A000364 -id:A000364 - cp's OEIS frontend

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

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)
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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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A. R. Kräuter, Über die Permanente gewisser zirkulärer Matrizen..., Séminaire Lotharingien de Combinatoire, B11b (1984), 11 pp.
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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N. E. Nørlund, Vorlesungen ueber Differenzenrechnung Springer 1924, p. 27.
N. E. Nörlund, Vorlesungen über Differenzenrechnung, Springer-Verlag, Berlin, 1924 [Annotated scanned copy of pages 144-151 and 456-463]
Jay Rosen, The Number of Product-Weighted Lead Codes for Ballots and Its Relation to the Ursell Functions of the Linear Ising Model, Journal of Combinatorial Theory, Vol. 20, No.3, May 1976, 377-384.
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
N. J. A. Sloane, My favorite integer sequences, in Sequences and their Applications (Proceedings of SETA '98).
R. P. Stanley, Permutations
Ross Street, Trees, permutations and the tangent function gives definition of Joyce trees and tremolo permutations, arXiv:math/0303267 [math.HO], 2003.
Ross Street, Surprising relationships connecting ploughing a field, mathematical trees, permutations, and trigonometry, Slides from a talk, July 15 2015, Macquarie University. ["There is a Web Page oeis.org by N. J. A. Sloane. It tells, from typing the first few terms of a sequence, whether that sequence has occurred somewhere else in Mathematics. Postgraduate student Daniel Steffen traced this down and found, to our surprise, that the sequence was related to the tangent function tan x. Ryan and Tam searched out what was known about this connection and discovered some apparently new results. We all found this a lot of fun and I hope you will too."]
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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

A002445 Denominators of Bernoulli numbers B_{2n}.

Original entry on oeis.org

1, 6, 30, 42, 30, 66, 2730, 6, 510, 798, 330, 138, 2730, 6, 870, 14322, 510, 6, 1919190, 6, 13530, 1806, 690, 282, 46410, 66, 1590, 798, 870, 354, 56786730, 6, 510, 64722, 30, 4686, 140100870, 6, 30, 3318, 230010, 498, 3404310, 6, 61410, 272118, 1410, 6, 4501770, 6, 33330, 4326, 1590, 642, 209191710, 1518, 1671270, 42

Offset: 0

N. J. A. Sloane

From the von Staudt-Clausen theorem, denominator(B_2n) = product of primes p such that (p-1)|2n.
Row products of A138239. - Mats Granvik, Mar 08 2008
Equals row products of even rows in triangle A143343. In triangle A080092, row products = denominators of B1, B2, B4, B6, ... . - Gary W. Adamson, Aug 09 2008
Julius Worpitzky's 1883 algorithm for generating Bernoulli numbers is shown in A028246. - Gary W. Adamson, Aug 09 2008
There is a relation between the Euler numbers E_n and the Bernoulli numbers B_{2*n}, for n>0, namely, B_{2*n} = A000367(n)/a(n) = ((-1)^n/(2*(1-2^{2*n}))) * Sum_{k = 0..n-1} (-1)^k*2^{2*k}*C(2*n,2*k)*A000364(n-k)*A000367(k)/a(k). (See Bucur, et al.) - L. Edson Jeffery, Sep 17 2012
a(n) is the product of all primes of the form (k + n)/(k - n). - Thomas Ordowski, Jul 24 2025

			B_{2n} = [ 1, 1/6, -1/30, 1/42, -1/30, 5/66, -691/2730, 7/6, -3617/510, ... ].

Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 136.
G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
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).
See A000367 for further references and links (there are a lot).

T. D. Noe, Table of n, a(n) for n = 0..10000
Amelia Bucur, José Luis López-Bonilla, and Jaime Robles-García, A note on the Namias identity for Bernoulli numbers, Journal of Scientific Research (Banaras Hindu University, Varanasi), Vol. 56 (2012), 117-120.
Suyuong Choi and Younghan Yoon, A decomposition of graph a-numbers, arXiv:2508.06855 [math.CO], 2025. See p. 13.
G. Everest, A. J. van der Poorten, Y. Puri and T. Ward, Integer Sequences and Periodic Points, Journal of Integer Sequences, Vol. 5 (2002), Article 02.2.3
Shizuo Kaji, Toshiaki Maeno, Koji Nuida, and Yasuhide Numata, Polynomial Expressions of Carries in p-ary Arithmetics, arXiv preprint arXiv:1506.02742 [math.CO], 2015.
Takao Komatsu, Florian Luca, and Claudio de J. Pita Ruiz V. , A note on the denominators of Bernoulli numbers, Proc. Japan Acad., 90, Ser. A (2014), p. 71-72.
Guo-Dong Liu, H. M. Srivastava, and Hai-Quing Wang, Some Formulas for a Family of Numbers Analogous to the Higher-Order Bernoulli Numbers, J. Int. Seq. 17 (2014) # 14.4.6
Hong-Mei Liu, Shu-Hua Qi, and Shu-Yan Ding, Some Recurrence Relations for Cauchy Numbers of the First Kind, JIS 13 (2010) # 10.3.8.
Romeo Meštrović, On a Congruence Modulo n^3 Involving Two Consecutive Sums of Powers, Journal of Integer Sequences, Vol. 17 (2014), 14.8.4.
Niels Nielsen, Traité élémentaire des nombres de Bernoulli, Gauthier-Villars, 1923, pp. 398.
Niels Erik Nörlund, Vorlesungen über Differenzenrechnung, Springer-Verlag, Berlin, 1924 [Annotated scanned copy of pages 144-151 and 456-463]
Ronald Orozco López, Solution of the Differential Equation y^(k)= e^(a*y), Special Values of Bell Polynomials and (k,a)-Autonomous Coefficients, Universidad de los Andes (Colombia 2021).
Simon Plouffe, The First 498 Bernoulli numbers [Project Gutenberg Etext]
Jan W. H. Swanepoel, A Short Simple Probabilistic Proof of a Well Known Identity and the Derivation of Related New Identities Involving the Bernoulli Numbers and the Euler Numbers, Integers (2025) Vol. 25, Art. No. A50. See p. 2.
Index entries for sequences related to Bernoulli numbers.

