A056450 -id:A056450 - cp's OEIS frontend

A164906 Duplicate of A056450.

1, 4, 4, 16, 16, 64, 64, 256, 256, 1024, 1024, 4096, 4096, 16384, 16384, 65536, 65536, 262144, 262144, 1048576, 1048576, 4194304, 4194304, 16777216, 16777216, 67108864, 67108864, 268435456, 268435456, 1073741824, 1073741824, 4294967296

Offset: 0

dead

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).

Original entry on oeis.org

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.
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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
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N. J. A. Sloane, Transforms
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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

A007581 a(n) = (2^n+1)*(2^n+2)/6.

Original entry on oeis.org

1, 2, 5, 15, 51, 187, 715, 2795, 11051, 43947, 175275, 700075, 2798251, 11188907, 44747435, 178973355, 715860651, 2863377067, 11453377195, 45813246635, 183252462251, 733008800427, 2932033104555, 11728128223915, 46912504507051, 187650001250987

Offset: 0

Simon Plouffe and N. J. A. Sloane

Number of palindromic structures using a maximum of four different symbols. - Marks R. Nester
Dimension of the universal embedding of the symplectic dual polar space DSp(2n,2) (conjectured by A. Brouwer, proved by P. Li). - J. Taylor (jt_cpp(AT)yahoo.com), Apr 02 2004.
Apart from initial term, same as A124303. - Valery A. Liskovets, Nov 16 2006
Hankel transform is := [1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...]. - Philippe Deléham, Dec 04 2008
a(n) is also the number of distinct solutions (avoiding permutations) to the equation: XOR(A,B,C)=0 where A,B,C are n-bit binary numbers. - Ramasamy Chandramouli, Jan 11 2009
The rank of the fundamental group of the Z_p^n - cobordism category in dimension 1+1 for the case p=2 (see paper below). The expression for any prime p is (p^(2n-1)+p^(n+1)-p^(n-1)+p^2-p-1)/(p^2-1). - Carlos Segovia Gonzalez, Dec 05 2012
The number of isomorphic classes of regular four coverings of a graph with respect to the identity automorphism (S. Hong and J. H. Kwak). - Carlos Segovia Gonzalez, Aug 01 2013
The density of a language with four letters (N. Moreira and R. Reis). - Carlos Segovia Gonzalez, Aug 01 2013

P. Li, On the Brouwer Conjecture for Dual Polar Spaces of Symplectic Type over GF(2). Preprint.
M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..200
Joerg Arndt and N. J. A. Sloane, Counting Words that are in "Standard Order"
Barnes, Jeffrey M.; Benkart, Georgia; Halverson, Tom McKay centralizer algebras. Proc. Lond. Math. Soc. (3) 112, No. 2, 375-414 (2016).
Georgia Benkart and Tom Halverson, McKay Centralizer Algebras, hal-02173744 [math.CO], 2020.
A. Blokhuis and A. E. Brouwer, The universal embedding dimension of the binary symplectic dual polar space, Discr. Math., 264 (2003), 3-11.
P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
Bruno Cisneros, Carlos Segovia, An approximation for the number of subgroups, arXiv:1805.04633 [math.GT], 2018.
B. N. Cooperstein and E. E. Shult, A note on embedding and generating dual polar spaces. Adv. Geom. 1 (2001), 37-48.
Yuanan Diao, Michael Finney, and Dawn Ray, The number of oriented rational links with a given deficiency number, arXiv:2007.02819 [math.GT], 2020. See p. 16.
A. M. Hinz, S. Klavžar, U. Milutinović, and C. Petr, The Tower of Hanoi - Myths and Maths, Birkhäuser 2013. See page 183. Book's website
S. Hong and J. H. Kwak, Regular fourfold covering with respect to the identity automorphism, J. Graph Theory, 17 (1993), 621-627.
Masashi Kosuda and Manabu Oura, Centralizer algebras of the primitive unitary reflection group of order 96, arXiv:1505.00318 [math.RT], 2015.
George S. Lueker, Improved Bounds on the Average Length of Longest Common Subsequences (Jul 22, 2005) (Fig.1).
N. Moreira and R. Reis, On the Density of Languages Representing Finite Set Partitions, Journal of Integer Sequences, Vol. 8 (2005), Article 05.2.8.
C. Segovia, Unexpected relations of cobordism categories with another [sic] subjects in mathematics, 2013.
C. Segovia, The classifying space of the 1+1 dimensional G-cobordism category, arXiv:1211.2144 [math.AT], 2012-2019.
C. Segovia, Numerical computations in cobordism categories, arXiv:1307.2850 [math.AT], 2013.
C. Segovia and M. Winklmeier, Combinatorial Computations in Cobordism Categories, arXiv preprint arXiv:1409.2067 [math.CO], 2014-2015.
C. Segovia and M. Winklmeier, On the density of certain languages with p^2 letters, Electronic Journal of Combinatorics 22(3) (2015), #P3.16.
Index entries for linear recurrences with constant coefficients, signature (7,-14,8).

