A000111 -id:A000111 - 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
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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

A000364 Euler (or secant or "Zig") numbers: e.g.f. (even powers only) sec(x) = 1/cos(x).

Original entry on oeis.org

1, 1, 5, 61, 1385, 50521, 2702765, 199360981, 19391512145, 2404879675441, 370371188237525, 69348874393137901, 15514534163557086905, 4087072509293123892361, 1252259641403629865468285, 441543893249023104553682821, 177519391579539289436664789665

Offset: 0

N. J. A. Sloane

Inverse Gudermannian gd^(-1)(x) = log(sec(x) + tan(x)) = log(tan(Pi/4 + x/2)) = arctanh(sin(x)) = 2 * arctanh(tan(x/2)) = 2 * arctanh(csc(x) - cot(x)). - Michael Somos, Mar 19 2011
a(n) is the number of downup permutations of [2n]. Example: a(2)=5 counts 4231, 4132, 3241, 3142, 2143. - David Callan, Nov 21 2011
a(n) is the number of increasing full binary trees on vertices {0,1,2,...,2n} for which the leftmost leaf is labeled 2n. - David Callan, Nov 21 2011
a(n) is the number of unordered increasing trees of size 2n+1 with only even degrees allowed and degree-weight generating function given by cosh(t). - Markus Kuba, Sep 13 2014
a(n) is the number of standard Young tableaux of skew shape (n+1,n,n-1,...,3,2)/(n-1,n-2,...2,1). - Ran Pan, Apr 10 2015
Since cos(z) has a root at z = Pi/2 and no other root in C with a smaller |z|, the radius of convergence of the e.g.f. (intended complex-valued) is Pi/2 = A019669 (see also A028296). - Stanislav Sykora, Oct 07 2016
All terms are odd. - Alois P. Heinz, Jul 22 2018
The sequence starting with a(1) is periodic modulo any odd prime p. The minimal period is (p-1)/2 if p == 1 mod 4 and p-1 if p == 3 mod 4 [Knuth & Buckholtz, 1967, Theorem 2]. - Allen Stenger, Aug 03 2020
Conjecture: taking the sequence [a(n) : n >= 1] modulo an integer k gives a purely periodic sequence with period dividing phi(k). For example, the sequence taken modulo 21 begins [1, 5, 19, 20, 16, 2, 1, 5, 19, 20, 16, 2, 1, 5, 19, 20, 16, 2, 1, 5, 19, ...] with an apparent period of 6 = phi(21)/2. - Peter Bala, May 08 2023

			G.f. = 1 + x + 5*x^2 + 61*x^3 + 1385*x^4 + 50521*x^5 + 2702765*x^6 + 199360981*x^7 + ...
sec(x) = 1 + 1/2*x^2 + 5/24*x^4 + 61/720*x^6 + ...
From _Gary W. Adamson_, Jul 18 2011: (Start)
The first few rows of matrix M are:
   1,  1,  0,  0,  0, ...
   4,  4,  4,  0,  0, ...
   9,  9,  9,  9,  0, ...
  16, 16, 16, 16, 16, ... (End)

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Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
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Index entries for "core" sequences.
Index entries for sequences related to boustrophedon transform.

Cf. A000111, A000182, A011248, A019669, A034947, A060075, A013525, A000816, A002436, A000464, A002105, A002439, A079144, A158690.
Essentially same as A028296 and A122045.
First column of triangle A060074.
Two main diagonals of triangle A060058 (as iterated sums of squares).
Absolute values of row sums of A160485. - Johannes W. Meijer, Jul 06 2009
Left edge of triangle A210108, see also A125053, A076552. Cf. A255881.
Bisection (even part) of A317139.
The sequences [(-k^2)^n*Euler(2*n, 1/k), n = 0, 1, ...] are: A000007 (k=1), A000364 (k=2), |A210657| (k=3), A000281 (k=4), A272158 (k=5), A002438 (k=6), A273031 (k=7).

series(sec(x),x,40): SERIESTOSERIESMULT(%): subs(x=sqrt(y),%): seriestolist(%);
# end of program
A000364_list := proc(n) local S,k,j; S[0] := 1;
for k from 1 to n do S[k] := k*S[k-1] od;
for k from  1 to n do
    for j from k to n do
        S[j] := (j-k)*S[j-1]+(j-k+1)*S[j] od od;
seq(S[j], j=1..n)  end:
A000364_list(16);  # Peter Luschny, Apr 02 2012
A000364 := proc(n)
    abs(euler(2*n)) ;
end proc: # R. J. Mathar, Mar 14 2013

Take[ Range[0, 32]! * CoefficientList[ Series[ Sec[x], {x, 0, 32}], x], {1, 32, 2}] (* Robert G. Wilson v, Apr 23 2006 *)
Table[Abs[EulerE[2n]], {n, 0, 30}] (* Ray Chandler, Mar 20 2007 *)
a[ n_] := If[ n < 0, 0, With[{m = 2 n}, m! SeriesCoefficient[ Sec[ x], {x, 0, m}]]]; (* Michael Somos, Nov 22 2013 *)
a[ n_] := If[ n < 0, 0, With[{m = 2 n + 1}, m! SeriesCoefficient[ InverseGudermannian[ x], {x, 0, m}]]]; (* Michael Somos, Nov 22 2013 *)
a[n_] := Sum[Sum[Binomial[k, m] (-1)^(n+k)/(2^(m-1)) Sum[Binomial[m, j]* (2j-m)^(2n), {j, 0, m/2}] (-1)^(k-m), {m, 0, k}], {k, 1, 2n}]; Table[ a[n], {n, 0, 16}] (* Jean-François Alcover, Jun 26 2019, after Vladimir Kruchinin *)
a[0] := 1; a[n_] := a[n] = -Sum[a[n - k]/(2 k)!, {k, 1, n}]; Map[(-1)^# (2 #)! a[#] &, Range[0, 16]] (* Oliver Seipel, May 18 2024 *)

a(n):=sum(sum(binomial(k,m)*(-1)^(n+k)/(2^(m-1))*sum(binomial(m,j)*(2*j-m)^(2*n),j,0,m/2)*(-1)^(k-m),m,0,k),k,1,2*n); /* Vladimir Kruchinin, Aug 05 2010 */

a[n]:=if n=0 then 1 else sum(sum((i-k)^(2*n)*binomial(2*k, i)*(-1)^(i+k+n), i, 0, k-1)/ (2^(k-1)), k, 1, 2*n); makelist(a[n], n, 0, 16); /* Vladimir Kruchinin, Oct 05 2012 */

{a(n)=local(CF=1+x*O(x^n));if(n<0,return(0), for(k=1,n,CF=1/(1-(n-k+1)^2*x*CF));return(Vec(CF)[n+1]))} \\ Paul D. Hanna Oct 07 2005

{a(n) = if( n<0, 0, (2*n)! * polcoeff( 1 / cos(x + O(x^(2*n + 1))), 2*n))}; /* Michael Somos, Jun 18 2002 */

{a(n) = my(A); if( n<0, 0, n = 2*n+1 ; A = x * O(x^n); n! * polcoeff( log(1 / cos(x + A) + tan(x + A)), n))}; /* Michael Somos, Aug 15 2007 */

{a(n)=polcoeff(sum(m=0, n, (2*m)!/2^m * x^m/prod(k=1, m, 1+k^2*x+x*O(x^n))), n)} \\ Paul D. Hanna, Sep 20 2012

list(n)=my(v=Vec(1/cos(x+O(x^(2*n+1)))));vector(n,i,v[2*i-1]*(2*i-2)!) \\ Charles R Greathouse IV, Oct 16 2012

a(n)=subst(bernpol(2*n+1),'x,1/4)*4^(2*n+1)*(-1)^(n+1)/(2*n+1) \\ Charles R Greathouse IV, Dec 10 2014

a(n)=abs(eulerfrac(2*n)) \\ Charles R Greathouse IV, Mar 23 2022

\\ Based on an algorithm of Peter Bala, cf. link in A110501.
upto(n) = my(v1, v2, v3); v1 = vector(n+1, i, 0); v1[1] = 1; v2 = vector(n, i, i^2); v3 = v1; for(i=2, n+1, for(j=2, i-1, v1[j] += v2[i-j+1]*v1[j-1]); v1[i] = v1[i-1]; v3[i] = v1[i]); v3 \\ Mikhail Kurkov, Aug 30 2025

from functools import lru_cache
from math import comb
@lru_cache(maxsize=None)
def A000364(n): return 1 if n == 0 else (1 if n % 2 else -1)*sum((-1 if i % 2 else 1)*A000364(i)*comb(2*n,2*i) for i in range(n)) # Chai Wah Wu, Jan 14 2022

# after Mikhail Kurkov, based on an algorithm of Peter Bala, cf. link in A110501.
def euler_list(len: int) -> list[int]:
    if len == 0: return []
    v1 = [1] + [0] * (len - 1)
    v2 = [i**2 for i in range(1, len + 1)]
    result = [0] * len
    result[0] = 1
    for i in range(1, len):
        for j in range(1, i):
            v1[j] += v2[i - j] * v1[j - 1]
        v1[i] = v1[i - 1]
        result[i] = v1[i]
    return result
print(euler_list(1000))  # Peter Luschny, Aug 30 2025

# Algorithm of L. Seidel (1877)
# n -> [a(0), a(1), ..., a(n-1)] for n > 0.
def A000364_list(len) :
    R = []; A = {-1:0, 0:1}; k = 0; e = 1
    for i in (0..2*len-1) :
        Am = 0; A[k + e] = 0; e = -e
        for j in (0..i) : Am += A[k]; A[k] = Am; k += e
        if e < 0 : R.append(A[-i//2])
    return R
A000364_list(17) # Peter Luschny, Mar 31 2012