Cf. A090801 (distinct numbers appearing as denominators of Bernoulli numbers)
B_n gives A027641/A027642. See A027641 for full list of references, links, formulas, etc.
See A000367 for numerators. Cf. A027762, A027641, A027642, A002882, A003245, A127187, A127188, A138239, A028246, A143343, A080092, A001897, A277087.
Cf. A160014 for a generalization.

[Denominator(Bernoulli(2*n)): n in [0..60]]; // Vincenzo Librandi, Nov 16 2014

A002445 := n -> mul(i,i=select(isprime,map(i->i+1,numtheory[divisors] (2*n)))): seq(A002445(n),n=0..40); # Peter Luschny, Aug 09 2011
# Alternative
N:= 1000: # to get a(0) to a(N)
A:= Vector(N,2):
for p in select(isprime,[seq(2*i+1,i=1..N)]) do
  r:= (p-1)/2;
  for n from r to N by r do
    A[n]:= A[n]*p
  od
od:
1, seq(A[n],n=1..N); # Robert Israel, Nov 16 2014

Take[Denominator[BernoulliB[Range[0,100]]],{1,-1,2}] (* Harvey P. Dale, Oct 17 2011 *)

a(n)=prod(p=2,2*n+1,if(isprime(p),if((2*n)%(p-1),1,p),1)) \\ Benoit Cloitre

A002445(n,P=1)=forprime(p=2,1+n*=2,n%(p-1)||P*=p);P \\ M. F. Hasler, Jan 05 2016

a(n) = denominator(bernfrac(2*n)); \\ Michel Marcus, Jul 16 2021

def A002445(n):
    if n == 0:
        return 1
    M = (i + 1 for i in divisors(2 * n))
    return prod(s for s in M if is_prime(s))
[A002445(n) for n in (0..57)] # Peter Luschny, Feb 20 2016

E.g.f: x/(exp(x) - 1); take denominators of even powers.
B_{2n}/(2n)! = 2*(-1)^(n-1)*(2*Pi)^(-2n) Sum_{k=1..inf} 1/k^(2n) (gives asymptotics) - Rademacher, p. 16, Eq. (9.1). In particular, B_{2*n} ~ (-1)^(n-1)*2*(2*n)!/ (2*Pi)^(2*n).
If n>=3 is prime,then a((n+1)/2)==(-1)^((n-1)/2)*12*|A000367((n+1)/2)|(mod n). - Vladimir Shevelev, Sep 04 2010
a(n) = denominator(-I*(2*n)!/(Pi*(1-2*n))*integral(log(1-1/t)^(1-2*n) dt, t=0..1)). - Gerry Martens, May 17 2011
a(n) = 2*denominator((2*n)!*Li_{2*n}(1)) for n > 0. - Peter Luschny, Jun 28 2012
a(n) = gcd(2!S(2n+1,2),...,(2n+1)!S(2n+1,2n+1)). Here S(n,k) is the Stirling number of the second kind. See the paper of Komatsu et al. - Istvan Mezo, May 12 2016
a(n) = 2*A001897(n) = A027642(2*n) = 3*A277087(n) for n>0. - Jonathan Sondow, Dec 14 2016

A013663 Decimal expansion of zeta(5).

