n) - A197870(n+1).

A322417 a(n) - 2*a(n-1) = period 2: repeat [3, 0] for n > 0, a(0)=5, a(1)=13.

Original entry on oeis.org

5, 13, 26, 55, 110, 223, 446, 895, 1790, 3583, 7166, 14335, 28670, 57343, 114686, 229375, 458750, 917503, 1835006, 3670015, 7340030, 14680063, 29360126, 58720255, 117440510, 234881023, 469762046, 939524095, 1879048190, 3758096383, 7516192766

Offset: 0

Paul Curtz, Dec 07 2018

a(n) mod 9 = period 6: repeat [5, 4, 8, 1, 2, 7]. See A177883(n+2).

a(n+1) mod 10 = period 4: repeat [3, 6, 5, 0].

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

Cf. A166920, A175805, A206372, A321483.

Cf. A177883.

a:=[13,26];; for n in [3..30] do a[n]:=a[n-2]+21*2^(n-2); od; Concatenation([5],a); # Muniru A Asiru, Dec 07 2018

a[0] = 5; a[1] = 13; a[n_] := a[n] = a[n - 2] + 21*2^(n - 2); Array[a, 30, 0] (* Amiram Eldar, Dec 07 2018 *)
LinearRecurrence[{2, 1, -2}, {5, 13, 26}, 31] (* Jean-François Alcover, Jan 28 2019 *)

Vec((5 + 3*x - 5*x^2) / ((1 - x)*(1 + x)*(1 - 2*x)) + O(x^40)) \\ Colin Barker, Dec 07 2018

a(n) = A166920(n) + A166920(n+1) + A166920(n+2) for n >= 2.
a(n) = a(n-2) + 21*2^(n-2) for n >= 2.
a(n) = a(n-1) + A321483(n) for n > 0.
From Colin Barker, Dec 07 2018: (Start)
G.f.: (5 + 3*x - 5*x^2) / ((1 - x)*(1 + x)*(1 - 2*x)).
a(n) = 7*2^n - 2 for n even.
a(n) = 7*2^n - 1 for n odd.
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) for n > 2.
(End)
a(2*n+1) = A206372(n). a(2*n+2) = 2*A206372(n) for n > 0.

First formula corrected by Jean-François Alcover, Feb 01 2019

A335444 Define the Fibonacci polynomials by F[1] = 1, F[2] = x; for n > 2, F[n] = xF[n-1] + F[n-2] (cf. A049310, A053119). Swamy's inequality states that F[n]^2 <= G[n] = (x^2 + 1)^2(x^2 + 2)^(n-3) for all n >= 3 and all real x. The sequence gives a triangle of the coefficients of the even exponents of G[n] - F[n]^2 read by rows.

Original entry on oeis.org

0, 2, 1, 3, 6, 2, 8, 19, 14, 3, 15, 52, 58, 26, 4, 32, 128, 192, 132, 42, 5, 63, 300, 558, 518, 253, 62, 6, 128, 679, 1496, 1742, 1152, 433, 86, 7, 255, 1506, 3801, 5294, 4413, 2248, 684, 114, 8, 512, 3292, 9308, 14999, 15040, 9680, 3992, 1018, 146, 9

Offset: 3

Petros Hadjicostas, Jun 10 2020

Swamy's (1966) inequality states that F[n]^2 <= G[n] for all real x and all integers n >= 3.
Row n >= 3 of this irregular table gives the coefficients of the even powers of the polynomial G[n] - F[n]^2 (with exponents in increasing order). The coefficients of the odd powers are zero, and they are thus omitted. The degree of G[n] - F[n]^2 is 2*n - 6, so row n >= 3 contains n - 2 terms.
To prove that the degree of G[n] - F[n]^2 is 2*n - 6, note that the first few terms of G[n] are x^(2*n-2) + 2*(n-2)*x^(2*n-4) + (2*n^2 - 10*n + 13)*x^(2*n-6) + ... while the first few terms of F[n]^2 are x^(2*n-2) + 2*(n-2)*x^(2*n-4) + (2*n^2 - 11*n + 16)*x^(2*n-6) + ..., so the leading term of the polynomial G[n] - F[n]^2 is (n-3)*x^(2*n-6).
Guilfoyle (1967) notes that F[n] = det(A_n), where A_n is the (n-1) X (n-1) matrix [[x, -1, 0, 0, ..., 0, 0, 0], [1, x, -1, 0, ..., 0, 0, 0], [0, 1, x, -1, ..., 0, 0, 0], ..., [0, 0, 0, 0, ..., 1, x, -1], [0, 0, 0, 0, ..., 0, 1, x]], and Swamy's original inequality follows from Hadamard's inequality.
Koshy (2019) writes Swamy's original inequality in the form x^(n-3)*F[n]^2 <= F[3]^2*F[4]^(n-3) for x >= 1, and gives a counterpart inequality for Lucas polynomials. Notice, however, that the original form of Swamy's inequality is true for all real x.