E.g.f.: Sum_{n >= 0} a(n) * x^(2*n) / (2*n)! = sec(x). - Michael Somos, Aug 15 2007
E.g.f.: Sum_{n >= 0} a(n) * x^(2*n+1) / (2*n+1)! = gd^(-1)(x). - Michael Somos, Aug 15 2007
E.g.f.: Sum_{n >= 0} a(n)*x^(2*n+1)/(2*n+1)! = 2*arctanh(cosec(x)-cotan(x)). - Ralf Stephan, Dec 16 2004
Pi/4 - [Sum_{k=0..n-1} (-1)^k/(2*k+1)] ~ (1/2)*[Sum_{k>=0} (-1)^k*E(k)/(2*n)^(2k+1)] for positive even n. [Borwein, Borwein, and Dilcher]
Also, for positive odd n, log(2) - Sum_{k = 1..(n-1)/2} (-1)^(k-1)/k ~ (-1)^((n-1)/2) * Sum_{k >= 0} (-1)^k*E(k)/n^(2*k+1), where E(k) is the k-th Euler number, by Borwein, Borwein, and Dilcher, Lemma 2 with f(x) := 1/(x + 1/2), h := 1/2 and then replace x with (n-1)/2. - Peter Bala, Oct 29 2016
Let M_n be the n X n matrix M_n(i, j) = binomial(2*i, 2*(j-1)) = A086645(i, j-1); then for n>0, a(n) = det(M_n); example: det([1, 1, 0, 0; 1, 6, 1, 0; 1, 15, 15, 1; 1, 28, 70, 28 ]) = 1385. - Philippe Deléham, Sep 04 2005
This sequence is also (-1)^n*EulerE(2*n) or abs(EulerE(2*n)). - Paul Abbott (paul(AT)physics.uwa.edu.au), Apr 14 2006
a(n) = 2^n * E_n(1/2), where E_n(x) is an Euler polynomial.
a(k) = a(j) (mod 2^n) if and only if k == j (mod 2^n) (k and j  are even). [Stern; see also Wagstaff and Sun]
E_k(3^(k+1)+1)/4 = (3^k/2)*Sum_{j=0..2^n-1} (-1)^(j-1)*(2j+1)^k*[(3j+1)/2^n] (mod 2^n) where k is even and [x] is the greatest integer function. [Sun]
a(n) ~ 2^(2*n+2)*(2*n)!/Pi^(2*n+1) as n -> infinity. [corrected by Vaclav Kotesovec, Jul 10 2021]
a(n) = Sum_{k=0..n} A094665(n, k)*2^(n-k). - Philippe Deléham, Jun 10 2004
Recurrence: a(n) = -(-1)^n*Sum_{i=0..n-1} (-1)^i*a(i)*binomial(2*n, 2*i). - Ralf Stephan, Feb 24 2005
O.g.f.: 1/(1-x/(1-4*x/(1-9*x/(1-16*x/(...-n^2*x/(1-...)))))) (continued fraction due to T. J. Stieltjes). - Paul D. Hanna, Oct 07 2005
a(n) = (Integral_{t=0..Pi} log(tan(t/2)^2)^(2n)dt)/Pi^(2n+1). - Logan Kleinwaks (kleinwaks(AT)alumni.princeton.edu), Mar 15 2007
From Peter Bala, Mar 24 2009: (Start)
Basic hypergeometric generating function: 2*exp(-t)*Sum {n >= 0} Product_{k = 1..n} (1-exp(-(4*k-2)*t))*exp(-2*n*t)/Product_{k = 1..n+1} (1+exp(-(4*k-2)*t)) = 1 + t + 5*t^2/2! + 61*t^3/3! + .... For other sequences with generating functions of a similar type see A000464, A002105, A002439, A079144 and A158690.
a(n) = 2*(-1)^n*L(-2*n), where L(s) is the Dirichlet L-function L(s) = 1 - 1/3^s + 1/5^s - + .... (End)
Sum_{n>=0} a(n)*z^(2*n)/(4*n)!! = Beta(1/2-z/(2*Pi),1/2+z/(2*Pi))/Beta(1/2,1/2) with Beta(z,w) the Beta function. - Johannes W. Meijer, Jul 06 2009
a(n) = Sum_(Sum_(binomial(k,m)*(-1)^(n+k)/(2^(m-1))*Sum_(binomial(m,j)*(2*j-m)^(2*n),j,0,m/2)*(-1)^(k-m),m,0,k),k,1,2*n), n>0. - Vladimir Kruchinin, Aug 05 2010
If n is prime, then a(n)==1 (mod 2*n). - Vladimir Shevelev, Sep 04 2010
From Peter Bala, Jan 21 2011: (Start)
(1)... a(n) = (-1/4)^n*B(2*n,-1),
where {B(n,x)}n>=1 = [1, 1+x, 1+6*x+x^2, 1+23*x+23*x^2+x^3, ...] is the sequence of Eulerian polynomials of type B - see A060187. Equivalently,
(2)... a(n) = Sum_{k = 0..2*n} Sum_{j = 0..k} (-1)^(n-j) *binomial(2*n+1,k-j)*(j+1/2)^(2*n).
We also have
(3)... a(n) = 2*A(2*n,i)/(1+i)^(2*n+1),
where i = sqrt(-1) and where {A(n,x)}n>=1 = [x, x + x^2, x + 4*x^2 + x^3, ...] denotes the sequence of Eulerian polynomials - see A008292. Equivalently,
(4)... a(n) = i*Sum_{k = 1..2*n} (-1)^(n+k)*k!*Stirling2(2*n,k) *((1+i)/2)^(k-1)
= i*Sum_{k = 1..2*n} (-1)^(n+k)*((1+i)/2)^(k-1) Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*j^(2*n).
Either this explicit formula for a(n) or (2) above may be used to obtain congruence results for a(n). For example, for prime p
(5a)... a(p) = 1 (mod p)
(5b)... a(2*p) = 5 (mod p)
and for odd prime p
(6a)... a((p+1)/2) = (-1)^((p-1)/2) (mod p)
(6b)... a((p-1)/2) = -1 + (-1)^((p-1)/2) (mod p).
(End)
a(n) = (-1)^n*2^(4*n+1)*(zeta(-2*n,1/4) - zeta(-2*n,3/4)). - Gerry Martens, May 27 2011
a(n) may be expressed as a sum of multinomials taken over all compositions of 2*n into even parts (Vella 2008): a(n) = Sum_{compositions 2*i_1 + ... + 2*i_k = 2*n} (-1)^(n+k)* multinomial(2*n, 2*i_1, ..., 2*i_k). For example, there are 4 compositions of the number 6 into even parts, namely 6, 4+2, 2+4 and 2+2+2, and hence a(3) = 6!/6! - 6!/(4!*2!) - 6!/(2!*4!) + 6!/(2!*2!*2!) = 61. A companion formula expressing a(n) as a sum of multinomials taken over the compositions of 2*n-1 into odd parts has been given by Malenfant 2011. - Peter Bala, Jul 07 2011
a(n) = the upper left term in M^n, where M is an infinite square production matrix; M[i,j] = A000290(i) = i^2, i >= 1 and 1 <= j <= i+1, and M[i,j] = 0, i >= 1 and j >= i+2 (see examples). - Gary W. Adamson, Jul 18 2011
E.g.f. A'(x) satisfies the differential equation A'(x)=cos(A(x)). - Vladimir Kruchinin, Nov 03 2011
From Peter Bala, Nov 28 2011: (Start)
a(n) = D^(2*n)(cosh(x)) evaluated at x = 0, where D is the operator cosh(x)*d/dx. a(n) = D^(2*n-1)(f(x)) evaluated at x = 0, where f(x) = 1+x+x^2/2! and D is the operator f(x)*d/dx.
Other generating functions: cosh(Integral_{t = 0..x} 1/cos(t)) dt = 1 + x^2/2! + 5*x^4/4! + 61*x^6/6! + 1385*x^8/8! + .... Cf. A012131.
A(x) := arcsinh(tan(x)) = log( sec(x) + tan(x) ) = x + x^3/3! + 5*x^5/5! + 61*x^7/7! + 1385*x^9/9! + .... A(x) satisfies A'(x) = cosh(A(x)).
B(x) := Series reversion( log(sec(x) + tan(x)) ) = x - x^3/3! + 5*x^5/5! - 61*x^7/7! + 1385*x^9/9! - ... = arctan(sinh(x)). B(x) satisfies B'(x) = cos(B(x)). (End)
HANKEL transform is A097476. PSUM transform is A173226. - Michael Somos, May 12 2012
a(n+1) - a(n) = A006212(2*n). - Michael Somos, May 12 2012
a(0) = 1 and, for n > 0, a(n) = (-1)^n*((4*n+1)/(2*n+1) - Sum_{k = 1..n} (4^(2*k)/2*k)*binomial(2*n,2*k-1)*A000367(k)/A002445(k)); see the Bucur et al. link. - L. Edson Jeffery, Sep 17 2012
O.g.f.: Sum_{n>=0} (2*n)!/2^n * x^n / Product_{k=1..n} (1 + k^2*x). - Paul D. Hanna, Sep 20 2012
From Sergei N. Gladkovskii, Oct 31 2011 to Oct 11 2013: (Start)
Continued fractions:
E.g.f.: (sec(x)) = 1+x^2/T(0), T(k) = 2(k+1)(2k+1) - x^2 + x^2*(2k+1)(2k+2)/T(k+1).
E.g.f.: 2/Q(0) where Q(k) = 1 + 1/(1 - x^2/(x^2 - 2*(k+1)*(2*k+1)/Q(k+1))).
G.f.: 1/Q(0) where Q(k) = 1 + x*k*(3*k-1) - x*(k+1)*(2*k+1)*(x*k^2+1)/Q(k+1).
E.g.f.: (2 + x^2 + 2*U(0))/(2 + (2 - x^2)*U(0)) where U(k)=  4*k + 4 + 1/( 1 + x^2/(2 - x^2 + (2*k+3)*(2*k+4)/U(k+1))).
E.g.f.: 1/cos(x) = 8*(x^2+1)/(4*x^2 + 8 - x^4*U(0)) where U(k) = 1 + 4*(k+1)*(k+2)/(2*k+3 - x^2*(2*k+3)/(x^2 - 8*(k+1)*(k+2)*(k+3)/U(k+1))).
G.f.: 1/U(0) where U(k) = 1 + x - x*(2*k+1)*(2*k+2)/(1 - x*(2*k+1)*(2*k+2)/U(k+1)).
G.f.: 1 + x/G(0) where G(k) = 1 + x - x*(2*k+2)*(2*k+3)/(1 - x*(2*k+2)*(2*k+3)/G(k+1)).
Let F(x) = sec(x^(1/2)) = Sum_{n>=0} a(n)*x^n/(2*n)!, then F(x)=2/(Q(0) + 1) where Q(k)= 1 - x/(2*k+1)/(2*k+2)/(1 - 1/(1 + 1/Q(k+1))).
G.f.: Q(0), where Q(k) = 1 - x*(k+1)^2/( x*(k+1)^2 - 1/Q(k+1)).
E.g.f.: 1/cos(x) = 1 + x^2/(2-x^2)*Q(0), where Q(k) = 1 - 2*x^2*(k+1)*(2*k+1)/( 2*x^2*(k+1)*(2*k+1)+ (12-x^2 + 14*k + 4*k^2)*(2-x^2 + 6*k + 4*k^2)/Q(k+1)). (End)
a(n) = Sum_{k=1..2*n} (Sum_{i=0..k-1} (i-k)^(2*n)*binomial(2*k,i)*(-1)^(i+k+n)) / 2^(k-1) for n>0, a(0)=1. - Vladimir Kruchinin, Oct 05 2012
It appears that a(n) = 3*A076552(n -1) + 2*(-1)^n for n >= 1. Conjectural congruences: a(2*n) == 5 (mod 60) for n >= 1 and a(2*n+1) == 1 (mod 60) for n >= 0. - Peter Bala, Jul 26 2013
From Peter Bala, Mar 09 2015: (Start)
O.g.f.: Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 - sqrt(-x)*(2*k + 1)) = Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 + x*(2*k + 1)^2).
O.g.f. is 1 + x*d/dx(log(F(x))), where F(x) = 1 + x + 3*x^2 + 23*x^3 + 371*x^4 + ... is the o.g.f. for A255881. (End)
Sum_(n >= 1, A034947(n)/n^(2d+1)) = a(d)*Pi^(2d+1)/(2^(2d+2)-2)(2d)! for d >= 0; see Allouche and Sondow, 2015. - Jonathan Sondow, Mar 21 2015
Asymptotic expansion: 4*(4*n/(Pi*e))^(2*n+1/2)*exp(1/2+1/(24*n)-1/(2880*n^3) +1/(40320*n^5)-...). (See the Luschny link.) - Peter Luschny, Jul 14 2015
a(n) = 2*(-1)^n*Im(Li_{-2n}(i)), where Li_n(x) is polylogarithm, i=sqrt(-1). - Vladimir Reshetnikov, Oct 22 2015
Limit_{n->infinity} ((2*n)!/a(n))^(1/(2*n)) = Pi/2. - Stanislav Sykora, Oct 07 2016
O.g.f.: 1/(1 + x - 2*x/(1 - 2*x/(1 + x - 12*x/(1 - 12*x/(1 + x - 30*x/(1 - 30*x/(1 + x - ... - (2*n - 1)*(2*n)*x/(1 - (2*n - 1)*(2*n)*x/(1 + x - ... ))))))))). - Peter Bala, Nov 09 2017
For n>0, a(n) = (-PolyGamma(2*n, 1/4) / 2^(2*n - 1) - (2^(2*n + 2) - 2) * Gamma(2*n + 1) * zeta(2*n + 1)) / Pi^(2*n + 1). - Vaclav Kotesovec, May 04 2020
a(n) ~ 2^(4*n + 3) * n^(2*n + 1/2) / (Pi^(2*n + 1/2) * exp(2*n)) * exp(Sum_{k>=1} bernoulli(k+1) / (k*(k+1)*2^k*n^k)). - Vaclav Kotesovec, Mar 05 2021
Peter Bala's conjectured congruences, a(2n) == 5 (mod 60) for n >= 1 and a(2n + 1) == 1 (mod 60), hold due to the results of Stern (mod 4) and Knuth & Buckholtz (mod 3 and 5). - Charles R Greathouse IV, Mar 23 2022

Typo in name corrected by Anders Claesson, Dec 01 2015

A000165 Double factorial of even numbers: (2n)!! = 2^n*n!.

Original entry on oeis.org

1, 2, 8, 48, 384, 3840, 46080, 645120, 10321920, 185794560, 3715891200, 81749606400, 1961990553600, 51011754393600, 1428329123020800, 42849873690624000, 1371195958099968000, 46620662575398912000, 1678343852714360832000, 63777066403145711616000

Offset: 0

N. J. A. Sloane

a(n) is also the size of the automorphism group of the graph (edge graph) of the n-dimensional hypercube and also of the geometric automorphism group of the hypercube (the two groups are isomorphic). This group is an extension of an elementary Abelian group (C_2)^n by S_n. (C_2 is the cyclic group with two elements and S_n is the symmetric group.) - Avi Peretz (njk(AT)netvision.net.il), Feb 21 2001
Then a(n) appears in the power series: sqrt(1+sin(y)) = Sum_{n>=0} (-1)^floor(n/2)*y^(n)/a(n) and sqrt((1+cos(y))/2) = Sum_{n>=0} (-1)^n*y^(2n)/a(2n). - Benoit Cloitre, Feb 02 2002
Appears to be the BinomialMean transform of A001907. See A075271. - John W. Layman, Sep 28 2002
Number of n X n monomial matrices with entries 0, +-1.
Also number of linear signed orders.
Define a "downgrade" to be the permutation d which places the items of a permutation p in descending order. This note concerns those permutations that are equal to their double-downgrades. The number of permutations of order 2n having this property are equinumerous with those of order 2n+1. a(n) = number of double-downgrading permutations of order 2n and 2n+1. - Eugene McDonnell (eemcd(AT)mac.com), Oct 27 2003
a(n) = (Integral_{x=0..Pi/2} cos(x)^(2*n+1) dx) where the denominators are b(n) = (2*n)!/(n!*2^n). - Al Hakanson (hawkuu(AT)excite.com), Mar 02 2004
1 + (1/2)x - (1/8)x^2 - (1/48)x^3 + (1/384)x^4 + ... = sqrt(1+sin(x)).
a(n)*(-1)^n = coefficient of the leading term of the (n+1)-th derivative of arctan(x), see Hildebrand link. - Reinhard Zumkeller, Jan 14 2006
a(n) is the Pfaffian of the skew-symmetric 2n X 2n matrix whose (i,j) entry is j for iDavid Callan, Sep 25 2006
a(n) is the number of increasing plane trees with n+1 edges. (In a plane tree, each subtree of the root is an ordered tree but the subtrees of the root may be cyclically rotated.) Increasing means the vertices are labeled 0,1,2,...,n+1 and each child has a greater label than its parent. Cf. A001147 for increasing ordered trees, A000142 for increasing unordered trees and A000111 for increasing 0-1-2 trees. - David Callan, Dec 22 2006
Hamed Hatami and Pooya Hatami prove that this is an upper bound on the cardinality of any minimal dominating set in C_{2n+1}^n, the Cartesian product of n copies of the cycle of size 2n+1, where 2n+1 is a prime. - Jonathan Vos Post, Jan 03 2007
This sequence and (1,-2,0,0,0,0,...) form a reciprocal pair under the list partition transform and associated operations described in A133314. - Tom Copeland, Oct 29 2007
a(n) = number of permutations of the multiset {1,1,2,2,...,n,n,n+1,n+1} such that between the two occurrences of i, there is exactly one entry >i, for i=1,2,...,n. Example: a(2) = 8 counts 121323, 131232, 213123, 231213, 232131, 312132, 321312, 323121. Proof: There is always exactly one entry between the two 1s (when n>=1). Given a permutation p in A(n) (counted by a(n)), record the position i of the first 1, then delete both 1s and subtract 1 from every entry to get a permutation q in A(n-1). The mapping p -> (i,q) is a bijection from A(n) to the Cartesian product [1,2n] X A(n-1). - David Callan, Nov 29 2007
Row sums of A028338. - Paul Barry, Feb 07 2009
a(n) is the number of ways to seat n married couples in a row so that everyone is next to their spouse. Compare A007060. - Geoffrey Critzer, Mar 29 2009
From Gary W. Adamson, Apr 21 2009: (Start)
Equals (-1)^n * (1, 1, 2, 8, 48, ...) dot (1, -3, 5, -7, 9, ...).
Example: a(4) = 384 = (1, 1, 2, 8, 48) dot (1, -3, 5, -7, 9) = (1, -3, 10, -56, 432). (End)
exp(x/2) = Sum_{n>=0} x^n/a(n). - Jaume Oliver Lafont, Sep 07 2009
Assuming n starts at 0, a(n) appears to be the number of Gray codes on n bits. It certainly is the number of Gray codes on n bits isomorphic to the canonical one. Proof: There are 2^n different starting positions for each code. Also, each code has a particular pattern of bit positions that are flipped (for instance, 1 2 1 3 1 2 1 for n=3), and these bit position patterns can be permuted in n! ways. - D. J. Schreffler (ds1404(AT)txstate.edu), Jul 18 2010
E.g.f. of 0,1,2,8,... is x/(1-2x/(2-2x/(3-8x/(4-8x/(5-18x/(6-18x/(7-... (continued fraction). - Paul Barry, Jan 17 2011
Number of increasing 2-colored trees with choice of two colors for each edge. In general, if we replace 2 with k we get the number of increasing k-colored trees. For example, for k=3 we get the triple factorial numbers. - Wenjin Woan, May 31 2011
a(n) = row sums of triangle A193229. - Gary W. Adamson, Jul 18 2011
Also the number of permutations of 2n (or of 2n+1) that are equal to their reverse-complements. (See the Egge reference.) Note that the double-downgrade described in the preceding comment (McDonnell) is equivalent to the reverse-complement. - Justin M. Troyka, Aug 11 2011
The e.g.f. can be used to form a generator, [1/(1-2x)] d/dx, for A000108, so a(n) can be applied to A145271 to generate the Catalan numbers. - Tom Copeland, Oct 01 2011
The e.g.f. of 1/a(n) is BesselI(0,sqrt(2*x)). See Abramowitz-Stegun (reference and link under A008277), p. 375, 9.6.10. - Wolfdieter Lang, Jan 09 2012
a(n) = order of the largest imprimitive group of degree 2n with n systems of imprimitivity (see [Miller], p. 203). - L. Edson Jeffery, Feb 05 2012
Row sums of triangle A208057. - Gary W. Adamson, Feb 22 2012
a(n) is the number of ways to designate a subset of elements in each n-permutation. a(n) = A000142(n) + A001563(n) + A001804(n) + A001805(n) + A001806(n) + A001807(n) + A035038(n) * n!. - Geoffrey Critzer, Nov 08 2012
For n>1, a(n) is the order of the Coxeter groups (also called Weyl groups) of types B_n and C_n. - Tom Edgar, Nov 05 2013
For m>0, k*a(m-1) is the m-th cumulant of the chi-squared probability distribution for k degrees of freedom. - Stanislav Sykora, Jun 27 2014
a(n) with 0 prepended is the binomial transform of A120765. - Vladimir Reshetnikov, Oct 28 2015
Exponential self-convolution of A001147. - Vladimir Reshetnikov, Oct 08 2016
Also the order of the automorphism group of the n-ladder rung graph. - Eric W. Weisstein, Jul 22 2017
a(n) is the order of the group O_n(Z) = {A in M_n(Z): A*A^T = I_n}, the group of n X n orthogonal matrices over the integers. - Jianing Song, Mar 29 2021
a(n) is the number of ways to tile a (3n,3n)-benzel or a (3n+1,3n+2)-benzel using left stones and two kinds of bones; see Defant et al., below. - James Propp, Jul 22 2023
a(n) is the number of labeled histories for a labeled topology with the modified lodgepole shape and n+1 cherry nodes. - Noah A Rosenberg, Jan 16 2025

			The following permutations and their reversals are all of the permutations of order 5 having the double-downgrade property:
  0 1 2 3 4
  0 3 2 1 4
  1 0 2 4 3
  1 4 2 0 3
G.f. = 1 + 2*x + 8*x^2 + 48*x^3 + 384*x^4 + 3840*x^5 + 46080*x^6 + 645120*x^7 + ...

