A056830 -id:A056830 - cp's OEIS frontend

A085815 Least k such that A056830(n) + k is prime, where A056830 = alternate digits 1 and 0.

2, 1, 1, 0, 3, 2, 11, 28, 13, 12, 29, 62, 1, 10, 1, 30, 27, 40, 73, 42, 63, 90, 29, 8, 61, 120, 37, 20, 41, 20, 1, 66, 283, 66, 27, 146, 1, 222, 1, 8, 77, 190, 173, 18, 1, 50, 149, 50, 29, 66, 31, 26, 23, 10, 29, 150, 99, 330, 81, 356, 53, 102, 7, 126, 123, 10, 227, 526, 117, 96

Offset: 0

Jason Earls, Jul 25 2003

			a(7)=28 because 1010101 + 28 = 1010129, a prime.

Harvey P. Dale, Table of n, a(n) for n = 0..500

Table[Module[{k=0,d=FromDigits[PadRight[{},n,{1,0}]]},While[!PrimeQ[d+k],k++];k],{n,0,70}] (* Harvey P. Dale, Jul 14 2025 *)

A086696 Least k such that A056830(n) + k is semiprime, where A056830 = alternate digits 1 and 0.

Original entry on oeis.org

3, 0, 5, 1, 1, 1, 1, 1, 5, 3, 2, 5, 6, 21, 2, 9, 2, 8, 4, 1, 5, 9, 34, 23, 4, 12, 5, 8, 17, 9, 10, 1, 12, 11, 10, 3, 0, 37, 26, 3, 20, 29, 16, 21, 38, 41, 18, 11, 6, 21, 8, 7, 2, 31, 30, 12, 38, 69, 2, 27

Offset: 1

Jason Earls, Jul 28 2003

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.
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).
Paul K. Stockmeyer, An Exploration of Sequence A000975, arXiv:1608.08245 [math.CO], 2016; Fib. Quart. 55 (2017) 174.
Eric Weisstein's World of Mathematics, Baguenaudier
A. K. Whitford, Binet's Formula Generalized, Fibonacci Quarterly, Vol. 15, No. 1, 1979, pp. 21, 24, 29.
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

A128174 Transform, (1,0,1,...) in every column.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1

Offset: 1

Gary W. Adamson, Feb 17 2007

Inverse of the triangle = a tridiagonal matrix with (1,1,1,...) in the superdiagonal, (0,0,0,...) in the main diagonal and (-1,-1,-1,...) in the subdiagonal.
Riordan array (1/(1-x^2), x) with inverse (1-x^2,x). - Paul Barry, Sep 10 2008
The position of 1's in this sequence is equivalent to A246705, and the position of 0's is equivalent to A246706. - Bernard Schott, Jun 05 2019

			First few rows of the triangle are:
  1;
  0, 1;
  1, 0, 1;
  0, 1, 0, 1;
  1, 0, 1, 0, 1; ...

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened

Cf. A004526 (row sums).

a128174 n k = a128174_tabl !! (n-1) !! (k-1)
a128174_row n = a128174_tabl !! (n-1)
a128174_tabl = iterate (\xs@(x:_) -> (1 - x) : xs) [1]
-- Reinhard Zumkeller, Aug 01 2014

[[(1+(-1)^(n-k))/2: k in [1..n]]: n in [1..12]]; // G. C. Greubel, Jun 05 2019

A128174 := proc(n,k)
    if k > n or k < 1 then
        0;
    else
        modp(k+n+1,2) ;
    end if;
end proc: # R. J. Mathar, Aug 06 2016

a128174[r_] := Table[If[EvenQ[n+k], 1, 0], {n, 1, r}, {k, 1, n}]
TableForm[a128174[5]] (* triangle *)
Flatten[a128174[10]] (* data *) (* Hartmut F. W. Hoft, Mar 15 2017 *)
Table[(1+(-1)^(n-k))/2, {n,1,12}, {k,1,n}]//Flatten (* G. C. Greubel, Sep 26 2017 *)

for(n=1,12, for(k=1,n, print1((1+(-1)^(n-k))/2, ", "))) \\ G. C. Greubel, Sep 26 2017