Cf. A056272, A056273, A007051, A000392, A056450, A028401, A060919.

A row of the array in A278984.

List([0..30],n->(2^n+1)*(2^n+2)/6); # Muniru A Asiru, Aug 09 2018

[(2^n+1)*(2^n+2)/6: n in [0..25]]; // Vincenzo Librandi, Aug 09 2018

A007581:=n->(2^n+1)*(2^n+2)/6; seq(A007581(n), n=0..50); # Wesley Ivan Hurt, Nov 25 2013

Table[(3*2^(n-1)+2^(2n-1)+1)/3,{n,0,30}] (* or *) LinearRecurrence[ {7,-14,8},{1,2,5},31] (* Harvey P. Dale, Jul 24 2011 *)
CoefficientList[Series[(1 - 5 x + 5 x^2) / ((1 - x) (1 - 2 x) (1 - 4 x)), {x, 0, 33}], x] (* Vincenzo Librandi, Aug 09 2018 *)

a(n)=(3*2^(n-1)+2^(2*n-1)+1)/3; \\ Charles R Greathouse IV, Jun 24 2011

a(n)=if(n==0,1,(2<<(2*n--))\/3+2^n); \\ Charles R Greathouse IV, Jun 24 2011

a(n) = (3*2^(n-1) + 2^(2*n-1) + 1)/3.
a(n) = Sum_{k=1..4} Stirling2(n, k). - Winston Yang (winston(AT)cs.wisc.edu), Aug 23 2000
Binomial transform of 3^n/6 + 1/2 + 0^n/3, i.e., of A007051 with an extra leading 1. a(n) = binomial(2^n+2, 2^n-1)/2^n. - Paul Barry, Jul 19 2003
a(n) = C(2+2^n, 3)/2^n = a(n-1) + 2^(n-1) + 4^(n-3/2) = A092055(n)/A000079(n). - Henry Bottomley, Feb 19 2004
Second binomial transform of A001045(n-1) + 0^n/2. G.f.: (1-5*x+5*x^2)/((1-x)*(1-2*x)*(1-4*x)). - Paul Barry, Apr 28 2004
a(n) is the top entry of the vector M^n*[1,1,1,1,0,0,0,...], where M is an infinite bidiagonal matrix with M(r,r)=r, r >= 1, as the main diagonal, M(r,r+1)=1, and the rest zeros. ([1,1,1,...] is a column vector and transposing gives the same in terms of a leftmost column term.) - Gary W. Adamson, Jun 24 2011
a(0)=1, a(1)=2, a(2)=5, a(n) = 7*a(n-1) - 14*a(n-2) + 8*a(n-3). - Harvey P. Dale, Jul 24 2011
E.g.f.: (exp(2*x) + 1/3*exp(4*x) + 2/3*exp(x))/2 = G(0)/2; G(k)=1 + (2^k)/(3 - 6/(2 + 4^k - 3*x*(8^k)/(3*x*(2^k) + (k+1)/G(k+1)))); (continued fraction). - Sergei N. Gladkovskii, Dec 08 2011

A032121 Number of reversible strings with n beads of 4 colors.

Original entry on oeis.org

1, 4, 10, 40, 136, 544, 2080, 8320, 32896, 131584, 524800, 2099200, 8390656, 33562624, 134225920, 536903680, 2147516416, 8590065664, 34359869440, 137439477760, 549756338176, 2199025352704, 8796095119360, 35184380477440, 140737496743936, 562949986975744

Offset: 0

Christian G. Bower

Also the number of 4-ary strings of length m = n+1 with number of 1's, 2's and 3's all even. Bijective proof, anyone? - Frank Ruskey, Jul 14 2002

			a(2) = 10 = |{000, 110,101,011, 220,202,022, 330,303,033}|.