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
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..100
Paul Barry, General Eulerian Polynomials as Moments Using Exponential Riordan Arrays, Journal of Integer Sequences, 16 (2013), #13.9.6.
Paul Barry, Generalized Eulerian Triangles and Some Special Production Matrices, arXiv:1803.10297 [math.CO], 2018.
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, and Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.
P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
CombOS - Combinatorial Object Server, Generate colored permutations
R. Coquereaux and J.-B. Zuber, Maps, immersions and permutations, arXiv preprint arXiv:1507.03163 [math.CO], 2015.
Colin Defant, Rupert Li, James Propp, and Benjamin Young, Tilings of Benzels via the Abacus Bijection, arXiv preprint, arXiv:2209.05717 [math.CO], 2022.
Eric S. Egge, Restricted symmetric permutations, Ann. Combin., 11 (2007), 405-434.
Peter C. Fishburn, Signed Orders, Choice Probabilities and Linear Polytopes, Journal of Mathematical Psychology, Volume 45, Issue 1, (2001), pp. 53-80.
Joël Gay and Vincent Pilaud, The weak order on Weyl posets, arXiv:1804.06572 [math.CO], 2018.
G. Gordon, The answer is 2^n*n! What is the question?, Amer. Math. Monthly, 106 (1999), 636-645.
Guo-Niu Han, Enumeration of Standard Puzzles
Guo-Niu Han, Enumeration of Standard Puzzles [Cached copy]
Hamed Hatami and Pooya Hatami, Perfect dominating sets in the Cartesian products of prime cycles, arXiv:math/0701018 [math.CO], 2006-2009.
Jason D. Hildebrand, Differentiating Arctan(x)
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 136
E. Lappo and N. A. Rosenberg, A lattice structure for ancestral configurations arising from the relationship between gene trees and species trees, Adv. Appl. Math. 343 (2024), 65-81.
L. C. Larson, The number of essentially different nonattacking rook arrangements, J. Recreat. Math., 7 (No. 3, 1974), circa pages 180-181. [Annotated scan of pages 180 and 181 only]
E. Lucas, Théorie des nombres (annotated scans of a few selected pages)
Eugene McDonnell, Magic Squares and Permutations, APL Quote Quad 7.3 (Fall 1976).
B. E. Meserve, Double Factorials, American Mathematical Monthly, 55 (1948), 425-426.
G. A. Miller, Groups formed by special matrices, Bull. Amer. Math. Soc. 24 (1918), 203-206.
R. Ondrejka, Tables of double factorials, Math. Comp., 24 (1970), 231.
Alexsandar Petojevic, The Function vM_m(s; a; z) and Some Well-Known Sequences, Journal of Integer Sequences, Vol. 5 (2002), Article 02.1.7.
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014.
R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976).
M. Z. Spivey and L. L. Steil, The k-Binomial Transforms and the Hankel Transform, J. Integ. Seqs. Vol. 9 (2006), #06.1.1.
Eric Weisstein's World of Mathematics, Double Factorial
Eric Weisstein's World of Mathematics, Graph Automorphism
Eric Weisstein's World of Mathematics, Ladder Rung Graph
Index to divisibility sequences
Index entries for sequences related to factorial numbers

Cf. A000142 (n!), A001147 ((2n-1)!!), A032184 (2^n*(n-1)!).
Cf. A000079, A000108, A000111, A001044, A001563, A001804, A001805, A001806.
Cf. A001807, A001813, A001907, A002454, A006882, A007060, A008277, A008544.
Cf. A010050, A019774, A028326, A028338, A035038, A039683, A047053, A047055.
Cf. A047058, A047657, A075271, A084947, A084948, A084949, A092605, A133314.
Cf. A145271, A193229, A208057.
This sequence gives the row sums in A060187, and (-1)^n*a(n) the alternating row sums in A039757.
Also row sums in A028338.
Column k=2 of A329070.

a000165 n = product [2, 4 .. 2 * n]  -- Reinhard Zumkeller, Mar 28 2015

[2^n*Factorial(n): n in [0..35]]; // Vincenzo Librandi, Apr 22 2011

I:=[2,8]; [1] cat [n le 2 select I[n]  else (3*n-1)*Self(n-1)-2*(n-1)^2*Self(n-2): n in [1..35] ]; // Vincenzo Librandi, Feb 19 2015

A000165 := proc(n) option remember; if n <= 1 then 1 else n*A000165(n-2); fi; end;
ZL:=[S, {a = Atom, b = Atom, S = Prod(X,Sequence(Prod(X,b))), X = Sequence(b,card >= 0)}, labelled]: seq(combstruct[count](ZL, size=n), n=0..17); # Zerinvary Lajos, Mar 26 2008
G(x):=(1-2*x)^(-1): f[0]:=G(x): for n from 1 to 29 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=0..17); # Zerinvary Lajos, Apr 03 2009
A000165 := proc(n) doublefactorial(2*n) ; end proc; seq(A000165(n),n=0..10) ; # R. J. Mathar, Oct 20 2009

Table[(2 n)!!, {n, 30}] (* Vladimir Joseph Stephan Orlovsky, Dec 13 2008 *)
(2 Range[0, 30])!! (* Harvey P. Dale, Jan 23 2015 *)
RecurrenceTable[{a[n] == 2 n*a[n-1], a[0] == 1}, a, {n,0,30}] (* Ray Chandler, Jul 30 2015 *)

a(n)=n!<Charles R Greathouse IV, Feb 11 2011

{a(n) = prod( k=1, n, 2*k)}; /* Michael Somos, Jan 04 2013 */

from math import factorial
def A000165(n): return factorial(n)<Chai Wah Wu, Jan 24 2023

[2^n*factorial(n) for n in range(31)] # G. C. Greubel, Jul 21 2024

E.g.f.: 1/(1-2*x).
a(n) = A001044(n)/A000142(n)*A000079(n) = Product_{i=0..n-1} (2*i+2) = 2^n*Pochhammer(1,n). - Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003
D-finite with recurrence a(n) = 2*n * a(n-1), n>0, a(0)=1. - Paul Barry, Aug 26 2004
This is the binomial mean transform of A001907. See Spivey and Steil (2006). - Michael Z. Spivey (mspivey(AT)ups.edu), Feb 26 2006
a(n) = Integral_{x>=0} x^n*exp(-x/2)/2 dx. - Paul Barry, Jan 28 2008
G.f.: 1/(1-2x/(1-2x/(1-4x/(1-4x/(1-6x/(1-6x/(1-.... (continued fraction). - Paul Barry, Feb 07 2009
a(n) = A006882(2*n). - R. J. Mathar, Oct 20 2009
From Gary W. Adamson, Jul 18 2011: (Start)
a(n) = upper left term in M^n, M = a production matrix (twice Pascal's triangle deleting the first "2", with the rest zeros; cf. A028326):
  2,  2,  0,  0,  0,  0, ...
  2,  4,  2,  0,  0,  0, ...
  2,  6,  6,  2,  0,  0, ...
  2,  8, 12,  8,  2,  0, ...
  2, 10, 20, 20, 10,  2, ...
  ... (End)
From Sergei N. Gladkovskii, Apr 11 2013, May 01 2013, May 24 2013, Sep 30 2013, Oct 27 2013: (Start)
Continued fractions:
G.f.: 1 + x*(Q(0) - 1)/(x+1) where Q(k) = 1 + (2*k+2)/(1-x/(x+1/Q(k+1))).
G.f.: 1/Q(0) where Q(k) = 1 + 2*k*x - 2*x*(k+1)/Q(k+1).
G.f.: G(0)/2 where G(k) = 1 + 1/(1 - x*(2*k+2)/(x*(2*k+2) + 1/G(k+1))).
G.f.: 1/Q(0) where Q(k) = 1 - x*(4*k+2) - 4*x^2*(k+1)^2/Q(k+1).
G.f.: R(0) where R(k) = 1 - x*(2*k+2)/(x*(2*k+2)-1/(1-x*(2*k+2)/(x*(2*k+2) -1/R(k+1)))). (End)
a(n) = (2n-2)*a(n-2) + (2n-1)*a(n-1), n>1. - Ivan N. Ianakiev, Aug 06 2013
From Peter Bala, Feb 18 2015: (Start)
Recurrence equation: a(n) = (3*n - 1)*a(n-1) - 2*(n - 1)^2*a(n-2) with a(1) = 2 and a(2) = 8.
The sequence b(n) = A068102(n) also satisfies this second-order recurrence. This leads to the generalized continued fraction expansion lim_{n -> oo} b(n)/a(n) = log(2) = 1/(2 - 2/(5 - 8/(8 - 18/(11 - ... - 2*(n - 1)^2/((3*n - 1) - ... ))))). (End)
From Amiram Eldar, Jun 25 2020: (Start)
Sum_{n>=0} 1/a(n) = sqrt(e) (A019774).
Sum_{n>=0} (-1)^n/a(n) = 1/sqrt(e) (A092605). (End)
Limit_{n->oo} a(n)^4 / (n * A134372(n)) = Pi. - Daniel Suteu, Apr 09 2022
a(n) = 1/([x^n] hypergeom([1], [1], x/2)). - Peter Luschny, Sep 13 2024
a(n) = Sum_{k=0..n} k!*(n-k)!*binomial(n,k)^2. - Ridouane Oudra, Jul 13 2025

A000182 Tangent (or "Zag") numbers: e.g.f. tan(x), also (up to signs) e.g.f. tanh(x).

Original entry on oeis.org

1, 2, 16, 272, 7936, 353792, 22368256, 1903757312, 209865342976, 29088885112832, 4951498053124096, 1015423886506852352, 246921480190207983616, 70251601603943959887872, 23119184187809597841473536, 8713962757125169296170811392, 3729407703720529571097509625856

Offset: 1

N. J. A. Sloane

Number of Joyce trees with 2n-1 nodes. Number of tremolo permutations of {0,1,...,2n}. - Ralf Stephan, Mar 28 2003
The Hankel transform of this sequence is A000178(n) for n odd = 1, 12, 34560, ...; example: det([1, 2, 16; 2, 16, 272, 16, 272, 7936]) = 34560. - Philippe Deléham, Mar 07 2004
a(n) is the number of increasing labeled full binary trees with 2n-1 vertices. Full binary means every non-leaf vertex has two children, distinguished as left and right; labeled means the vertices are labeled 1,2,...,2n-1; increasing means every child has a label greater than its parent. - David Callan, Nov 29 2007
From Micha Hofri (hofri(AT)wpi.edu), May 27 2009: (Start)
a(n) was found to be the number of permutations of [2n] which when inserted in order, to form a binary search tree, yield the maximally full possible tree (with only one single-child node).
The e.g.f. is sec^2(x)=1+tan^2(x), and the same coefficients can be manufactured from the tan(x) itself, which is the e.g.f. for the number of trees as above for odd number of nodes. (End)
a(n) is the number of increasing strict binary trees with 2n-1 nodes. For more information about increasing strict binary trees with an associated permutation, see A245894. - Manda Riehl, Aug 07 2014
For relations to alternating permutations, Euler and Bernoulli polynomials, zigzag numbers, trigonometric functions, Fourier transform of a square wave, quantum algebras, and integrals over and in n-dimensional hypercubes and over Green functions, see Hodges and Sukumar. For further discussion on the quantum algebra, see the later Hodges and Sukumar reference and the paper by Hetyei presenting connections to the general combinatorial theory of Viennot on orthogonal polynomials, inverse polynomials, tridiagonal matrices, and lattice paths (thereby related to continued fractions and cumulants). - Tom Copeland, Nov 30 2014
The Zigzag Hankel transform is A000178. That is, A000178(2*n - k) = det( [a(i+j - k)]{i,j = 1..n} ) for n>0 and k=0,1. - _Michael Somos, Mar 12 2015
a(n) is the number of standard Young tableaux of skew shape (n,n,n-1,n-2,...,3,2)/(n-1,n-2,n-3,...,2,1). - Ran Pan, Apr 10 2015
For relations to the Sheffer Appell operator calculus and a Riccati differential equation for generating the Meixner-Pollaczek and Krawtchouk orthogonal polynomials, see page 45 of the Feinsilver link and Rzadkowski. - Tom Copeland, Sep 28 2015
For relations to an elliptic curve, a Weierstrass elliptic function, the Lorentz formal group law, a Lie infinitesimal generator, and the Eulerian numbers A008292, see A155585. - Tom Copeland, Sep 30 2015
Absolute values of the alternating sums of the odd-numbered rows (where the single 1 at the apex of the triangle is counted as row #1) of the Eulerian triangle, A008292. The actual alternating sums alternate in sign, e.g., 1, -2, 16, -272, etc. (Even-numbered rows have alternating sums always 0.) - Gregory Gerard Wojnar, Sep 28 2018
The sequence is periodic modulo any odd prime p. The minimal period is (p-1)/2 if p == 1 mod 4 and p-1 if p == 3 mod 4 [Knuth & Buckholtz, 1967, Theorem 1]. - Allen Stenger, Aug 03 2020
From Peter Bala, Dec 24 2021: (Start)
Conjectures:
1) The sequence taken modulo any integer k eventually becomes periodic with period dividing phi(k).
2) The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k, except when p = 2, n = 1 and k = 1 or 2.
3) For i >= 1 define a_i(n) = a(n+i). The Gauss congruences a_i(n*p^k) == a_i(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k. If true, then for each i >= 1 the expansion of exp(Sum_{n >= 1} a_i(n)*x^n/n) has integer coefficients. For an example, see A262145.(End)

			tan(x) = x + 2*x^3/3! + 16*x^5/5! + 272*x^7/7! + ... = x + 1/3*x^3 + 2/15*x^5 + 17/315*x^7 + 62/2835*x^9 + O(x^11).
tanh(x) = x - 1/3*x^3 + 2/15*x^5 - 17/315*x^7 + 62/2835*x^9 - 1382/155925*x^11 + ...
(sec x)^2 = 1 + x^2 + 2/3*x^4 + 17/45*x^6 + ...
a(3)=16 because we have: {1, 3, 2, 5, 4}, {1, 4, 2, 5, 3}, {1, 4, 3, 5, 2},
  {1, 5, 2, 4, 3}, {1, 5, 3, 4, 2}, {2, 3, 1, 5, 4}, {2, 4, 1, 5, 3},
  {2, 4, 3, 5, 1}, {2, 5, 1, 4, 3}, {2, 5, 3, 4, 1}, {3, 4, 1, 5, 2},
  {3, 4, 2, 5, 1}, {3, 5, 1, 4, 2}, {3, 5, 2, 4, 1}, {4, 5, 1, 3, 2},
  {4, 5, 2, 3, 1}. - _Geoffrey Critzer_, May 19 2013

Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 88.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 111.
H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 69.
L. M. Milne-Thompson, Calculus of Finite Differences, 1951, p. 148 (the numbers |C^{2n-1}|).
J. W. Milnor and J. D. Stasheff, Characteristic Classes, Princeton, 1974, p. 282.
S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 444.
H. Rademacher, Topics in Analytic Number Theory, Springer, 1973, Chap. 1, p. 20.
L. Seidel, Über eine einfache Entstehungsweise der Bernoullischen Zahlen und einiger verwandten Reihen, Sitzungsberichte der mathematisch-physikalischen Classe der königlich bayerischen Akademie der Wissenschaften zu München, volume 7 (1877), 157-187.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
E. van Fossen Conrad, Some continued fraction expansions of elliptic functions, PhD thesis, The Ohio State University, 2002, p. 28.
J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, pp. 267-268.