[[(1+(-1)^(n-k))/2 for k in (1..n)] for n in (1..12)] # G. C. Greubel, Jun 05 2019

A lower triangular matrix transform, (1, 0, 1, ...) in every column; n terms of (1, 0, 1, ...) in odd rows; n terms of (0, 1, 0, ...) in even rows.
T(n,k) = [k<=n]*(1+(-1)^(n-k))/2. - Paul Barry, Sep 10 2008
With offset n=1, k=0: Sum_{k=0..n} {T(n,k)*x^k} = A000035(n), A004526(n+1), A000975(n), A033113(n), A033114(n), A033115(n), A033116(n), A033117(n), A033118(n), A033119(n), A056830(n+1) for x=0,1,2,3,4,5,6,7,8,9,10 respectively. - Philippe Deléham, Oct 17 2011
T(n+1,1) = 1 - T(n,1); T(n+1,k) = T(n,k-1), 1 < k <= n+1. - Reinhard Zumkeller, Aug 01 2014

A094028 Expansion of 1/((1-x)(1-100x)).

Original entry on oeis.org

1, 101, 10101, 1010101, 101010101, 10101010101, 1010101010101, 101010101010101, 10101010101010101, 1010101010101010101, 101010101010101010101, 10101010101010101010101, 1010101010101010101010101, 101010101010101010101010101, 10101010101010101010101010101

Offset: 0

Paul Barry, Apr 22 2004

Regarded as binary numbers and converted to decimal, these become 1,5,21,85,... the partial sums of 4^n (see A002450).
Partial sums of 100^n.
Odd terms of A056830. - Alexandre Wajnberg, May 31 2005
101 is the only term that is prime, since (100^k-1)/99 = (10^k+1)/11 * (10^k-1)/9. When k is odd and not 1, (10^k+1)/11 is an integer > 1 and thus (100^k-1)/99 is nonprime. When k is even and greater than 2, (100^k-1)/99 has the prime factor 101 and is nonprime. - Felix Fröhlich, Oct 17 2015
Previous comment is the answer to the problem A1 proposed during the 50th Putnam Competition in 1989 (link). - Bernard Schott, Mar 24 2023

			From _Omar E. Pol_, Dec 13 2008: (Start)
=======================
n ....... a(n)
0 ........ 1
1 ....... 101
2 ...... 10101
3 ..... 1010101
4 .... 101010101
5 ... 10101010101
======================
(End)

Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 60.
Stephen Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Robert Price, Table of n, a(n) for n = 0..999
Kiran S. Kedlaya, The 50th William Lowell Putnam Mathematical Competition, Problem A1, Dec 02 1989.
J. V. Leyendekkers and A.G. Shannon, Modular Rings and the Integer 3, Notes on Number Theory & Discrete Mathematics, 17 (2011), pp. 47-51.
Robert Price, Comments on A094028 concerning Elementary Cellular Automata, Feb 21 2016.
Eric Weisstein's World of Mathematics, Elementary Cellular Automaton.
Stephen Wolfram, A New Kind of Science.
Index to Elementary Cellular Automata.
Index entries for sequences related to cellular automata.
Index entries for linear recurrences with constant coefficients, signature (101,-100).
Index to sequences related to Olympiads and other Mathematical Competitions.

Bisection of A147759. [Omar E. Pol, Nov 13 2008]
Cf. A002450, A056830, A094027.
Cf. similar sequences of the form (k^n-1)/(k-1) listed in A269025.

[1+100*(100^n-1)/99 : n in [0..15]]; // Wesley Ivan Hurt, Oct 17 2015

A094028:=n->1+100*(100^n-1)/99: seq(A094028(n), n=0..15); # Wesley Ivan Hurt, Oct 17 2015

CoefficientList[Series[1/((1-x)(1-100x)),{x,0,20}],x] (* or *) Table[ FromDigits[ PadRight[{},2n-1,{1,0}]],{n,20}] (* or *) LinearRecurrence[ {101,-100},{1,101},20] (* or *) NestList[100#+1&,1,20] (* Harvey P. Dale, Apr 27 2015 *)