Colin Barker, Table of n, a(n) for n = 0..1000
Christian Barrientos, Sarah Minion, Series-Parallel Operations with Alpha-Graphs, Theory and Applications of Graphs (2019) Vol. 6, Issue 1, Article 4.
C. G. Bower, Transforms (2)
Index entries for linear recurrences with constant coefficients, signature (4,4,-16).

Column 4 of A277504.

Cf. A000302 (oriented), A032087(n>1) (chiral), A056450 (achiral).

k = 4; Table[(k^n + k^Ceiling[n/2])/2, {n, 0, 30}] (* Robert A. Russell, Nov 25 2017 *)
LinearRecurrence[{4, 4, -16}, {1, 4, 10}, 31] (* Robert A. Russell, Nov 10 2018 *)
CoefficientList[Series[1/4 E^(-2 x) (-1 + 3 E^(4 x) + 2 E^(6 x)), {x, 0, 20}], x]*Table[n!, {n, 0, 20}] (* Stefano Spezia, Nov 12 2018 *)

Vec((1-10*x^2) / ((1 - 2*x)*(1 + 2*x)*(1 - 4*x)) + O(x^40)) \\ Colin Barker, Nov 25 2017

"BIK" (reversible, indistinct, unlabeled) transform of 4, 0, 0, 0, ...
a(n) = (4^m+3*2^m+(-2)^m)/8, where m = n+1. - Frank Ruskey, Jul 14 2002
G.f.: (1-10x^2) / ((1-4x)*(1-4x^2)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009; corrected by R. J. Mathar, Sep 16 2009 [Adapted to offset 0 by Robert A. Russell, Nov 10 2018]
From Colin Barker, Nov 25 2017: (Start)
a(n) = 2^(n-2) * (3 + (-1)^(1+n) + 2^(1+n)).
a(n) = 4*a(n-1) + 4*a(n-2) - 16*a(n-3) for n>2.
(End)
a(n) = (4^n + 4^floor((n+1)/2)) / 2 = (A000302(n) + A056450(n)) / 2. - Robert A. Russell and Danny Rorabaugh, Jun 22 2018
E.g.f.: (1/4)*exp(-2*x)*(- 1 + 3*exp(4*x) + 2*exp(6*x)). - Stefano Spezia, Nov 12 2018

a(0) = 1 prepended by Robert A. Russell, Nov 10 2018

A213173 a(n) = 4^floor(n/2), Powers of 4 repeated.

Original entry on oeis.org

1, 1, 4, 4, 16, 16, 64, 64, 256, 256, 1024, 1024, 4096, 4096, 16384, 16384, 65536, 65536, 262144, 262144, 1048576, 1048576, 4194304, 4194304, 16777216, 16777216, 67108864, 67108864, 268435456, 268435456, 1073741824, 1073741824, 4294967296, 4294967296

Offset: 0

Philippe Deléham, Apr 14 2013

1, followed by A056450. - Joerg Arndt, Sep 17 2013
Binomial transform of A084567.
Unsigned version of A164111.
Also, number of walks of length n on the star S_4 = K_1,4 starting at a specific vertex of degree 1. - Sean A. Irvine, Jun 03 2025

G. C. Greubel, Table of n, a(n) for n = 0..1000
Sean A. Irvine, Walks on Graphs.
Wikipedia, Star (graph theory)
Index entries for linear recurrences with constant coefficients, signature (0,4).

Cf. A000302, A056450, A164111.