Seiichi Manyama, Table of n, a(n) for n = 1..243 (terms 1..100 from N. J. A. Sloane)
V. E. Adler and A. B. Shabat, Volterra chain and Catalan numbers, arXiv:1810.13198 [nlin.SI], 2018.
J. L. Arregui, Tangent and Bernoulli numbers related to Motzkin and Catalan numbers by means of numerical triangles, arXiv:math/0109108 [math.NT], 2001.
Étienne Bellin, Arthur Blanc-Renaudie, Emmanuel Kammerer, and Igor Kortchemski, Uniform attachment with freezing, arXiv:2308.00493 [math.PR], 2023. See p. 13.
Beáta Bényi, Miguel Méndez, José L. Ramírez, and Tanay Wakhare, Restricted r-Stirling Numbers and their Combinatorial Applications, arXiv:1811.12897 [math.CO], 2018.
Richard P. Brent and David Harvey, Fast computation of Bernoulli, Tangent and Secant numbers, arXiv preprint arXiv:1108.0286 [math.CO], 2011.
F. C. S. Brown, T. M. A. Fink and K. Willbrand, On arithmetic and asymptotic properties of up-down numbers, arXiv:math/0607763 [math.CO], 2006.
K.-W. Chen, Algorithms for Bernoulli numbers and Euler numbers, J. Integer Sequences, 4 (2001), #01.1.6.
Suyuong Choi and Younghan Yoon, A decomposition of graph a-numbers, arXiv:2508.06855 [math.CO], 2025. See p. 13.
Bishal Deb and Alan D. Sokal, Classical continued fractions for some multivariate polynomials generalizing the Genocchi and median Genocchi numbers, arXiv:2212.07232 [math.CO], 2022. See p. 11.
D. Dumont, Interprétations combinatoires des nombres de Genocchi, Duke Math. J., 41 (1974), 305-318.
D. Dumont and G. Viennot, A combinatorial interpretation of the Seidel generation of Genocchi numbers, Preprint, Annotated scanned copy.
A. L. Edmonds and S. Klee, The combinatorics of hyperbolized manifolds, arXiv preprint arXiv:1210.7396 [math.CO], 2012. - From _N. J. A. Sloane_, Jan 02 2013
P. Feinsilver, Lie algebras, representations, and analytic semigroups through dual vector fields [broken link]
C. J. Fewster and D. Siemssen, Enumerating Permutations by their Run Structure, arXiv preprint arXiv:1403.1723 [math.CO], 2014.
P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 144.
Dominique Foata and Guo-Niu Han, Doubloons and new q-tangent numbers, Quart. J. Math. 62 (2) (2011) 417-432.
Dominique Foata and Guo-Niu Han, Tree Calculus for Bivariable Difference Equations, Journal of Difference Equations and Applications, 20 (2014), 1453-1488; arXiv:1304.2484 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 02 2013
Dominique Foata and Guo-Niu Han, Seidel Triangle Sequences and Bi-Entringer Numbers, European Journal of Combinatorics, 42 (2014), 243-260.
Ghislain R. Franssens, On a Number Pyramid Related to the Binomial, Deleham, Eulerian, MacMahon and Stirling number triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.4.1.
M.-P. Grosset and A. P. Veselov, Bernoulli numbers and solitons, arXiv:math/0503175 [math.GM], 2005.
Christian Günther and Kai-Uwe Schmidt, L^q norms of Fekete and related polynomials, arXiv:1602.01750 [math.NT], 2016.
Guo-Niu Han and Jing-Yi Liu, Combinatorial proofs of some properties of tangent and Genocchi numbers, European Journal of Combinatorics, Vol. 71 (2018), pp. 99-110; arXiv preprint, arXiv:1707.08882 [math.CO], 2017-2018.
Yanjun Han and Jonathan Niles-Weed, Approximate independence of permutation mixtures, arXiv:2408.09341 [math.ST], 2024. See p. 12.
G. Hetyei, Meixner polynomials of the second kind and quantum algebras representing su(1,1), arXiv preprint arXiv:0909.4352 [math.QA], 2009.
A. Hodges and C. Sukumar, Bernoulli, Euler, permutations and quantum algebras, Proc. Royal Soc. A (2007) 463, 2401-2414.
A. Hodges and C. Sukumar, Quantum algebras and parity-dependent spectra, Proc. Royal Soc. A (2007) 463, 2415-2427.
Jan Hubička, Matěj Konečný, Štěpán Vodseďálek, and Andy Zucker, Counting big Ramsey degrees of the homogeneous and universal K_4-free graph, arXiv:2505.22620 [math.CO], 2025. See pp. 2, 4.
Hsien-Kuei Hwang and Emma Yu Jin, Asymptotics and statistics on Fishburn matrices and their generalizations, arXiv:1911.06690 [math.CO], 2019.
Svante Janson, Euler-Frobenius numbers and rounding, arXiv preprint arXiv:1305.3512 [math.PR], 2013
Donald E. Knuth and Thomas J. Buckholtz, Computation of tangent, Euler and Bernoulli numbers, Math. Comp. 21 1967 663-688.
D. E. Knuth and Thomas J. Buckholtz, Computation of tangent, Euler and Bernoulli numbers, Math. Comp. 21 1967 663-688. [Annotated scanned copy]
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Johann Heinrich Lambert, Mémoire sur quelques propriétés remarquables des quantités transcendentes circulaires et logarithmiques, Histoire de l'Académie Royale des Sciences et des Belles-Lettres de Berlin 1761 (Berlin: Haude et Spener, 1768) pp. 265-322. [broken link]
F. Luca and P. Stanica, On some conjectures on the monotonicity of some arithmetical sequences, J. Combin. Number Theory 4 (2012) 1-10.
Peter Luschny, Approximation, inclusion and asymptotics of the Euler numbers
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G. Rzadkowski, Bernoulli numbers and solitons-revisited, Jrn. Nonlinear Math. Physics, 1711, pp. 121-126. (Added by Tom Copeland, Sep 29 2015)
Raphael Schumacher, Rapidly Convergent Summation Formulas involving Stirling Series, arXiv preprint arXiv:1602.00336 [math.NT], 2016.
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N. J. A. Sloane, Rough notes on Genocchi numbers
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R. P. Stanley, Permutations
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Bao-Xuan Zhu, Analytic approaches to monotonicity and log-behavior of combinatorial sequences, arXiv preprint arXiv:1309.5693 [math.CO], 2013.
Index entries for "core" sequences
Index entries for sequences related to boustrophedon transform
Index entries for sequences related to Bernoulli numbers.

A350972 is essentially the same sequence.
a(n)=2^(n-1)*A002105(n). Apart from signs, 2^(2n-2)*A001469(n) = n*a(n).
Cf. A001469, A002430, A036279, A000364 (secant numbers), A000111 (secant-tangent numbers), A024283, A009764. First diagonal of A059419 and of A064190.
Cf. A009006, A009725, A029584, A012509, A009123, A009567.
Equals A002425(n) * 2^A101921(n).
Equals leftmost column of A162005. - Johannes W. Meijer, Jun 27 2009
Cf. A258880, A258901. Cf. A002105, A080635, A234797.
Cf. A019538, A110501.

series(tan(x),x,40);
with(numtheory): a := n-> abs(2^(2*n)*(2^(2*n)-1)*bernoulli(2*n)/(2*n));
A000182_list := proc(n) local T,k,j; T[1] := 1;
for k from 2 to n do T[k] := (k-1)*T[k-1] od;
   for k from 2 to n do
       for j from k to n do
           T[j] := (j-k)*T[j-1]+(j-k+2)*T[j] od od;
seq(T[j], j=1..n)  end:
A000182_list(15);  # Peter Luschny, Apr 02 2012

Table[ Sum[2^(2*n + 1 - k)*(-1)^(n + k + 1)*k!*StirlingS2[2*n + 1, k], {k, 1, 2*n + 1}], {n, 0, 7}] (* Victor Adamchik, Oct 05 2005 *)
v[1] = 2; v[n_] /; n >= 2 := v[n] = Sum[ Binomial[2 n - 3, 2 k - 2] v[k] v[n - k], {k, n - 1}]; Table[ v[n]/2, {n, 15}] (* Zerinvary Lajos, Jul 08 2009 *)
Rest@ Union[ Range[0, 29]! CoefficientList[ Series[ Tan[x], {x, 0, 30}], x]] (* Harvey P. Dale, Oct 19 2011; modified by Robert G. Wilson v, Apr 02 2012 *)
t[1, 1] = 1; t[1, 0] = 0; t[n_ /; n > 1, m_] := t[n, m] = m*(m+1)*Sum[t[n-1, k], {k, m-1, n-1}]; a[n_] := t[n, 1]; Table[a[n], {n, 1, 15}]  (* Jean-François Alcover, Jan 02 2013, after A064190 *)
a[ n_] := If[ n < 1, 0, With[{m = 2 n - 1}, m! SeriesCoefficient[ Tan[x], {x, 0, m}]]]; (* Michael Somos, Mar 12 2015 *)
a[ n_] := If[ n < 1, 0, ((-16)^n - (-4)^n) Zeta[1 - 2 n]]; (* Michael Somos, Mar 12 2015 *)
Table[2 PolyGamma[2n - 1, 1/2]/Pi^(2n), {n, 1, 10}] (* Vladimir Reshetnikov, Oct 18 2015 *)
a[ n_] := a[n] = If[ n < 2, Boole[n == 1], Sum[Binomial[2 n - 2, 2 k - 1] a[k] a[n - k], {k, n - 1}]]; (* Michael Somos, Aug 02 2018 *)
a[n_] := (2^(2*n)*(2^(2*n) - 1)*Abs[BernoulliB[2*n]])/(2*n); a /@  Range[20]  (* Stan Wagon, Nov 21 2022 *)

a(n):=sum(sum(binomial(k,r)*sum(sum(binomial(l,j)/2^(j-1)*sum((-1)^(n)*binomial(j,i)*(j-2*i)^(2*n),i,0,floor((j-1)/2))*(-1)^(l-j),j,1,l)*(-1)^l*binomial(r+l-1,r-1),l,1,2*n)*(-1)^(1-r),r,1,k)/k,k,1,2*n); /* Vladimir Kruchinin, Aug 23 2010 */

a[n]:=if n=1 then 1 else 2*sum(sum(binomial(2*j,j+k)*(-4*k^2)^(n-1)*(-1)^k/(4^j),k,1,j),j,1,n-1);
makelist(a[n],n,1,30); /* Tani Akinari, Sep 20 2023 */

{a(n) = if( n<1, 0, ((-4)^n - (-16)^n) * bernfrac(2*n) / (2*n))};

{a(n) = my(an); if( n<2, n==1, an = vector(n, m, 1); for( m=2, n, an[m] = sum( k=1, m-1, binomial(2*m - 2, 2*k - 1) * an[k] * an[m-k])); an[n])}; /* Michael Somos */

{a(n) = if( n<1, 0, (2*n - 1)! * polcoeff( tan(x + O(x^(2*n + 2))), 2*n - 1))}; /* Michael Somos */

{a(n) = my(X=x+x*O(x^n),Egf); Egf = x*sum(m=0,n, prod(k=1,m, tanh(2*k*X))); (n-1)!*polcoeff(Egf,n)} /* Paul D. Hanna, May 11 2010 */

/* Continued Fraction for the e.g.f. tan(x), from Paul D. Hanna: */
{a(n)=local(CF=1+O(x)); for(i=1, n, CF=1/(2*(n-i+1)-1-x^2*CF)); (2*n-1)!*polcoeff(x*CF, 2*n-1)}

/* O.g.f. Sum_{n>=1} a(n)*x^n, from Paul D. Hanna Feb 05 2013: */
{a(n)=polcoeff( x+2*x*sum(m=1, n, x^m*prod(k=1, m, (2*k-1)^2/(1+(2*k-1)^2*x +x*O(x^n))) ), n)}

# The objective of this implementation is efficiency.
# n -> [0, a(1), a(2), ..., a(n)] for n > 0.
def A000182_list(n):
    T = [0 for i in range(1, n+2)]
    T[1] = 1
    for k in range(2, n+1):
        T[k] = (k-1)*T[k-1]
    for k in range(2, n+1):
        for j in range(k, n+1):
            T[j] = (j-k)*T[j-1]+(j-k+2)*T[j]
    return T
print(A000182_list(100)) # Peter Luschny, Aug 07 2011

from sympy import bernoulli
def A000182(n): return abs(((2-(2<<(m:=n<<1)))*bernoulli(m)<Chai Wah Wu, Apr 14 2023

# Algorithm of L. Seidel (1877)
# n -> [a(1), ..., a(n)] for n >= 1.
def A000182_list(len) :
    R = []; A = {-1:0, 0:1}; k = 0; e = 1
    for i in (0..2*len-1) :
        Am = 0; A[k + e] = 0; e = -e
        for j in (0..i) : Am += A[k]; A[k] = Am; k += e
        if e > 0 : R.append(A[i//2])
    return R
A000182_list(15) # Peter Luschny, Mar 31 2012