A094028(n):=1+100*(100^n-1)/99$
makelist(A094028(n),n,0,30); /* Martin Ettl, Nov 06 2012 */

a(n) = 1+100*(100^n-1)/99 \\ Felix Fröhlich, Oct 17 2015

Vec(1/((1-x)*(1-100*x)) + O(x^100)) \\ Altug Alkan, Oct 17 2015

G.f.: 1/((1-x)*(1-100*x)).
a(n) = 1 + 100*(100^n-1)/99. - N. J. A. Sloane, Apr 20 2008
a(n) = 100^(n+1)/99 - 1/99.
a(n) = A094027(2*n+1).
a(n) = 100*a(n-1) + 1, a(0) = 1. - Philippe Deléham, Feb 22 2014
a(n) = 101*a(n-1) - 100*a(n-2) for n > 1. - Wesley Ivan Hurt, Oct 17 2015
a(n) = (100^(n+1) - 1)/99. - Bernard Schott, Apr 15 2021
E.g.f.: exp(x)*(100*exp(99*x) - 1)/99. - Elmo R. Oliveira, Mar 06 2025

A030141 Numbers in which parity of the decimal digits alternates.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 21, 23, 25, 27, 29, 30, 32, 34, 36, 38, 41, 43, 45, 47, 49, 50, 52, 54, 56, 58, 61, 63, 65, 67, 69, 70, 72, 74, 76, 78, 81, 83, 85, 87, 89, 90, 92, 94, 96, 98, 101, 103, 105, 107, 109, 121, 123, 125, 127, 129

Offset: 1

Patrick De Geest

An alternating integer is a positive integer for which, in base-10, the parity of its digits alternates.
The number of terms < 10^n (n>=0): 1, 10, 55, 280, 1405, 7030, 35155, ..., . - Robert G. Wilson v, Apr 01 2011
The number of terms between 10^n and 10^(n+1) is 9 * 5^n for n>=0. For n>=0, number of terms < 10^n is 1 + 9 * (5^n-1)/4. - Franklin T. Adams-Watters, Apr 01 2011
A228710(a(n)) = 1. - Reinhard Zumkeller, Aug 31 2013

			121 is alternating and in the sequence because its consecutive digits are odd-even-odd, 1 being odd and 2 even. Of course, 1234567890 is also alternating.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
45th International Mathematical Olympiad (45th IMO), Problem #6 and Solution, Mathematics Magazine, 78 (2005), pp. 247, 250, 251.
Index entries for 10-automatic sequences.

Complement: A228709.
Subsequences: A030142, A030143, A030144, A030147, A030152, A062285.
Cf. A110303, A110304, A110305, A056830, A103181, A228722, A228723.

a030141 n = a030141_list !! (n-1)
a030141_list = filter ((== 1) . a228710) [0..]
-- Reinhard Zumkeller, Aug 31 2013

fQ[n_] := Block[{m = Mod[ IntegerDigits@ n, 2]}, m == Split[m, UnsameQ][[1]]]; Select[ Range[0, 130], fQ] (* Robert G. Wilson v, Apr 01 2011 *)
Select[Range[0,150],FreeQ[Differences[Boole[EvenQ[IntegerDigits[#]]]],0]&] (* Harvey P. Dale, Jul 19 2025 *)

is(n,d=digits(n))=for(i=2,#d, if((d[i]-d[i-1])%2==0, return(0))); 1 \\ Charles R Greathouse IV, Jul 08 2022

from itertools import count
def A030141_gen(startvalue=0): # generator of terms >= startvalue
    return filter(lambda n:all(int(a)+int(b)&1 for a, b in zip(str(n),str(n)[1:])),count(max(startvalue,0)))
A030141_list = list(islice(A030141_gen(),30)) # Chai Wah Wu, Jul 12 2022

from itertools import chain, count, islice
def altgen(seed, digits):
    allowed = "02468" if seed in "13579" else "13579"
    if digits == 1: yield from allowed; return
    for f in allowed: yield from (f + r for r in altgen(f, digits-1))
def agen(): yield from chain(range(10), (int(f+r) for d in count(2) for f in "123456789" for r in altgen(f, d-1)))
print(list(islice(agen(), 65))) # Michael S. Branicky, Jul 12 2022

Offset corrected by Reinhard Zumkeller, Aug 31 2013

A033113 Base-3 digits are, in order, the first n terms of the periodic sequence with initial period 1,0.