LinearRecurrence[{0, 4}, {1, 1}, 40] (* T. D. Noe, Apr 17 2013 *)
CoefficientList[Series[(1 + x)/(1 - 4*x^2), {x,0,50}], x] (* G. C. Greubel, Apr 30 2017 *)
With[{p4=4^Range[0,30]},Riffle[p4,p4]] (* Harvey P. Dale, Mar 17 2022 *)

a(n)=4^(n\2) \\ Charles R Greathouse IV, Oct 03 2016

O.g.f.: (1+x)/(1-4*x^2).
a(n) = A016116(n)^2.
a(n) = 4*a(n-2) with a(0) = a(1) = 1.
a(n) = a(n-1)*a(n-2)/a(n-3) = 4^A004526(n).
a(n) = sum(A152815(n,k)*3^k, 0<=k<=n). - Philippe Deléham, Apr 22 2013

A056455 Palindromes using exactly four different symbols.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 24, 24, 240, 240, 1560, 1560, 8400, 8400, 40824, 40824, 186480, 186480, 818520, 818520, 3498000, 3498000, 14676024, 14676024, 60780720, 60780720, 249401880, 249401880, 1016542800, 1016542800, 4123173624, 4123173624, 16664094960, 16664094960

Offset: 1

Marks R. Nester

M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Index entries for linear recurrences with constant coefficients, signature (1,9,-9,-26,26,24,-24).

Cf. A056450, A000919.

k=4; Table[k! StirlingS2[Ceiling[n/2],k],{n,1,30}] (* Robert A. Russell, Sep 25 2018 *)

a(n) = 4!*stirling((n+1)\2, 4, 2); \\ Altug Alkan, Sep 25 2018

a(n) = 4! * Stirling2( [(n+1)/2], 4).
G.f.: 24*x^7/((1-x)*(1-2*x)*(1+2*x)*(1-2*x^2)*(1-3*x^2)). - Colin Barker, May 05 2012
G.f.: k!(x^(2k-1)+x^(2k))/Product_{i=1..k}(1-ix^2), where k=4 is the number of symbols. - Robert A. Russell, Sep 25 2018

A056486 a(n) = (9*2^n + (-2)^n)/4 for n>0.

Original entry on oeis.org

4, 10, 16, 40, 64, 160, 256, 640, 1024, 2560, 4096, 10240, 16384, 40960, 65536, 163840, 262144, 655360, 1048576, 2621440, 4194304, 10485760, 16777216, 41943040, 67108864, 167772160, 268435456, 671088640, 1073741824, 2684354560, 4294967296, 10737418240

Offset: 1

Marks R. Nester

Old name was: "Number of periodic palindromes using a maximum of four different symbols".

Number of necklaces with n beads and 4 colors that are the same when turned over and hence have reflection symmetry. - Herbert Kociemba, Nov 24 2016

			G.f. = 4*x + 10*x^2 + 16*x^3 + 40*x^4 + 64*x^5 + 160*x^6 + 256*x^7 + 640*x^8 + ...
For example, aaabbb is not a (finite) palindrome but it is a periodic palindrome.

M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,4).

Cf. A029744, A038754, A056450.

Column 4 of A284855.

[(9*2^n + (-2)^n)/4 : n in [1..50]]; // Wesley Ivan Hurt, Nov 24 2016

A056486:=n->(9*2^n + (-2)^n)/4: seq(A056486(n), n=1..50); # Wesley Ivan Hurt, Nov 24 2016

CoefficientList[Series[-1+(1+4*x+6*x^2)/(1-4*x^2),{x,0,30}],x] (* Herbert Kociemba, Nov 24 2016 *)
k=4; Table[(k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2]) / 2, {n, 1, 30}] (* Robert A. Russell, Sep 21 2018 *)

a(n) = (9*2^n+(-2)^n)/4; \\ Altug Alkan, Sep 21 2018

[2^(n-2)*(9+(-1)^n) for n in range(1,51)] # G. C. Greubel, Mar 23 2024

a(n) = 4^((n+1)/2) for n odd, a(n) = 4^(n/2)*5/2 for n even.
From Colin Barker, Jul 08 2012: (Start)
a(n) = 4*a(n-2).
G.f.: 2*x*(2+5*x)/((1-2*x)*(1+2*x)). (End)
G.f.: -1 + (1+4*x+6*x^2)/(1-4*x^2). - Herbert Kociemba, Nov 24 2016
E.g.f.: 5*sinh(x)^2 + 2*sinh(2*x). - Ilya Gutkovskiy, Nov 24 2016
a(n) = ( 4^floor((n+1)/2) + 4^ceiling((n+1)/2) )/2. - Robert A. Russell, Sep 21 2018

Better name from Ralf Stephan, Jul 18 2013

A113249 Corresponds to m = 3 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4a(n-4) + (2m)^2a(n-3) - 4a(n-1), a(m,0) = -1, a(m,1) = 4, a(m,2) = -13 + 6(m-1) + 3(m-1)^2, a(m,3) = (-8+m^2)^2.