E.g.f.: log(sec x) = Sum_{n > 0} a(n)*x^(2*n)/(2*n)!.
E.g.f.: tan x = Sum_{n >= 0} a(n+1)*x^(2*n+1)/(2*n+1)!.
E.g.f.: (sec x)^2 = Sum_{n >= 0} a(n+1)*x^(2*n)/(2*n)!.
2/(exp(2x)+1) = 1 + Sum_{n>=1} (-1)^(n+1) a(n) x^(2n-1)/(2n-1)! = 1 - x + x^3/3 - 2*x^5/15 + 17*x^7/315 - 62*x^9/2835 + ...
a(n) = 2^(2*n) (2^(2*n) - 1) |B_(2*n)| / (2*n) where B_n are the Bernoulli numbers (A000367/A002445 or A027641/A027642).
Asymptotics: a(n) ~ 2^(2*n+1)*(2*n-1)!/Pi^(2*n).
Sum[2^(2*n + 1 - k)*(-1)^(n + k + 1)*k!*StirlingS2[2*n + 1, k], {k, 1, 2*n + 1}]. - Victor Adamchik, Oct 05 2005
a(n) = abs[c(2*n-1)] where c(n)= 2^(n+1) * (1-2^(n+1)) * Ber(n+1)/(n+1) = 2^(n+1) * (1-2^(n+1)) * (-1)^n * Zeta(-n) = [ -(1+EN(.))]^n = 2^n * GN(n+1)/(n+1) = 2^n * EP(n,0) = (-1)^n * E(n,-1) = (-2)^n * n! * Lag[n,-P(.,-1)/2] umbrally = (-2)^n * n! * C{T[.,P(.,-1)/2] + n, n} umbrally for the signed Euler numbers EN(n), the Bernoulli numbers Ber(n), the Genocchi numbers GN(n), the Euler polynomials EP(n,t), the Eulerian polynomials E(n,t), the Touchard / Bell polynomials T(n,t), the binomial function C(x,y) = x!/[(x-y)!*y! ] and the polynomials P(j,t) of A131758. - Tom Copeland, Oct 05 2007
a(1) = A094665(0,0)*A156919(0,0) and a(n) = Sum_{k=1..n-1} 2^(n-k-1)*A094665(n-1, k)*A156919(k,0) for n = 2, 3, .., see A162005. - Johannes W. Meijer, Jun 27 2009
G.f.: 1/(1-1*2*x/(1-2*3*x/(1-3*4*x/(1-4*5*x/(1-5*6*x/(1-... (continued fraction). - Paul Barry, Feb 24 2010
From Paul Barry, Mar 29 2010: (Start)
G.f.: 1/(1-2x-12x^2/(1-18x-240x^2/(1-50x-1260x^2/(1-98x-4032x^2/(1-162x-9900x^2/(1-... (continued fraction);
coefficient sequences given by 4*(n+1)^2*(2n+1)*(2n+3) and 2(2n+1)^2 (see Van Fossen Conrad reference). (End)
E.g.f.: x*Sum_{n>=0} Product_{k=1..n} tanh(2*k*x) = Sum_{n>=1} a(n)*x^n/(n-1)!. - Paul D. Hanna, May 11 2010 [corrected by Paul D. Hanna, Sep 28 2023]
a(n) = (-1)^(n+1)*Sum_{j=1..2*n+1} j!*Stirling2(2*n+1,j)*2^(2*n+1-j)*(-1)^j for n >= 0. Vladimir Kruchinin, Aug 23 2010: (Start)
If n is odd such that 2*n-1 is prime, then a(n) == 1 (mod (2*n-1)); if n is even such that 2*n-1 is prime, then a(n) == -1 (mod (2*n-1)). - Vladimir Shevelev, Sep 01 2010
Recursion: a(n) = (-1)^(n-1) + Sum_{i=1..n-1} (-1)^(n-i+1)*C(2*n-1,2*i-1)* a(i). - Vladimir Shevelev, Aug 08 2011
E.g.f.: tan(x) = Sum_{n>=1} a(n)*x^(2*n-1)/(2*n-1)! = x/(1 - x^2/(3 - x^2/(5 - x^2/(7 - x^2/(9 - x^2/(11 - x^2/(13 -...))))))) (continued fraction from J. H. Lambert - 1761). - Paul D. Hanna, Sep 21 2011
From Sergei N. Gladkovskii, Oct 31 2011 to Oct 09 2013: (Start)
Continued fractions:
E.g.f.: (sec(x))^2 = 1+x^2/(x^2+U(0)) where U(k) = (k+1)*(2k+1) - 2x^2 + 2x^2*(k+1)*(2k+1)/U(k+1).
E.g.f.: tan(x) = x*T(0) where T(k) = 1-x^2/(x^2-(2k+1)*(2k+3)/T(k+1)).
E.g.f.: tan(x) = x/(G(0)+x) where G(k) = 2*k+1 - 2*x + x/(1 + x/G(k+1)).
E.g.f.: tanh(x) = x/(G(0)-x) where G(k) = k+1 + 2*x - 2*x*(k+1)/G(k+1).
E.g.f.: tan(x) = 2*x - x/W(0) where W(k) = 1 + x^2*(4*k+5)/((4*k+1)*(4*k+3)*(4*k+5) - 4*x^2*(4*k+3) + x^2*(4*k+1)/W(k+1)).
E.g.f.: tan(x) = x/T(0) where T(k) = 1 - 4*k^2 + x^2*(1 - 4*k^2)/T(k+1).
E.g.f.: tan(x) = -3*x/(T(0)+3*x^2) where T(k)= 64*k^3 + 48*k^2 - 4*k*(2*x^2 + 1) - 2*x^2 - 3 - x^4*(4*k -1)*(4*k+7)/T(k+1).
G.f.: 1/G(0) where G(k) = 1 - 2*x*(2*k+1)^2 - x^2*(2*k+1)*(2*k+2)^2*(2*k+3)/G(k+1).
G.f.: 2*Q(0) - 1 where Q(k) = 1 + x^2*(4*k + 1)^2/(x + x^2*(4*k + 1)^2 - x^2*(4*k + 3)^2*(x + x^2*(4*k + 1)^2)/(x^2*(4*k + 3)^2 + (x + x^2*(4*k + 3)^2)/Q(k+1) )).
G.f.: (1 - 1/G(0))*sqrt(-x), where G(k) = 1 + sqrt(-x) - x*(k+1)^2/G(k+1).
G.f.: Q(0), where Q(k) = 1 - x*(k+1)*(k+2)/( x*(k+1)*(k+2) - 1/Q(k+1)). (End)
O.g.f.: x + 2*x*Sum_{n>=1} x^n * Product_{k=1..n} (2*k-1)^2 / (1 + (2*k-1)^2*x). - Paul D. Hanna, Feb 05 2013
a(n) = (-4)^n*Li_{1-2*n}(-1). - Peter Luschny, Jun 28 2012
a(n) = (-4)^n*(4^n-1)*Zeta(1-2*n). - Jean-François Alcover, Dec 05 2013
Asymptotic expansion: 4*((2*(2*n-1))/(Pi*e))^(2*n-1/2)*exp(1/2+1/(12*(2*n-1))-1/(360*(2*n-1)^3)+1/(1260*(2*n-1)^5)-...). (See Luschny link.) - Peter Luschny, Jul 14 2015
From Peter Bala, Sep 11 2015: (Start)
The e.g.f. A(x) = tan(x) satisfies the differential equation A''(x) = 2*A(x)*A'(x) with A(0) = 0 and A'(0) = 1, leading to the recurrence a(0) = 0, a(1) = 1, else a(n) = 2*Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i) for the aerated sequence [0, 1, 0, 2, 0, 16, 0, 272, ...].
Note, the same recurrence, but with the initial conditions a(0) = 1 and a(1) = 1, produces the sequence n! and with a(0) = 1/2 and a(1) = 1 produces A080635. Cf. A002105, A234797. (End)
a(n) = 2*polygamma(2*n-1, 1/2)/Pi^(2*n). - Vladimir Reshetnikov, Oct 18 2015
a(n) = 2^(2n-2)*|p(2n-1,-1/2)|, where p_n(x) are the shifted row polynomials of A019538. E.g., a(2) = 2 = 2^2 * |1 + 6(-1/2) + 6(-1/2)^2|. - Tom Copeland, Oct 19 2016
From Peter Bala, May 05 2017: (Start)
With offset 0, the o.g.f. A(x) = 1 + 2*x + 16*x^2 + 272*x^3 + ... has the property that its 4th binomial transform 1/(1 - 4*x) A(x/(1 - 4*x)) has the S-fraction representation 1/(1 - 6*x/(1 - 2*x/(1 - 20*x/(1 - 12*x/(1 - 42*x/(1 - 30*x/(1 - ...))))))), where the coefficients [6, 2, 20, 12, 42, 30, ...] in the partial numerators of the continued fraction are obtained from the sequence [2, 6, 12, 20, ..., n*(n + 1), ...] by swapping adjacent terms. Compare with the S-fraction associated with A(x) given above by Paul Barry.
A(x) = 1/(1 + x - 3*x/(1 - 4*x/(1 + x - 15*x/(1 - 16*x/(1 + x - 35*x/(1 - 36*x/(1 + x - ...))))))), where the unsigned coefficients in the partial numerators [3, 4, 15, 16, 35, 36,...] come in pairs of the form 4*n^2 - 1, 4*n^2 for n = 1,2,.... (End)
a(n) = Sum_{k = 1..n-1} binomial(2*n-2, 2*k-1) * a(k) * a(n-k), with a(1) = 1. - Michael Somos, Aug 02 2018
a(n) = 2^(2*n-1) * |Euler(2*n-1, 0)|, where Euler(n,x) are the Euler polynomials. - Daniel Suteu, Nov 21 2018 (restatement of one of Copeland's 2007 formulas.)
x - Sum_{n >= 1} (-1)^n*a(n)*x^(2*n)/(2*n)! = x - log(cosh(x)). The series reversion of x - log(cosh(x)) is (1/2)*x - (1/2)*log(2 - exp(x)) = Sum_{n >= 0} A000670(n)*x^(n+1)/(n+1)!. - Peter Bala, Jul 11 2022
For n > 1, a(n) = 2*Sum_{j=1..n-1} Sum_{k=1..j} binomial(2*j,j+k)*(-4*k^2)^(n-1)*(-1)^k/(4^j). - Tani Akinari, Sep 20 2023
a(n) = A110501(n) * 4^(n-1) / n (Han and Liu, 2018). - Amiram Eldar, May 17 2024

A001250 Number of alternating permutations of order n.

Original entry on oeis.org

1, 1, 2, 4, 10, 32, 122, 544, 2770, 15872, 101042, 707584, 5405530, 44736512, 398721962, 3807514624, 38783024290, 419730685952, 4809759350882, 58177770225664, 740742376475050, 9902996106248192, 138697748786275802, 2030847773013704704, 31029068327114173810

Offset: 0

N. J. A. Sloane

For n>1, a(n) is the number of permutations of order n with the length of longest run equal 2.
Boustrophedon transform of the Euler numbers (A000111). [Berry et al., 2013] - N. J. A. Sloane, Nov 18 2013
Number of inversion sequences of length n where all consecutive subsequences i,j,k satisfy i >= j < k or i < j >= k. a(4) = 10: 0010, 0011, 0020, 0021, 0022, 0101, 0102, 0103, 0112, 0113. - Alois P. Heinz, Oct 16 2019

			1 + x + 2*x^2 + 4*x^3 + 10*x^4 + 32*x^5 + 122*x^6 + 544*x^7 + 2770*x^8 + ...
From _Gus Wiseman_, Jun 21 2021: (Start)
The a(0) = 1 through a(4) = 10 permutations:
  ()  (1)  (1,2)  (1,3,2)  (1,3,2,4)
           (2,1)  (2,1,3)  (1,4,2,3)
                  (2,3,1)  (2,1,4,3)
                  (3,1,2)  (2,3,1,4)
                           (2,4,1,3)
                           (3,1,4,2)
                           (3,2,4,1)
                           (3,4,1,2)
                           (4,1,3,2)
                           (4,2,3,1)
(End)

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 261.
F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 262.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Alois P. Heinz, Table of n, a(n) for n = 0..500 (terms n=1..100 from Max Alekseyev)
Max A. Alekseyev, On the number of permutations with bounded run lengths, arXiv:1205.4581 [math.CO], 2012-2013.
Désiré André, Sur les permutations alternées, J. Math. Pur. Appl., 7 (1881), 167-184.
Désiré André, Étude sur les maxima, minima et séquences des permutations, Ann. Sci. Ecole Norm. Sup., 3, no. 1 (1884), 121-135.
Désiré André, Mémoire sur les permutations quasi-alternées, Journal de mathématiques pures et appliquées 5e série, tome 1 (1895), 315-350.
Désiré André, Mémoire sur les séquences des permutations circulaires, Bulletin de la S. M. F., tome 23 (1895), pp. 122-184.
Stefano Barbero, Umberto Cerruti, and Nadir Murru, Some combinatorial properties of the Hurwitz series ring arXiv:1710.05665 [math.NT], 2017.
D. Berry, J. Broom, D. Dixon, and A. Flaherty, Umbral Calculus and the Boustrophedon Transform, 2013.
C. K. Cook, M. R. Bacon, and R. A. Hillman, Higher-order Boustrophedon transforms for certain well-known sequences, Fib. Q., 55(3) (2017), 201-208.
C. Davis, Problem 4755, Amer. Math. Monthly, 64 (1957) 596; solution by W. J. Blundon, 65 (1958), 533-534.
Chandler Davis, Problem 4755: A Permutation Problem, Amer. Math. Monthly, 64 (1957) 596; solution by W. J. Blundon, 65 (1958), 533-534. [Denoted by P_n in solution.] [Annotated scanned copy]
S. Kitaev, Multi-avoidance of generalized patterns, Discrete Math., 260 (2003), 89-100. (See p. 100.)
S. T. Thompson, Problem E754: Skew Ordered Sequences, Amer. Math. Monthly, 54 (1947), 416-417. [Annotated scanned copy]
Eric Weisstein's World of Mathematics, Alternating Permutation

Cf. A000111. A diagonal of A010094.
Cf. A001251, A001252, A001253, A010026, A211318, A260786.
The version for permutations of prime indices is A345164.
The version for compositions is A025047, ranked by A345167.
The version for patterns is A345194.
A049774 counts permutations avoiding adjacent (1,2,3).
A344614 counts compositions avoiding adjacent (1,2,3) and (3,2,1).
A344615 counts compositions avoiding the weak adjacent pattern (1,2,3).
A344654 counts partitions without a wiggly permutation, ranked by A344653.
A345170 counts partitions with a wiggly permutation, ranked by A345172.
A345192 counts non-wiggly compositions, ranked by A345168.
Cf. A000041, A003242, A032020, A056986, A261962, A325534, A325535, A335452, A344652, A344740, A345165.
Row sums of A104345.

a001250 n = if n == 1 then 1 else 2 * a000111 n
-- Reinhard Zumkeller, Sep 17 2014

# With Eulerian polynomials:
A := (n, x) -> `if`(n<2, 1/2/(1+I)^(1-n), add(add((-1)^j*binomial(n+1, j)*(m+1-j)^n, j=0..m)*x^m, m=0..n-1)):
A001250 := n -> 2*(I-1)^(1-n)*exp(I*(n-1)*Pi/2)*A(n,I);
seq(A001250(i), i=0..22); # Peter Luschny, May 27 2012
# second Maple program:
b:= proc(u, o) option remember;
      `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
    end:
a:= n-> `if`(n<2, 1, 2)*b(n, 0):
seq(a(n), n=0..30);  # Alois P. Heinz, Nov 29 2015

a[n_] := 4*Abs[PolyLog[-n, I]]; a[0] = a[1] = 1; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Jan 09 2016, after M. F. Hasler *)
Table[Length[Select[Permutations[Range[n]],And@@(!(OrderedQ[#]||OrderedQ[Reverse[#]])&/@Partition[#,3,1])&]],{n,8}] (* Gus Wiseman, Jun 21 2021 *)
a[0]:=1; a[1]:=1; a[n_]:=a[n]=1/(n (n-1)) Sum[a[n-1-k] a[k] k, {k,1, n-1}]; Join[{a[0], a[1]}, Map[2 #! a[#]&, Range[2,24]]] (* Oliver Seipel, May 27 2024 *)

{a(n) = local(v=[1], t); if( n<0, 0, for( k=2, n+3, t=0; v = vector( k, i, if( i>1, t += v[k+1 - i]))); v[3])} /* Michael Somos, Feb 03 2004 */