Original entry on oeis.org

1, 3, 10, 30, 91, 273, 820, 2460, 7381, 22143, 66430, 199290, 597871, 1793613, 5380840, 16142520, 48427561, 145282683, 435848050, 1307544150, 3922632451, 11767897353, 35303692060, 105911076180, 317733228541, 953199685623

Offset: 1

Clark Kimberling

Written in base 3, this yields A056830. - M. F. Hasler, Oct 05 2018

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 906
R. J. Mathar, Counting Walks on Finite Graphs, Nov 2020, Section 6.
Index entries for linear recurrences with constant coefficients, signature (3,1,-3).

Cf. A003462, A039300.

[Round(3^(n+1)/8): n in [1..30]]; // Vincenzo Librandi, Jun 25 2011

a[0]:=0: a[1]:=1: for n from 2 to 50 do a[n]:=2*a[n-1]+3*a[n-2]+1 od: seq(a[n], n=1..33);# Zerinvary Lajos, Dec 14 2008
g:=x*(1/(1-3*x)/(1-x))/(1+x): gser:=series(g, x=0, 43): seq(coeff(gser, x, n), n=1..30);# Zerinvary Lajos, Jan 11 2009
A033113 := proc(n) (9*3^(n-1)-(-1)^n-2)/8 ; end proc: # R. J. Mathar, Jan 08 2011

Join[{a=1,b=3},Table[c=2*b+3*a+1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2011 *)
Module[{nn=30,d},d=PadRight[{},nn,{1,0}];Table[FromDigits[Take[d,n],3],{n,nn}]] (* or *) LinearRecurrence[{3,1,-3},{1,3,10},30] (* Harvey P. Dale, May 24 2014 *)

a(n)=3^n*3\8 \\ Simplified by M. F. Hasler, Oct 06 2018

A033113(n)=3^(n+1)>>3 \\ M. F. Hasler, Oct 05 2018

a(n) = A039300(n)-1.
a(n)+a(n+1) = A003462(n+1).
a(n) = 3*a(n-1) + a(n-2) -3*a(n-3). - R. J. Mathar, Jun 28 2010
From Paul Barry, Nov 12 2003: (Start)
G.f.: x/((1-x)*(1+x)*(1-3*x)).
a(n) = 2*a(n-1) + 3*a(n-2) + 1.
Partial sums of A015518. (End)
E.g.f.: (1/2)*exp(x)*(sinh(x))^2. - Paul Barry, Mar 12 2003
a(n) = Sum_{k=0..floor(n/2)} C(n+2, 2k+2)*4^k. - Paul Barry, Aug 24 2003
a(n) = Sum_{k=0..floor(n/2)} 3^(n-2*k); a(n) = Sum_{k=0..n} Sum_{j=0..k} (-1)^(j+k)*3^j. - Paul Barry, Nov 12 2003
Convolution of A000244 and A059841 (3^n and periodic{1, 0}). a(n) = Sum_{k=0..n} (1 + (-1)^(n-k))*3^k/2. - Paul Barry, Jul 19 2004
a(n) = round(3^(n+1)/8) = floor((3^(n+1)-1)/8) = ceiling((3^(n+1)-3)/8) = round((3^(n+1)-3)/8). a(n) = a(n-2) + 3^(n-1), n > 2. - Mircea Merca, Dec 27 2010
a(n) = floor((3^(n+1))/4) / 2 = A081251(n)/2, n >= 1. - Wolfdieter Lang, Apr 13 2012

A032858 Numbers whose base-3 representation Sum_{i=0..m} d(i)*3^i has d(m) > d(m-1) < d(m-2) > ...