Original entry on oeis.org

-1, 4, 11, 1, 59, 484, -1009, 6241, -2761, 13924, 87251, 57121, 49139, 4072324, -7165609, 35058241, 10350959, 30492484, 559712411, 973502401, -1957852501, 30450948004, -41421000289, 174055005601, 241428053159, 9658565284, 2872244917091, 11300885699041, -25300162140061

Offset: 0

Creighton Dement, Oct 20 2005

Conjecture: a(m, 2*n+1) is a perfect square for all m, n. Disregarding signs and/or initial term, we have: m = 0 (A000302), m = 1 (A097948), m = 2 (A056450), m = 3 (a(n)), m = 4 (A113250), m = 5 (A113251), m = 6 (A113252), m = 7 (A113253), m = 8 (A113254), m = 9 (A113255), m = 10 (A113256).

In this case, a(2n+1) = b(n)^2 where b(n) = Re((2+sqrt(-5))^(n+1)) satisfies the recurrence b(n) = 4*b(n-1) - 9*b(n-2) with b(0)=2, b(1)=-1. - Robert Israel, Oct 23 2017

			a(3, 13) = 93161710957356599364/((-2+i*sqrt(5))^14*(2+i*sqrt(5))^14) = 4072324 = 2^2*1009^2.

Colin Barker, Table of n, a(n) for n = 0..1000
Robert Munafo, Sequences Related to Floretions
Index entries for linear recurrences with constant coefficients, signature (-4,0,36,81).

Cf. A000302, A097948, A056450, A113250, A113251, A113252, A113253, A113254, A113255, A113256.

Cf. A176333, A190967.

f:= gfun:-rectoproc({a(n) = 81*a(n-4)+36*a(n-3)-4*a(n-1),a(0) = -1, a(1) = 4, a(2) = 11, a(3) = 1},a(n),remember):
map(f, [$0..30]); # Robert Israel, Oct 23 2017

LinearRecurrence[{-4, 0, 36, 81}, {-1, 4, 11, 1}, 29] (* Jean-François Alcover, Sep 25 2017 *)

Vec(-(1 - 27*x^2 - 81*x^3) / ((1 - 3*x)*(1 + 3*x)*(1 + 4*x + 9*x^2)) + O(x^30)) \\ Colin Barker, May 19 2019

G.f.: (-1+27*x^2+81*x^3)/((-3*x+1)*(3*x+1)*(9*x^2+4*x+1)).
a(2k+1) = (2*A176333(k)-3*A190967(k))^2. - Robert Israel, Oct 23 2017
a(n) = -4*a(n-1) + 36*a(n-3) + 81*a(n-4) for n>3. - Colin Barker, May 19 2019
a(n) = 3^(n+1)*(1 - (-1)^n + 2*cos(arccos(-2/3)*(n+1)))/4. - Eric Simon Jacob, Aug 06 2023

A113250 Expansion of g.f. -(1 - 48x^2 - 256x^3) / ((1 - 4x)(1 + 4x)(1 + 4x + 16x^2)).

Original entry on oeis.org

-1, 4, 32, 64, -256, 4096, -4096, 16384, 131072, 262144, -1048576, 16777216, -16777216, 67108864, 536870912, 1073741824, -4294967296, 68719476736, -68719476736, 274877906944, 2199023255552, 4398046511104, -17592186044416, 281474976710656, -281474976710656

Offset: 0

Creighton Dement, Nov 18 2005

Previous name was: Corresponds to m = 4 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4*a(n-4) + (2*m)^2*a(n-3) - 4*a(m-1), a(m,0) = -1, a(m,1) = 4, a(m,2) = -13 + 6*(m-1) + 3*(m-1)^2, a(m,3) = (-8+m^2)^2.