{a(n) = if( n<0, 0, n! * polcoeff( (tan(x + x * O(x^n)) + 1 / cos(x + x * O(x^n)))^2, n))} /* Michael Somos, Feb 05 2011 */

A001250(n)=sum(m=0,n\2,my(k);(-1)^m*sum(j=0,k=n+1-2*m,binomial(k,j)*(-1)^j*(k-2*j)^(n+1))/k>>k)*2-(n==1)  \\ M. F. Hasler, May 19 2012

A001250(n)=4*abs(polylog(-n,I))-(n==1)  \\ M. F. Hasler, May 20 2012

my(x='x+O('x^66), egf=1+2*(tan(x)+1/cos(x))-2-x); Vec(serlaplace(egf)) /* Joerg Arndt, May 28 2012 */

from itertools import accumulate, islice
def A001250_gen(): # generator of terms
    yield from (1,1)
    blist = (0,2)
    while True:
        yield (blist := tuple(accumulate(reversed(blist),initial=0)))[-1]
A001250_list = list(islice(A001250_gen(),40)) # Chai Wah Wu, Jun 09-11 2022

from sympy import bernoulli, euler
def A001250(n): return 1 if n<2 else abs(((1<Chai Wah Wu, Nov 13 2024

# Algorithm of L. Seidel (1877)
def A001250_list(n) :
    R = [1]; A = {-1:0, 0:2}; k = 0; e = 1
    for i in (0..n) :
        Am = 0; A[k + e] = 0; e = -e
        for j in (0..i) : Am += A[k]; A[k] = Am; k += e
        if i > 1 : R.append(A[-i//2] if i%2 == 0 else A[i//2])
    return R
A001250_list(22) # Peter Luschny, Mar 31 2012

a(n) = coefficient of x^(n-1)/(n-1)! in power series expansion of (tan(x) + sec(x))^2 = (tan(x)+1/cos(x))^2.
a(n) = coefficient of x^n/n! in power series expansion of 2*(tan(x) + sec(x)) - 2 - x. - Michael Somos, Feb 05 2011
For n>1, a(n) = 2 * A000111(n). - Michael Somos, Mar 19 2011
a(n) = 4*|Li_{-n}(i)| - [n=1] = Sum_{m=0..n/2} (-1)^m*2^(1-k)*Sum_{j=0..k} binomial(k,j)*(-1)^j*(k-2*j)^(n+1)/k - [n=1], where k = k(m) = n+1-2*m and [n=1] equals 1 if n=1 and zero else; Li denotes the polylogarithm (and i^2 = -1). - M. F. Hasler, May 20 2012
From Sergei N. Gladkovskii, Jun 18 2012: (Start)
Let E(x) = 2/(1-sin(x))-1 (essentially the e.g.f.), then
E(x) = -1 + 2*(-1/x + 1/(1-x)/x - x^3/((1-x)*((1-x)*G(0) + x^2))) where G(k) = (2*k+2)*(2*k+3)-x^2+(2*k+2)*(2*k+3)*x^2/G(k+1); (continued fraction, Euler's 1st kind, 1-step).
E(x) = -1 + 2*(-1/x + 1/(1-x)/x - x^3/((1-x)*((1-x)*G(0) + x^2))) where G(k) = 8*k + 6 - x^2/(1 + (2*k+2)*(2*k+3)/G(k+1)); (continued fraction, Euler's 2nd kind, 2-step).
E(x) = (tan(x) + sec(x))^2 = -1 + 2/(1-x*G(0)) where G(k) = 1 - x^2/(2*(2*k+1)*(4*k+3) - 2*x^2*(2*k+1)*(4*k+3)/(x^2 - 4*(k+1)*(4*k+5)/G(k+1))); (continued fraction, 3rd kind, 3-step).
(End)
G.f.: conjecture: 2*T(0)/(1-x) -1, where T(k) = 1 - x^2*(k+1)*(k+2)/(x^2*(k+1)*(k+2) - 2*(1-x*(k+1))*(1-x*(k+2))/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 19 2013
a(n) ~ 2^(n+3) * n! / Pi^(n+1). - Vaclav Kotesovec, Sep 06 2014
a(n) = Sum_{k=0..n-1} A109449(n-1,k)*A000111(k). - Reinhard Zumkeller, Sep 17 2014

Edited by Max Alekseyev, May 04 2012

a(0)=1 prepended by Alois P. Heinz, Nov 29 2015

A001286 Lah numbers: a(n) = (n-1)*n!/2.

Original entry on oeis.org

1, 6, 36, 240, 1800, 15120, 141120, 1451520, 16329600, 199584000, 2634508800, 37362124800, 566658892800, 9153720576000, 156920924160000, 2845499424768000, 54420176498688000, 1094805903679488000, 23112569077678080000, 510909421717094400000

Offset: 2

N. J. A. Sloane

Number of surjections from {1,...,n} to {1,...,n-1}. - Benoit Cloitre, Dec 05 2003
First Eulerian transform of 0,1,2,3,4,... . - Ross La Haye, Mar 05 2005
With offset 0 : determinant of the n X n matrix m(i,j)=(i+j+1)!/i!/j!. - Benoit Cloitre, Apr 11 2005
These numbers arise when expressing n(n+1)(n+2)...(n+k)[n+(n+1)+(n+2)+...+(n+k)] as sums of squares: n(n+1)[n+(n+1)] = 6(1+4+9+16+ ... + n^2), n(n+1)(n+2)(n+(n+1)+(n+2)) = 36(1+(1+4)+(1+4+9)+...+(1+4+9+16+ ... + n^2)), n(n+1)(n+2)(n+3)(n+(n+1)+(n+2)+(n+3)) = 240(...), ... . - Alexander R. Povolotsky, Oct 16 2006
a(n) is the number of edges in the Hasse diagram for the weak Bruhat order on the symmetric group S_n. For permutations p,q in S_n, q covers p in the weak Bruhat order if p,q differ by an adjacent transposition and q has one more inversion than p. Thus 23514 covers 23154 due to the transposition that interchanges the third and fourth entries. Cf. A002538 for the strong Bruhat order. - David Callan, Nov 29 2007
a(n) is also the number of excedances in all permutations of {1,2,...,n} (an excedance of a permutation p is a value j such p(j)>j). Proof: j is exceeded (n-1)! times by each of the numbers j+1, j+2, ..., n; now, Sum_{j=1..n} (n-j)(n-1)! = n!(n-1)/2. Example: a(3)=6 because the number of excedances of the permutations 123, 132, 312, 213, 231, 321 are 0, 1, 1, 1, 2, 1, respectively. - Emeric Deutsch, Dec 15 2008
(-1)^(n+1)*a(n) is the determinant of the n X n matrix whose (i,j)-th element is 0 for i = j, is j-1 for j>i, and j for j < i. - Michel Lagneau, May 04 2010
Row sums of the triangle in A030298. - Reinhard Zumkeller, Mar 29 2012
a(n) is the total number of ascents (descents) over all n-permutations. a(n) = Sum_{k=1..n} A008292(n,k)*k. - Geoffrey Critzer, Jan 06 2013
For m>=4, a(m-2) is the number of Hamiltonian cycles in a simple graph with m vertices which is complete, except for one edge. Proof: think of distinct round-table seatings of m persons such that persons "1" and "2" may not be neighbors; the count is (m-3)(m-2)!/2. See also A001710. - Stanislav Sykora, Jun 17 2014
Popularity of left (right) children in treeshelves. Treeshelves are ordered binary (0-1-2) increasing trees where every child is connected to its parent by a left or a right link. Popularity is the sum of a certain statistic (number of left children, in this case) over all objects of size n. See A278677, A278678 or A278679 for more definitions and examples. See A008292 for the distribution of the left (right) children in treeshelves. - Sergey Kirgizov, Dec 24 2016

			G.f. = x^2 + 6*x^3 + 36*x^4 + 240*x^5 + 1800*x^6 + 15120*x^7 + 141120*x^8 + ...
a(10) = (1+2+3+4+5+6+7+8+9)*(1*2*3*4*5*6*7*8*9) = 16329600. - _Reinhard Zumkeller_, May 15 2010

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 90, ex. 4.
Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 156.
A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 44.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
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 = 2..100
Yasmin Aguillon et al., On Parking Functions and The Tower of Hanoi, arXiv:2206.00541 [math.CO], 2022.
Jean-Luc Baril, Sergey Kirgizov, and Vincent Vajnovszki, Patterns in treeshelves, arXiv:1611.07793 [cs.DM], 2016.
Kyle Celano, Jennifer Elder, Kimberly P. Hadaway, Pamela E. Harris, Jeremy L. Martin, Amanda Priestley, and Gabe Udell, Statistics on L-interval parking functions, arXiv:2507.07243 [math.CO], 2025. See p. 7.
Eldar Fischer, Johann A. Makowsky, and Vsevolod Rakita, MC-finiteness of restricted set partition functions, arXiv:2302.08265 [math.CO], 2023.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 399.
Jennifer Elder, Pamela E. Harris, Jan Kretschmann, and J. Carlos Martínez Mori, Boolean intervals in the weak order of S_n, arXiv:2306.14734 [math.CO], 2023.
Lucas Chaves Meyles, Pamela E. Harris, Richter Jordaan, Gordon Rojas Kirby, Sam Sehayek, and Ethan Spingarn, Unit-Interval Parking Functions and the Permutohedron, arXiv:2305.15554 [math.CO], 2023.
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Sandi Klavžar, Uroš Milutinović and Ciril Petr, Hanoi graphs and some classical numbers, Expo. Math. 23 (2005), no. 4, 371-378.
Siegfried Lehr, Jeffrey Shallit, and John Tromp, On the vector space of the automatic reals, Theoret. Comput. Sci. 163 (1996), no. 1-2, 193-210.
P. A. Piza, Kummer numbers, Mathematics Magazine, 21 (1947/1948), 257-260.
P. A. Piza, Kummer numbers, Mathematics Magazine, 21 (1947/1948), 257-260. [Annotated scanned copy]
Eric Weisstein's World of Mathematics, Bruhat Graph.
Eric Weisstein's World of Mathematics, Edge Count.
Eric Weisstein's World of Mathematics, Permutation Ascent.

Cf. A001710, A052609, A062119, A075181, A060638, A060608, A060570, A060612, A135218, A019538, A053495, A051683, A213168, A278677, A278678, A278679, A008292.
A002868 is an essentially identical sequence.
Column 2 of |A008297|.
Third column (m=2) of triangle |A111596(n, m)|: matrix product of |S1|.S2 Stirling number matrices.
Cf. also A000110, A000111.
Cf. A001113, A001620, A091725, A099285.

a001286 n = sum[1..n-1] * product [1..n-1]
-- Reinhard Zumkeller, Aug 01 2011

[(n-1)*Factorial(n)/2: n in [2..25]]; // Vincenzo Librandi, Sep 09 2016

seq(sum(mul(j,j=3..n), k=2..n), n=2..21); # Zerinvary Lajos, Jun 01 2007

Table[Sum[n!, {i, 2, n}]/2, {n, 2, 20}] (* Zerinvary Lajos, Jul 12 2009 *)
nn=20;With[{a=Accumulate[Range[nn]],t=Range[nn]!},Times@@@Thread[{a,t}]] (* Harvey P. Dale, Jan 26 2013 *)
Table[(n - 1) n! / 2, {n, 2, 30}] (* Vincenzo Librandi, Sep 09 2016 *)

A001286(n):=(n-1)*n!/2$
makelist(A001286(n),n,1,30); /* Martin Ettl, Nov 03 2012 */

a(n)=(n-1)*n!/2 \\ Charles R Greathouse IV, Nov 20 2012

from _future_ import division
A001286_list = [1]
for n in range(2,100):
    A001286_list.append(A001286_list[-1]*n*(n+1)//(n-1)) # Chai Wah Wu, Apr 11 2018

[(n-1)*factorial(n)/2 for n in range(2, 21)] # Zerinvary Lajos, May 16 2009

a(n) = Sum_{i=0..n-1} (-1)^(n-i-1) * i^n * binomial(n-1,i). - Yong Kong (ykong(AT)curagen.com), Dec 26 2000 [corrected by Amiram Eldar, May 02 2022]
E.g.f.: x^2/[2(1-x)^2]. - Ralf Stephan, Apr 02 2004
a(n+1) = (-1)^(n+1)*det(M_n) where M_n is the n X n matrix M_(i,j)=max(i*(i+1)/2,j*(j+1)/2). - Benoit Cloitre, Apr 03 2004
Row sums of table A051683. - Alford Arnold, Sep 29 2006
5th binomial transform of A135218: (1, 1, 1, 25, 25, 745, 3145, ...). - Gary W. Adamson, Nov 23 2007
If we define f(n,i,x) = Sum_{k=i..n} Sum_{j=i..k} binomial(k,j)*Stirling1(n,k)*Stirling2(j,i)*x^(k-j) then a(n)=(-1)^n*f(n,2,-2), (n>=2). - Milan Janjic, Mar 01 2009
a(n) = A000217(n-1)*A000142(n-1). - Reinhard Zumkeller, May 15 2010
a(n) = (n+1)!*Sum_{k=1..n-1} 1/(k^2+3*k+2). - Gary Detlefs, Sep 14 2011
Sum_{n>=2} 1/a(n) = 2*(2 - exp(1) - gamma + Ei(1)) = 1.19924064599..., where gamma = A001620 and Ei(1) = A091725. - Ilya Gutkovskiy, Nov 24 2016
a(n+1) = a(n)*n*(n+1)/(n-1). - Chai Wah Wu, Apr 11 2018
Sum_{n>=2} (-1)^n/a(n) = 2*(gamma - Ei(-1)) - 2/e, where e = A001113 and Ei(-1) = -A099285. - Amiram Eldar, May 02 2022

A001586 Generalized Euler numbers, or Springer numbers.