			Triangle T(n,k) (with rows n >= 3 and columns k = 0..n-3) begins:
   0;
   2,   1;
   3,   6,   2;
   8,  19,  14,   3;
  15,  52,  58,  26,  4;
  32, 128, 192, 132, 42, 5;
  ...

Thomas Koshy, Fibonacci and Lucas numbers with Applications, Vol. 2, Wiley, 2019; see p. 33. [Vol. 1 was published in 2001.]
D. S. Mitrinovic, Analytic Inequalities, Springer-Verlag, 1970; see p. 232, Sect. 3.3.38.

Richard Guilfoyle, Comment to the solution of Problem E1846, Amer. Math. Monthly, 74(5), 1967, 593. [It is pointed out that the inequality is a special case of Hadamard's inequality.]
M. N. S. Swamy, Problem E1846 proposed for solution, Amer. Math. Monthly, 73(1) (1966), 81.
M. N. S. Swamy and R. E. Giudici, Solution to Problem E1846, Amer. Math. Monthly, 74(5), 1967, 592-593.
M. N. S. Swamy and Joel Pitcain, Comment to Problem E1846, Amer. Math. Monthly, 75(3) (1968), 295. [It is pointed out that I^{n-1}*F[n](x) = U_{n-1}(I*x/2), where U_{n-1}(cos(t)) = sin(n*t)/sin(t) and I = sqrt(-1); Cf. A049310 and A053119, but with different notation.]
Wikipedia, Fibonacci polynomials.
Wikipedia, Hadamard's inequality.

Cf. A002620, A045623, A049310, A051890, A053119, A128932 (similar), A166920.

lista(nn) = {my(f=vector(nn)); my(g=vector(nn)); my(h=vector(nn)); f[1]=1; f[2]=x; g[1]=0; g[2]=0; for(n=3, nn, g[n] = (x^2+1)^2*(x^2+2)^(n-3)); for(n=3, nn, f[n] = x*f[n-1]+f[n-2]); for(n=1, nn, h[n] = g[n]-f[n]^2); for(n=3, nn, for(k=0, n-3, print1(polcoef(h[n], 2*k, x), ", ")); print(); ); }

T(n,0) = 2^(n-3) - (1 - (-1)^n)/2 = A166920(n-3) for n >= 3.
T(n,1) = 2^(n-4)*(n + 1) - floor(n/2)*ceiling(n/2) = A045623(n-2) - A002620(n) for n >= 4.
T(n, n-4) = 2*(n^2 - 7*n + 13) = A051890(n-3) for n >= 4.
T(n, n-3) = n - 3 for n >= 3.

cp's OEIS Frontend

A000975 a(2n) = 2*a(2n-1), a(2n+1) = 2*a(2n)+1 (also a(n) is the n-th number without consecutive equal binary digits).

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A014551 Jacobsthal-Lucas numbers.

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A166956 a(n) = 2^n +(-1)^n - 2.

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A320933 a(n) = 2^n - floor((n+3)/2).

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A166918 Triangle T(n,k) read by rows: T(n,0) = n mod 2. T(n,k) = 2^(k-1), 0

A000975 a(2n) = 2a(2n-1), a(2n+1) = 2a(2n)+1 (also a(n) is the n-th number without consecutive equal binary digits).