Conjecture: a(m, 2*n+1) is a perfect square for all m (see A113249). Initial terms factored (without regards to sign): 1, 4, (2)^5, (2)^6,(2)^8, (2)^12, (2)^12, (2)^14, (2)^17, (2)^18, (2)^20, (2)^24, (2)^24, (2)^26, (2)^29, (2)^30, (2)^32, (2)^36.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-4,0,64,256).

Cf. A000302, A097948, A056450, A113249, A113251, A113252, A113253, A113254, A113255, A113256.

LinearRecurrence[{-4, 0, 64, 256}, {-1, 4, 32, 64}, 25] (* Robert P. P. McKone, Aug 25 2023 *)
CoefficientList[Series[-(1-48x^2-256x^3)/((1-4x)(1+4x)(1+4x+16x^2)),{x,0,30}],x] (* Harvey P. Dale, Aug 27 2025 *)

Vec(-(1 - 48*x^2 - 256*x^3) / ((1 - 4*x)*(1 + 4*x)*(1 + 4*x + 16*x^2)) + O(x^25)) \\ Colin Barker, May 19 2019

G.f.: -(1 - 48*x^2 - 256*x^3) / ((1 - 4*x)*(1 + 4*x)*(1 + 4*x + 16*x^2)). Corrected by Colin Barker, May 19 2019

New name using g.f. from Joerg Arndt, Aug 25 2023

A113251 Corresponds to m = 5 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4a(n-4) + (2m)^2a(n-3) - 4a(m-1), a(m,0) = -1, a(m,1) = 4, a(m,2) = -13 + 6(m-1) + 3(m-1)^2, a(m,3) = (-8+m^2)^2.

Original entry on oeis.org

-1, 4, 59, 289, -1381, 13924, 10079, 2209, 520439, 7628644, -23994301, 149401729, 490531859, 406344964, -1681645081, 149155846849, -249406479121, 1083427010884, 9530848465739, 30158362505569, -168169798384501, 2302905921914404, -239007146013841, 2988025311585889

Offset: 0

Creighton Dement, Nov 18 2005

Conjecture: a(m, 2*n+1) is a perfect square for all m,n (see A113249).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-4,0,100,625).

Cf. A000302, A097948, A056450, A113249, A113250, A113252, A113253, A113254, A113255, A113256.

with(gfun): seriestolist(series((-1+75*x^2+625*x^3)/((5*x+1)*(1-5*x)*(25*x^2+4*x+1)), x=0,25));

LinearRecurrence[{-4,0,100,625},{-1,4,59,289},40] (* Harvey P. Dale, Jul 05 2021 *)

Vec(-(1 - 75*x^2 - 625*x^3) / ((1 - 5*x)*(1 + 5*x)*(1 + 4*x + 25*x^2)) + O(x^30)) \\ Colin Barker, May 20 2019

G.f.: (-1+75*x^2+625*x^3) / ((5*x+1)*(1-5*x)*(25*x^2+4*x+1)).
a(n) = -4*a(n-1) + 100*a(n-3) + 625*a(n-4) for n>3. - Colin Barker, May 20 2019
a(n) = 5^(n+1)*(1 - (-1)^n + 2*cos(arccos(-2/5)*(n+1)))/4. - Eric Simon Jacob, Jul 29 2023

cp's OEIS Frontend

A164906 Duplicate of A056450.

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A007581 a(n) = (2^n+1)*(2^n+2)/6.

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A032121 Number of reversible strings with n beads of 4 colors.

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A213173 a(n) = 4^floor(n/2), Powers of 4 repeated.

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A056455 Palindromes using exactly four different symbols.

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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).

A113249 Corresponds to m = 3 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4a(n-4) + (2m)^2a(n-3) - 4a(n-1), a(m,0) = -1, a(m,1) = 4, a(m,2) = -13 + 6(m-1) + 3(m-1)^2, a(m,3) = (-8+m^2)^2.

A113250 Expansion of g.f. -(1 - 48x^2 - 256x^3) / ((1 - 4x)(1 + 4x)(1 + 4x + 16x^2)).

A113251 Corresponds to m = 5 in a family of 4th-order linear recurrence sequences given by a(m,n) = m^4a(n-4) + (2m)^2a(n-3) - 4a(m-1), a(m,0) = -1, a(m,1) = 4, a(m,2) = -13 + 6(m-1) + 3(m-1)^2, a(m,3) = (-8+m^2)^2.