Original entry on oeis.org

0, 1, 2, 3, 6, 7, 10, 11, 19, 20, 23, 30, 33, 34, 57, 60, 61, 69, 70, 91, 92, 100, 101, 104, 172, 173, 181, 182, 185, 208, 209, 212, 273, 276, 277, 300, 303, 304, 312, 313, 516, 519, 520, 543, 546, 547, 555, 556, 624, 627, 628, 636, 637

Offset: 1

Clark Kimberling

Every other base-3 digit must be strictly less than its neighbors. - M. F. Hasler, Oct 05 2018

The terms can be generated in the following way: if A(n) are the terms with n digits in base 3, the terms with n+2 digits are obtained by prefixing them with '10' and with '20', and prefixing '21' to those starting with a digit '2'. It is easy to prove that #A(n) = A000045(n+2), since from the above we have #A(n+2) = 2*#A(n) + #A(n-1) = #A(n) + #A(n+1). (The #A(n-1) numbers starting with '2' are #A(n-2) numbers prefixed with '20' and #A(n-3) prefixed with '21'.) - M. F. Hasler, Oct 05 2018

			The base-3 representation of the initial terms is 0, 1, 2, 10, 20, 21, 101, 102, 201, 202, 212, 1010, 1020, 1021, 2010, 2020, 2021, 2120, 2121, 10101, 10102, ...

M. F. Hasler, Table of n, a(n) for n = 1..5000

Cf. A032859 .. A032865 for base-4 .. 10 variants.
Cf. A000975 (or A056830 in binary) for the base-2 analog.
Cf. A306105 for these terms written in base 3.

sdQ[n_]:=Module[{s=Sign[Differences[IntegerDigits[n, 3]]]}, s==PadRight[{}, Length[s], {-1, 1}]]; Select[Range[0, 700], sdQ] (* Vincenzo Librandi, Oct 06 2018 *)

is(n,b=3)=!for(i=2,#n=digits(n,b),(n[i-1]-n[i])*(-1)^i>0||return) \\ M. F. Hasler, Oct 05 2018

a(A000071(n+3)) = floor(3^(n+1)/8) = A033113(n). - M. F. Hasler, Oct 05 2018

Definition edited, cross-references and a(1) = 0 inserted by M. F. Hasler, Oct 05 2018

A068876 Smallest n-digit prime with property that digits alternate in parity.

Original entry on oeis.org

2, 23, 101, 2129, 10103, 210101, 1010129, 21010127, 101010167, 2101010147, 10101010163, 210101010187, 1010101010341, 21010101010147, 101010101010323, 2101010101010141, 10101010101010141, 210101010101010323, 1010101010101010143

Offset: 1

Amarnath Murthy, Mar 19 2002

			a(4) = 2129 as 2, 1, 2 and 9 have even and odd parity alternately.

Robert Israel, Table of n, a(n) for n = 1..700

Cf. A030144, A068877, A056830.

alp:= proc(n) local L,d;
L:= convert(n,base,10);
d:= nops(L);
if d::even then L:= L + map(op, [[0,1]$(d/2)]) else L:= L + map(op, [[0,1]$((d-1)/2),[0]]) fi;
nops(convert(L mod 2, set))=1
end proc:
f:= proc(d) local s;
  if d::even then s:= 2*10^(d-1)+(10^d-1)/99-1
  else s:= (10^(d+1)-1)/99-1
  fi;
  do s:= nextprime(s);
     if alp(s) then return s fi
  od
end proc:
seq(f(d),d=1..20); # Robert Israel, Aug 14 2018

fQ[n_] := Block[{m = Mod[ IntegerDigits@ n, 2]}, m == Split[m, UnsameQ][[1]]]; f[n_] := Block[{c = 1 + 100 (100^Ceiling[n/2 - 1] - 1)/99, k}, k = If[ OddQ@ n, c, 2*10^(n - 1) + c]; k = NextPrime[k - 1]; While[ !fQ@ k, k = NextPrime@ k]; k]; Array[f, 21] (* Robert G. Wilson v, Apr 01 2011 *)

concat = lambda x: Integer(''.join(map(str,x)),base=10)
def A068876(n):
    dd = {0:range(0,10,2), 1: range(1,10,2)}
    for d0 in [1..9]:
        if n % 2 == 0 and d0 % 2 == 1: continue # optimization
        ds = [dd[(d0+1+i) % 2] for i in range(n-1)]
        for dr in cartesian_product(ds):
            c = concat([d0]+dr)
            if is_prime(c): return c  # D. S. McNeil, Apr 02 2011

a(9)-a(13) corrected and a(14)-a(19) from Donovan Johnson, Apr 01 2011

A144864 a(n) = (4*16^(n-1)-1)/3.