Original entry on oeis.org

1, 1, 3, 11, 57, 361, 2763, 24611, 250737, 2873041, 36581523, 512343611, 7828053417, 129570724921, 2309644635483, 44110959165011, 898621108880097, 19450718635716001, 445777636063460643, 10784052561125704811, 274613643571568682777, 7342627959965776406281

Offset: 0

N. J. A. Sloane

From Peter Bala, Feb 02 2011: (Start)
The Springer numbers were originally considered by Glaisher (see references). They are a type B analog of the zigzag numbers A000111 for the group of signed permutations.
COMBINATORIAL INTERPRETATIONS
Several combinatorial interpretations of the Springer numbers are known:
1) a(n) gives the number of Weyl chambers in the principal Springer cone of the Coxeter group B_n of symmetries of an n dimensional cube. An example can be found in [Arnold - The Calculus of snakes...].
2) Arnold found an alternative combinatorial interpretation of the Springer numbers in terms of snakes. Snakes are a generalization of alternating permutations to the group of signed permutations. A signed permutation is a sequence (x_1,x_2,...,x_n) of integers such that {|x_1|,|x_2|,...,|x_n|} = {1,2,...,n}. They form a group, the hyperoctahedral group of order 2^n*n! = A000165(n), isomorphic to the group of symmetries of the n dimensional cube. A snake of type B_n is a signed permutation (x_1,x_2,...,x_n) such that 0 < x_1  > x_2 < ... x_n. For example, (3,-4,-2,-5,1,-6) is a snake of type B_6. a(n) gives the number of snakes of type B_n [Arnold]. The cases n=2 and n=3 are given in the Example section below.
3) The Springer numbers also arise in the study of the critical points of functions; they count the topological types of odd functions with 2*n critical values [Arnold, Theorem 35].
4) Let F_n be the set of plane rooted forests satisfying the following conditions:
... each root has exactly one child, and each of the other internal nodes has exactly two (ordered) children,
... there are n nodes labeled by integers from 1 to n, but some leaves can be non-labeled (these are called empty leaves), and labels are increasing from each root down to the leaves. Then a(n) equals the cardinality of F_n. An example and proof are given in [Verges, Theorem 4.5].
OTHER APPEARANCES OF THE SPRINGER NUMBERS
1) Hoffman has given a connection between Springer numbers, snakes and the successive derivatives of the secant and tangent functions.
2) For integer N the quarter Gauss sums Q(N) are defined by ... Q(N) := Sum_{r = 0..floor(N/4)} exp(2*Pi*I*r^2/N). In the cases N = 1 (mod 4) and N = 3 (mod 4) an asymptotic series for Q(N) as N -> inf that involves the Springer numbers has been given by Evans et al., see 1.32 and 1.33.
For a sequence of polynomials related to the Springer numbers see A185417. For a table to recursively compute the Springer numbers see A185418.
(End)
Similar to the way in which the signed Euler numbers A122045 are 2^n times the value of the Euler polynomials at 1/2, the generalized signed Euler numbers A188458 can be seen as 2^n times the value of generalized Euler polynomials at 1/2. These are the Swiss-Knife polynomials A153641. A recursive definition of these polynomials is given in A081658. - Peter Luschny, Jul 19 2012
a(n) is the number of reverse-complementary updown permutations of [2n]. For example, the updown permutation 241635 is reverse-complementary because its complement is 536142, which is the same as its reverse, and a(2)=3 counts 1324, 2413, 3412. - David Callan, Nov 29 2012
a(n) = |2^n G(n,1/2;-1)|, a specialization of the Appell sequence of polynomials umbrally formed by G(n,x;t) = (G(.,0;t) + x)^n from the Grassmann polynomials G(n,0;t) of A046802 enumerating the cells of the positive Grassmannians. - Tom Copeland, Oct 14 2015
Named "Springer numbers" after the Dutch mathematician Tonny Albert Springer (1926-2011). - Amiram Eldar, Jun 13 2021

			a(2) = 3: The three snakes of type B_2 are
  (1,-2), (2,1), (2,-1).
a(3) = 11: The 11 snakes of type B_3 are
  (1,-2,3), (1,-3,2), (1,-3,-2),
  (2,1,3), (2,-1,3), (2,-3,1), (2,-3,-1),
  (3,1,2), (3,-1,2), (3,-2,1), (3,-2,-1).

V. I. Arnold, Springer numbers and Morsification spaces. J. Algebraic Geom., Vol. 1, No. 2 (1992), pp. 197-214.
J. W. L. Glaisher, "On the coefficients in the expansions of cos x/cos 2x and sin x/cos 2x", Quart. J. Pure and Applied Math., Vol. 45 (1914), pp. 187-222.
J. W. L. Glaisher, On the Bernoullian function, Q. J. Pure Appl. Math., Vol. 29 (1898), pp. 1-168.
Ulrike Sattler, Decidable classes of formal power series with nice closure properties, Diplomarbeit im Fach Informatik, Univ. Erlangen - Nürnberg, Jul 27 1994.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Tonny Albert Springer, Remarks on a combinatorial problem, Nieuw Arch. Wisk., Vol. 19, No. 3 (1971), pp. 30-36.

Matthew House, Table of n, a(n) for n = 0..432 (terms 0..100 from T. D. Noe)
Vladimir Igorevich Arnol'd, The calculus of snakes and the combinatorics of Bernoulli, Euler and Springer numbers of Coxeter groups, Uspekhi Mat. nauk., Vol. 47, No. 1 (1992), pp. 3-45; English version, Russian Math. Surveys, Vol. 47 (1992), pp. 1-51.
Marilena Barnabei, Flavio Bonetti, Niccolò Castronuovo and Matteo Silimbani, Ascending runs in permutations and valued Dyck paths, Ars Mathematica Contemporanea, Vol. 16, No. 2 (2019), pp. 445-463.
Paul Barry, A Note on Three Families of Orthogonal Polynomials defined by Circular Functions, and Their Moment Sequences, Journal of Integer Sequences, Vol. 15 (2012), Article 12.7.2.
Paul Barry, On a transformation of Riordan moment sequences, arXiv:1802.03443 [math.CO], 2018.
William Y. C. Chen, Neil J. Y. Fan and Jeffrey Y. T. Jia, Labeled Ballot Paths and the Springer Numbers, arXiv:1009.2233 [math.CO], Sep 12 2010. [From _Jonathan Vos Post_, Sep 13 2010]
Shaoshi Chen, Hanqian Fang, Sergey Kitaev, and Candice X.T. Zhang, Patterns in Multi-dimensional Permutations, arXiv:2411.02897 [math.CO], 2024. See p. 12.
Suyoung Choi, Boram Park and Hanchul Park, The Betti numbers of real toric varieties associated to Weyl chambers of type B, arXiv preprint arXiv:1602.05406 [math.AT], 2016.
Chak-On Chow and Shi-Mei Ma, Counting signed permutations by their alternating runs, Discrete Mathematics, Vol. 323 (28 May 2014), pp. 49-57.
D. Dumont, Further triangles of Seidel-Arnold type and continued fractions related to Euler and Springer numbers Adv. Appl. Math., Vol. 16, No. 3 (1995), pp. 275-296.
Ronald Evans, Marvin Minei and Bennet Yee, Incomplete higher order Gauss sums, J. Math. Anal. Appl., Vol. 281, No. 2 (2003), pp. 454-476. See 1.32 and 1.33.
Dominique Foata and Guo-Niu Han, Multivariable Tangent and Secant q-derivative Polynomials, arXiv:1304.2486 [math.CO], 2013; alternative link.
J. W. L. Glaisher, On a set of coefficients analogous to the Eulerian numbers, Proc. London Math. Soc., 31 (1899), 216-235.
Tian Han, Sergey Kitaev, and Philip B. Zhang, Distribution of maxima and minima statistics on alternating permutations, Springer numbers, and avoidance of flat POPs, arXiv:2408.12865 [math.CO], 2024. See p. 2.
Christopher R. H. Hanusa, Alejandro H. Morales, and Martha Yip, Column convex matrices, G-cyclic orders, and flow polytopes, arXiv:2107.07326 [math.CO], 2021.
Michael E. Hoffman, Derivative Polynomials, Euler Polynomials, and Associated Integer Sequences, The Electronic Journal of Combinatorics, Vol. 6 (1999), #R21 (see Th. 3.1).
Matthieu Josuat-Vergès, Enumeration of snakes and cycle-alternating permutations, arXiv:1011.0929 [math.CO], 2010.
Matthieu Josuat-Vergès, J.-C. Novelli and J.-Y. Thibon, The algebraic combinatorics of snakes, arXiv preprint arXiv:1110.5272 [math.CO], 2011.
Emanuele Munarini, Two-Parameter Identities for q-Appell Polynomials, Journal of Integer Sequences, Vol. 26 (2023), Article 23.3.1.
Igor Pak and Andrew Soffer, On Higman's k(U_n(F_q)) conjecture, arXiv preprint arXiv:1507.00411 [math.CO], 2015.
Daniel Shanks, Generalized Euler and class numbers, Math. Comp., Vol. 21, No. 100 (1967), pp. 689-694; Vol. 22, No. 103 (1968), p. 699. [Annotated scanned copy]
Daniel Shanks, Generalized Euler and class numbers. Math. Comp., Vol. 21, No. 100 (1967), pp. 689-694.
Daniel Shanks, Corrigenda to: "Generalized Euler and class numbers", Math. Comp. 22, No. 103 (1968), p. 699.
Alan D. Sokal, The Euler and Springer numbers as moment sequences, arXiv:1804.04498 [math.CO], 2018.
Zhi-Hong Sun, Congruences involving Bernoulli polynomials, Discr. Math., Vol. 308, No. 1 (2007), pp. 71-112.
Zhi-Wei Sun, Conjectures involving arithmetical sequences, Number Theory: Arithmetic in Shangri-La (eds., S. Kanemitsu, H.-Z. Li and J.-Y. Liu), Proc. the 6th China-Japan Sem. Number Theory (Shanghai, August 15-17, 2011), World Sci., Singapore, 2013, pp. 244-258; see Conjectures involving combinatorial sequences, arXiv preprint arXiv:1208.2683 [math.CO], 2012.
M. S. Tokmachev, Correlations Between Elements and Sequences in a Numerical Prism, Bulletin of the South Ural State University, Ser. Mathematics. Mechanics. Physics, Vol. 11, No. 1 (2019), pp. 24-33.
Andrei Vieru, Agoh's conjecture: its proof, its generalizations, its analogues, arXiv preprint arXiv:1107.2938 [math.NT], 2011.
Younghan Yoon, Cohomology of type B real permutohedral varieties, arXiv:2501.09496 [math.AT], 2025. See p. 2.

Cf. A007836, A079858, A185417, A185418, A212435.
Row 2 of A349271.
Bisections are A000281 and A000464. Overview in A349264.
Related polynomials are given in A098432, A081658 and A153641.
Cf. A046802.

a := proc(n) local k; (-1)^iquo(n,2)*add(2^k*binomial(n,k)*euler(k),k=0..n) end; # Peter Luschny, Jul 08 2009
a := n -> (-1)^(n+iquo(n,2))*2^(3*n+1)*(Zeta(0,-n,1/8) - Zeta(0,-n,5/8)):
seq(a(n),n=0..21); # Peter Luschny, Mar 11 2015

n=21; CoefficientList[Series[1/(Cos[x]-Sin[x]), {x, 0, n}], x] * Table[k!, {k, 0, n}] (* Jean-François Alcover, May 18 2011 *)
Table[Abs[Numerator[EulerE[n,1/4]]],{n,0,35}] (* Harvey P. Dale, May 18 2011 *)

{a(n) = if(n<0, 0, n! * polcoeff( 1 / (cos(x + x * O(x^n)) - sin(x + x * O(x^n))), n))}; /* Michael Somos, Feb 03 2004 */

{a(n) = my(an); if(n<2, n>=0, an = vector(n+1, m, 1); for(m=2, n, an[m+1] = 2*an[m] + an[m-1] + sum(k=0, m-3, binomial(m-2, k) * (an[k+1] * an[m-1-k] + 2*an[k+2] * an[m-k] - an[k+3] * an[m-1-k]))); an[n+1])}; /* Michael Somos, Feb 03 2004 */

/* Explicit formula by Peter Bala: */
{a(n)=((1+I)/2)^n*sum(k=0,n,((1-I)/(1+I))^k*sum(j=0,k,(-1)^(k-j)*binomial(n+1,k-j)*(2*j+1)^n))}

@CachedFunction
def p(n,x) :
    if n == 0 : return 1
    w = -1 if n%2 == 0 else 0
    v =  1 if n%2 == 0 else -1
    return v*add(p(k,0)*binomial(n,k)*(x^(n-k)+w) for k in range(n)[::2])
def A001586(n) : return abs(2^n*p(n, 1/2))
[A001586(n) for n in (0..21)] # Peter Luschny, Jul 19 2012

E.g.f.: 1/(cos(x) - sin(x)).
Values at 1 of polynomials Q_n() defined in A104035. - N. J. A. Sloane, Nov 06 2009
a(n) = numerator of abs(Euler(n,1/4)). - N. J. A. Sloane, Nov 07 2009
Let B_n(x) = Sum_{k=0.. n*(n-1)/2} b(n,k)*x^k, where b(n,k) is number of n-node acyclic digraphs with k arcs, cf. A081064; then a(n) = |B_n(-2)|. - Vladeta Jovovic, Jan 25 2005
G.f. A(x) = y satisfies y'^2 = 2y^4 - y^2, y''y = y^2 + 2y'^2. - Michael Somos, Feb 03 2004
a(n) = (-1)^floor(n/2) Sum_{k=0..n} 2^k C(n,k) Euler(k). - Peter Luschny, Jul 08 2009
From Peter Bala, Feb 02 2011: (Start)
(1)... a(n) = ((1 + i)/2)^n*B(n,(1 - i)/(1 + i)), where i = sqrt(-1) and {B(n,x)}n>=0 = [1, 1 + x, 1 + 6*x + x^2, 1 + 23*x + 23*x^2 + x^3, ...] is the sequence of type B Eulerian polynomials - see A060187.
This yields the explicit formula
(2)... a(n) = ((1 + i)/2)^n*Sum_{k = 0..n} ((1 - i)/(1 + i))^k * Sum_{j = 0..k} (-1)^(k-j)*binomial(n+1,k-j)*(2*j + 1)^n.
The result (2) can be used to find congruences satisfied by the Springer numbers. For example, for odd prime p
(3)
... a(p) = 1 (mod p) when p = 4*n + 1
... a(p) = -1 (mod p) when p = 4*n + 3.
(End)
E.g.f.: 1/Q(0) where Q(k) = 1 - x/((2k+1)-x*(2k+1)/(x+(2k+2)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 19 2011
E.g.f.: 2/U(0) where U(k) = 1 + 1/(1 + x/(2*k + 1 -x - (2*k+1)/(2 - x/(x+ (2*k+2)/U(k+1))))); (continued fraction, 5-step). - Sergei N. Gladkovskii, Sep 24 2012
E.g.f.: 1/G(0) where G(k) = 1 - x/(4*k+1 - x*(4*k+1)/(4*k+2 + x + x*(4*k+2)/(4*k+3 - x - x*(4*k+3)/(x + (4*k+4)/G(k+1) )))); (continued fraction, 3rd kind, 5-step). - Sergei N. Gladkovskii, Oct 02 2012
G.f.: 1/G(0) where G(k) = 1 - x*(2*k+1) - 2*x^2*(k+1)*(k+1)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 11 2013
a(n) = | 2*4^n*lerchphi(-1, -n, 1/4) |. - Peter Luschny, Apr 27 2013
a(n) ~ 4 * n^(n+1/2) * (4/Pi)^n / (sqrt(Pi)*exp(n)). - Vaclav Kotesovec, Oct 07 2013
G.f.: T(0)/(1-x), where T(k) = 1 - 2*x^2*(k+1)^2/( 2*x^2*(k+1)^2 - (1-x-2*x*k)*(1-3*x-2*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 15 2013
a(n) = (-1)^C(n+1,2)*2^(3*n+1)*(Zeta(-n,1/8)-Zeta(-n,5/8)), where Zeta(a,z) is the generalized Riemann zeta function. - Peter Luschny, Mar 11 2015
E.g.f. A(x) satisfies: A(x) = exp( Integral A(x)/A(-x) dx ). - Paul D. Hanna, Feb 04 2017
E.g.f. A(x) satisfies: A'(x) = A(x)^2/A(-x). - Paul D. Hanna, Feb 04 2017

More terms from Vladeta Jovovic, Jan 25 2005

A155585 a(n) = 2^n*E(n, 1) where E(n, x) are the Euler polynomials.