Original entry on oeis.org

1, 21, 341, 5461, 87381, 1398101, 22369621, 357913941, 5726623061, 91625968981, 1466015503701, 23456248059221, 375299968947541, 6004799503160661, 96076792050570581, 1537228672809129301, 24595658764946068821, 393530540239137101141, 6296488643826193618261, 100743818301219097892181

Offset: 1

Artur Jasinski, Sep 23 2008

Old name was: A144863, read as binary numbers, converted to base 10.
All numbers in this sequence for n>1 are congruent to 5 mod 16. - Artur Jasinski, Sep 25 2008
From Omar E. Pol, Sep 10 2011: (Start)
It appears that this is a bisection of A002450.
It appears that this is a bisection of A084241.
It appears that this is a bisection of A153497.
It appears that this is a bisection of A088556, if n>=2.
(End)
All of the above is trivially true. - Joerg Arndt, Aug 19 2014
The aerated sequence (b(n))n>=1 = [1, 0, 21, 0, 341, 0, 5461, 0, 87381, ...] is a fourth-order linear divisibility sequence; that is, a(n) divides a(m) whenever n divides m. It is the case P1 = 0, P2 = -9, Q = -4 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Aug 26 2022

Vincenzo Librandi, Table of n, a(n) for n = 1..500
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
Index entries for linear recurrences with constant coefficients, signature (17,-16).

Cf. A001025, A002450, A013776, A056830, A084241, A088556, A094028, A098704, A131865, A135576, A144863, A153497.

Third quadrisection of Jacobsthal numbers A001045; the other quadrisections are A195156 (first), A139792 (second), and A141060 (fourth).

[16^n/12-1/3: n in [1..20]]; // Vincenzo Librandi, Aug 03 2011

Table[1/3 (-1 + 16^(n - 1)) + 16^(n - 1), {n, 1, 17}] (* Artur Jasinski, Sep 25 2008 *)
LinearRecurrence[{17,-16},{1,21},20] (* Harvey P. Dale, Jun 29 2022 *)

vector(66,n,(4*16^(n-1)-1)/3) \\ Joerg Arndt, Aug 19 2014

a(n) = 16^n/12 - 1/3; a(n) = 16*a(n-1) + 5, a(1)=1. - Artur Jasinski, Sep 25 2008
G.f.: x*(1+4*x) / ( (16*x-1)*(x-1) ). - R. J. Mathar, Jan 06 2011
a(n)=b such that Integral_{x=-Pi/2..Pi/2} (-1)^(n+1)*2^(2*n-3)*(cos((2*n-1)*x))/(5/4+sin(x)) dx = c+b*log(3). - Francesco Daddi, Aug 02 2011
a(n) = (2^(4*n-2)-1)/3. - Klaus Purath, Jan 31 2021
From Jianing Song, Aug 30 2022: (Start)
a(n) = A001045(4*n-2).
a(n+1) - a(n) = 10*A013776(n-1) = 20*A001025(n-1) for n >= 1.
a(n) = 10*A098704(n) + 1 = 20*A131865(n-2) + 1 for n >= 2. (End)
E.g.f.: (exp(16*x) - 4*exp(x) + 3)/12. - Stefano Spezia, Apr 18 2024

New name from Joerg Arndt, Aug 19 2014

cp's OEIS Frontend

A085815 Least k such that A056830(n) + k is prime, where A056830 = alternate digits 1 and 0.

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A086696 Least k such that A056830(n) + k is semiprime, where A056830 = alternate digits 1 and 0.

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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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A128174 Transform, (1,0,1,...) in every column.

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A094028 Expansion of 1/((1-x)*(1-100*x)).

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A030141 Numbers in which parity of the decimal digits alternates.

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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).

A094028 Expansion of 1/((1-x)(1-100x)).