Original entry on oeis.org

1, 1, 0, -2, 0, 16, 0, -272, 0, 7936, 0, -353792, 0, 22368256, 0, -1903757312, 0, 209865342976, 0, -29088885112832, 0, 4951498053124096, 0, -1015423886506852352, 0, 246921480190207983616, 0, -70251601603943959887872, 0, 23119184187809597841473536, 0

Offset: 0

Paul D. Hanna, Jan 24 2009

Previous name was: a(n) = Sum_{k=0..n-1} (-1)^(k)*C(n-1,k)*a(n-1-k)*a(k) for n>0 with a(0)=1.
Factorials have a similar recurrence: f(n) = Sum_{k=0..n-1} C(n-1,k)*f(n-1-k)*f(k), n > 0.
Related to A102573: letting T(q,r) be the coefficient of n^(r+1) in the polynomial 2^(q-n)/n times Sum_{k=0..n} binomial(n,k)*k^q, then A155585(x) = Sum_{k=0..x-1} T(x,k)*(-1)^k. See Mathematica code below. - John M. Campbell, Nov 16 2011
For the difference table and the relation to the Seidel triangle see A239005. - Paul Curtz, Mar 06 2014
From Tom Copeland, Sep 29 2015: (Start)
Let z(t) = 2/(e^(2t)+1) = 1 + tanh(-t) = e.g.f.(-t) for this sequence = 1 - t + 2*t^3/3! - 16*t^5/5! + ... .
dlog(z(t))/dt = -z(-t), so the raising operators that generate Appell polynomials associated with this sequence, A081733, and its reciprocal, A119468, contain z(-d/dx) = e.g.f.(d/dx) as the differential operator component.
dz(t)/dt = z*(z-2), so the assorted relations to a Ricatti equation, the Eulerian numbers A008292, and the Bernoulli numbers in the Rzadkowski link hold.
From Michael Somos's formula below (drawing on the Edwards link), y(t,1)=1 and x(t,1) = (1-e^(2t))/(1+e^(2t)), giving z(t) = 1 + x(t,1). Compare this to the formulas in my list in A008292 (Sep 14 2014) with a=1 and b=-1,
A) A(t,1,-1) = A(t) = -x(t,1) = (e^(2t)-1)/(1+e^(2t)) = tanh(t) = t + -2*t^3/3! + 16*t^5/5! + -272*t^7/7! + ... = e.g.f.(t) - 1 (see A000182 and A000111)
B) Ainv(t) = log((1+t)/(1-t))/2 = tanh^(-1)(t) = t + t^3/3 + t^5/5 + ..., the compositional inverse of A(t)
C) dA/dt = (1-A^2), relating A(t) to a Weierstrass elliptic function
D) ((1-t^2)d/dt)^n t evaluated at t=0, a generator for the sequence A(t)
F) FGL(x,y)= (x+y)/(1+xy) = A(Ainv(x) + Ainv(y)), a related formal group law corresponding to the Lorentz FGL (Lorentz transformation--addition of parallel velocities in special relativity) and the Atiyah-Singer signature and the elliptic curve (1-t^2)*s = t^3 in Tate coordinates according to the Lenart and Zainoulline link and the Buchstaber and Bunkova link (pp. 35-37) in A008292.
A133437 maps the reciprocal odd natural numbers through the refined faces of associahedra to a(n).
A145271 links the differential relations to the geometry of flow maps, vector fields, and thereby formal group laws. See Mathworld for links of tanh to other geometries and statistics.
Since the a(n) are related to normalized values of the Bernoulli numbers and the Riemann zeta and Dirichlet eta functions, there are links to Witten's work on volumes of manifolds in two-dimensional quantum gauge theories and the Kervaire-Milnor formula for homotopy groups of hyperspheres (see my link below).
See A101343, A111593 and A059419 for this and the related generator (1 + t^2) d/dt and associated polynomials. (End)
With the exception of the first term (1), entries are the alternating sums of the rows of the Eulerian triangle, A008292. - Gregory Gerard Wojnar, Sep 29 2018

			E.g.f.: 1 + x - 2*x^3/3! + 16*x^5/5! - 272*x^7/7! + 7936*x^9/9! -+ ... = exp(x)/cosh(x).
O.g.f.: 1 + x - 2*x^3 + 16*x^5 - 272*x^7 + 7936*x^9 - 353792*x^11 +- ...
O.g.f.: 1 + x/(1+2*x) + 2!*x^2/((1+2*x)*(1+4*x)) + 3!*x^3/((1+2*x)*(1+4*x)*(1+6*x)) + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..200
P. Barry, Riordan Arrays, Orthogonal Polynomials as Moments, and Hankel Transforms, J. Int. Seq. 14 (2011) # 11.2.2, example 28.
Tom Copeland, The Elliptic Lie Triad: Ricatti and KdV Equations, Infinigens, and Elliptic Genera
Tom Copeland, The Kervaire-Milnor formula
H. M. Edwards, A normal form for elliptic curves, Bull. Amer. Math. Soc. (N.S.) 44 (2007), no. 3, 393-422. see section 12, pp. 410-411.
Wenjie Fang, Hsien-Kuei Hwang, and Mihyun Kang, Phase transitions from exp(n^(1/2)) to exp(n^(2/3)) in the asymptotics of banded plane partitions, arXiv:2004.08901 [math.CO], 2020.
G. Rzadkowski, Bernoulli numbers and solitons-revisited, Journal of Nonlinear Math. Physics, 1711, pp. 121-126.

Cf. A001057, A009006, A102573.
Equals row sums of A119879. - Johannes W. Meijer, Apr 20 2011
Cf. A081733, A119468, A008292, A000182, A000111, A101343, A111593, A059419, A133437, A145271.
(-1)^n*a(n) are the alternating row sums of A123125. - Wolfdieter Lang, Jul 12 2017

A155585 := n -> 2^n*euler(n, 1): # Peter Luschny, Jan 26 2009
a := proc(n) option remember; `if`(n::even, 0^n, -(-1)^n - add((-1)^k*binomial(n,k) *a(n-k), k = 1..n-1)) end: # Peter Luschny, Jun 01 2016
# Or via the recurrence of the Fubini polynomials:
F := proc(n) option remember; if n = 0 then return 1 fi;
expand(add(binomial(n, k)*F(n-k)*x, k = 1..n)) end:
a := n -> (-2)^n*subs(x = -1/2, F(n)):
seq(a(n), n = 0..30); # Peter Luschny, May 21 2021

a[m_] := Sum[(-2)^(m - k) k! StirlingS2[m, k], {k, 0, m}] (* Peter Luschny, Apr 29 2009 *)
poly[q_] :=  2^(q-n)/n*FunctionExpand[Sum[Binomial[n, k]*k^q, {k, 0, n}]]; T[q_, r_] :=  First[Take[CoefficientList[poly[q], n], {r+1, r+1}]]; Table[Sum[T[x, k]*(-1)^k, {k, 0, x-1}], {x, 1, 16}] (* John M. Campbell, Nov 16 2011 *)
f[n_] := (-1)^n 2^(n+1) PolyLog[-n, -1]; f[0] = -f[0]; Array[f, 27, 0] (* Robert G. Wilson v, Jun 28 2012 *)

a(n)=if(n==0,1,sum(k=0,n-1,(-1)^(k)*binomial(n-1,k)*a(n-1-k)*a(k)))

a(n)=local(X=x+x*O(x^n));n!*polcoeff(exp(X)/cosh(X),n)

a(n)=polcoeff(sum(m=0,n,m!*x^m/prod(k=1,m,1+2*k*x+x*O(x^n))),n) \\ Paul D. Hanna, Jul 20 2011

{a(n) = local(A); if( n<0, 0, A = x * O(x^n); n! * polcoeff( 1 + sinh(x + A) / cosh(x + A), n))} /* Michael Somos, Jan 16 2012 */

a(n)=local(A=1+x);for(i=1,n,A=sum(k=0,n,intformal(subst(A,x,-x)+x*O(x^n))^k/k!));n!*polcoeff(A,n)
for(n=0,30,print1(a(n),", ")) \\ Paul D. Hanna, Nov 25 2013

from sympy import bernoulli
def A155585(n): return (((2<<(m:=n+1))-2)*bernoulli(m)<>1) if n&1 else (0 if n else 1) # Chai Wah Wu, Apr 14 2023

def A155585(n) :
    if n == 0 : return 1
    return add(add((-1)^(j+1)*binomial(n+1,k-j)*j^n for j in (0..k)) for k in (1..n))
[A155585(n) for n in (0..26)] # Peter Luschny, Jul 23 2012

def A155585_list(n): # Akiyama-Tanigawa algorithm
    A = [0]*(n+1); R = []
    for m in range(n+1) :
        d = divmod(m+3, 4)
        A[m] = 0 if d[1] == 0 else (-1)^d[0]/2^(m//2)
        for j in range(m, 0, -1) :
            A[j - 1] = j * (A[j - 1] - A[j])
        R.append(A[0])
    return R
A155585_list(30) # Peter Luschny, Mar 09 2014

E.g.f.: exp(x)*sech(x) = exp(x)/cosh(x). (See A009006.) - Paul Barry, Mar 15 2006
Sequence of absolute values is A009006 (e.g.f. 1+tan(x)).
O.g.f.: Sum_{n>=0} n! * x^n / Product_{k=1..n} (1 + 2*k*x). - Paul D. Hanna, Jul 20 2011
a(n) = 2^n*E_{n}(1) where E_{n}(x) are the Euler polynomials. - Peter Luschny, Jan 26 2009
a(n) = EL_{n}(-1) where EL_{n}(x) are the Eulerian polynomials. - Peter Luschny, Aug 03 2010
a(n+1) = (4^n-2^n)*B_n(1)/n, where B_{n}(x) are the Bernoulli polynomials (B_n(1) = B_n for n <> 1). - Peter Luschny, Apr 22 2009
G.f.: 1/(1-x+x^2/(1-x+4*x^2/(1-x+9*x^2/(1-x+16*x^2/(1-...))))) (continued fraction). - Paul Barry, Mar 30 2010
G.f.: -log(x/(exp(x)-1))/x = Sum_{n>=0} a(n)*x^n/(2^(n+1)*(2^(n+1)-1)*n!). - Vladimir Kruchinin, Nov 05 2011
E.g.f.: exp(x)/cosh(x) = 2/(1+exp(-2*x)) = 2/(G(0) + 1); G(k) = 1 - 2*x/(2*k + 1 - x*(2*k+1)/(x - (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 10 2011
E.g.f. is x(t,1) + y(t,1) where x(t,a) and y(t,a) satisfy y(t,a)^2 = (a^2 - x(t,a)^2) / (1 - a^2 * x(t,a)^2) and dx(t,a) / dt = y(t,a) * (1 - a * x(t,a)^2) and are the elliptic functions of Edwards. - Michael Somos, Jan 16 2012
E.g.f.: 1/(1 - x/(1+x/(1 - x/(3+x/(1 - x/(5+x/(1 - x/(7+x/(1 - x/(9+x/(1 +...))))))))))), a continued fraction. - Paul D. Hanna, Feb 11 2012
E.g.f. satisfies: A(x) = Sum_{n>=0} Integral( A(-x) dx )^n / n!. - Paul D. Hanna, Nov 25 2013
a(n) = -2^(n+1)*Li_{-n}(-1). - Peter Luschny, Jun 28 2012
a(n) = Sum_{k=1..n} Sum_{j=0..k} (-1)^(j+1)*binomial(n+1,k-j)*j^n for n > 0. - Peter Luschny, Jul 23 2012
From Sergei N. Gladkovskii, Oct 25 2012 to Dec 16 2013: (Start)
Continued fractions:
G.f.: 1 + x/T(0) where T(k) = 1 + (k+1)*(k+2)*x^2/T(k+1).
E.g.f.: exp(x)/cosh(x) = 1 + x/S(0) where S(k) = (2*k+1) + x^2/S(k+1).
E.g.f.: 1 + x/(U(0)+x) where U(k) = 4*k+1 - x/(1 + x/(4*k+3 - x/(1 + x/U(k+1)))).
E.g.f.: 1 + tanh(x) = 4*x/(G(0)+2*x) where G(k) = 1 - (k+1)/(1 - 2*x/(2*x + (k+1)^2/G(k+1)));
G.f.: 1 + x/G(0) where G(k) = 1 + 2*x^2*(2*k+1)^2 - x^4*(2*k+1)*(2*k+2)^2*(2*k+3)/G(k+1) (due to Stieltjes).
E.g.f.: 1 + x/(G(0) + x) where G(k) = 1 - 2*x/(1 + (k+1)/G(k+1)).
G.f.: 2 - 1/Q(0) where Q(k) = 1 + x*(k+1)/( 1 - x*(k+1)/Q(k+1)).
G.f.: 2 - 1/Q(0) where Q(k) = 1 + x*k^2 + x/(1 - x*(k+1)^2/Q(k+1)).
G.f.: 1/Q(0) where Q(k) = 1 - 2*x + x*(k+1)/(1-x*(k+1)/Q(k+1)).
G.f.: 1/Q(0) where Q(k) = 1 - x*(k+1)/(1 + x*(k+1)/Q(k+1)).
E.g.f.: 1 + x*Q(0) where Q(k) = 1 - x^2/( x^2 + (2*k+1)*(2*k+3)/Q(k+1)).
G.f.: 2 - T(0)/(1+x) where T(k) = 1 - x^2*(k+1)^2/(x^2*(k+1)^2 + (1+x)^2/T(k+1)).
E.g.f.: 1/(x - Q(0)) where Q(k) = 4*k^2 - 1 + 2*x + x^2*(2*k-1)*(2*k+3)/Q(k+1). (End)
G.f.: 1 / (1 - b(1)*x / (1 - b(2)*x / (1 - b(3)*x / ... ))) where b = A001057. - Michael Somos, Jan 03 2013
From Paul Curtz, Mar 06 2014: (Start)
a(2n) = A000007(n).
a(2n+1) = (-1)^n*A000182(n+1).
a(n) is the binomial transform of A122045(n).
a(n) is the row sum of A081658. For fractional Euler numbers see A238800.
a(n) + A122045(n) = 2, 1, -1, -2, 5, 16, ... = -A163982(n).
a(n) - A122045(n) = -A163747(n).
a(n) is the Akiyama-Tanigawa transform applied to 1, 0, -1/2, -1/2, -1/4, 0, ... = A046978(n+3)/A016116(n). (End)
a(n) = 2^(2*n+1)*(zeta(-n,1/2) - zeta(-n, 1)), where zeta(a, z) is the generalized Riemann zeta function. - Peter Luschny, Mar 11 2015
a(n) = 2^(n + 1)*(2^(n + 1) - 1)*Bernoulli(n + 1, 1)/(n + 1). (From Bill Gosper, Oct 28 2015) - N. J. A. Sloane, Oct 28 2015 [See the above comment from Peter Luschny, Apr 22 2009.]
a(n) = -(n mod 2)*((-1)^n + Sum_{k=1..n-1} (-1)^k*C(n,k)*a(n-k)) for n >= 1. - Peter Luschny, Jun 01 2016
a(n) = (-2)^n*F_{n}(-1/2), where F_{n}(x) is the Fubini polynomial. - Peter Luschny, May 21 2021

New name from Peter Luschny, Mar 12 2015

cp's OEIS Frontend

A218529 Binomial transform of -1, -1, 1, 2, -5, -16, ... (signed variant of A000111).

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A261880 Array of higher-order differences of the sequence (-1)^n*A000111(n) read by downward antidiagonals.

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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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A000364 Euler (or secant or "Zig") numbers: e.g.f. (even powers only) sec(x) = 1/cos(x).

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A000165 Double factorial of even numbers: (2n)!! = 2^n*n!.

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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).