Original entry on oeis.org

1, 0, 3, 6, 9, 2, 7, 7, 5, 5, 1, 4, 3, 3, 6, 9, 9, 2, 6, 3, 3, 1, 3, 6, 5, 4, 8, 6, 4, 5, 7, 0, 3, 4, 1, 6, 8, 0, 5, 7, 0, 8, 0, 9, 1, 9, 5, 0, 1, 9, 1, 2, 8, 1, 1, 9, 7, 4, 1, 9, 2, 6, 7, 7, 9, 0, 3, 8, 0, 3, 5, 8, 9, 7, 8, 6, 2, 8, 1, 4, 8, 4, 5, 6, 0, 0, 4, 3, 1, 0, 6, 5, 5, 7, 1, 3, 3, 3, 3

Offset: 1

N. J. A. Sloane

In a widely distributed May 2011 email, Wadim Zudilin gave a rebuttal to v1 of Kim's 2011 preprint: "The mistake (unfixable) is on p. 6, line after eq. (3.3). 'Without loss of generality' can be shown to work only for a finite set of n_k's; as the n_k are sufficiently large (and N is fixed), the inequality for epsilon is false." In a May 2013 email, Zudilin extended his rebuttal to cover v2, concluding that Kim's argument "implies that at least one of zeta(2), zeta(3), zeta(4) and zeta(5) is irrational, which is trivial." - Jonathan Sondow, May 06 2013

General: zeta(2*s + 1) = (A000364(s)/A331839(s)) * Pi^(2*s + 1) * Product_{k >= 1} (A002145(k)^(2*s + 1) + 1)/(A002145(k)^(2*s + 1) - 1), for s >= 1. - Dimitris Valianatos, Apr 27 2020

			1/1^5 + 1/2^5 + 1/3^5 + 1/4^5 + 1/5^5 + 1/6^5 + 1/7^5 + ... =
1 + 1/32 + 1/243 + 1/1024 + 1/3125 + 1/7776 + 1/16807 + ... = 1.036927755143369926331365486457...

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 262.
Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 811.

Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Michael J. Dancs and Tian-Xiao He, An Euler-type formula for zeta(2k+1), Journal of Number Theory, Volume 118, Issue 2, June 2006, Pages 192-199.
Robert J. Harley, Zeta(3), Zeta(5), .., Zeta(99) 10000 digits (txt, 400 KB).
Yong-Cheol Kim, zeta(5) is irrational, arXiv:1105.0730 [math.CA], 2011. [Jonathan Vos Post, May 4, 2011].
Simon Plouffe, Computation of Zeta(5)
Simon Plouffe, Zeta(5), the sum(1/n**5, n=1..infinity) to 512 digits
Simon Plouffe, Other interesting computations at numberworld.org.
Zhi-Wei Sun, A curious hypergeometric series related to Riemann's zeta function, Question 486353 at MathOverflow, Jan. 21, 2025.
Chuanan Wei, Some fast convergent series for the mathematical constants zeta(4) and zeta(5), arXiv:2303.07887 [math.CO], 2023.
Wikipedia, Zeta constant
Wadim Zudilin, One of the numbers ζ(5), ζ(7), ζ(9), ζ(11) is irrational, Russ. Math. Surv., 56 (2001), 774-776.
Index entries for constants related to zeta

Cf. A013661, A002117, A013662, A013664, A013665, A013666, A013667, A013668, A013669, A013670, A013671, A013672, A013673, A013674, A013675, A013676, A013677, A013678, A293904.
Cf. A243264, A255323.
Cf. A023872, A023873, A248882, A255050, A255052, A057528, A260404.

RealDigits[Zeta[5], 10, 100][[1]] (* Alonso del Arte, Jan 13 2012 *)

PARI
```
zeta(5) \\ Michel Marcus, Apr 17 2016
```

From Peter Bala, Dec 04 2013: (Start)
Definition: zeta(5) = Sum_{n >= 1} 1/n^5.
zeta(5) = 2^5/(2^5 - 1)*(Sum_{n even} n^5*p(n)*p(1/n)/(n^2 - 1)^6 ), where p(n) = n^2 + 3. See A013667, A013671 and A013675. (End)
zeta(5) = Sum_{n >= 1} (A010052(n)/n^(5/2)) = Sum_{n >= 1} ((floor(sqrt(n)) - floor(sqrt(n-1)))/n^(5/2)). - Mikael Aaltonen, Feb 22 2015
zeta(5) = Product_{k>=1} 1/(1 - 1/prime(k)^5). - Vaclav Kotesovec, Apr 30 2020
From Artur Jasinski, Jun 27 2020: (Start)
zeta(5) = (-1/30)*Integral_{x=0..1} log(1-x^4)^5/x^5.
zeta(5) = (1/24)*Integral_{x=0..infinity} x^4/(exp(x)-1).
zeta(5) = (2/45)*Integral_{x=0..infinity} x^4/(exp(x)+1).
zeta(5) = (1/(1488*zeta(1/2)^5))*(-5*Pi^5*zeta(1/2)^5 + 96*zeta'(1/2)^5 - 240*zeta(1/2)*zeta'(1/2)^3*zeta''(1/2) + 120*zeta(1/2)^2*zeta'(1/2)*zeta''(1/2)^2 + 80*zeta(1/2)^2*zeta'(1/2)^2*zeta'''(1/2)- 40*zeta(1/2)^3*zeta''(1/2)*zeta'''(1/2) - 20*zeta(1/2)^3*zeta'(1/2)*zeta''''(1/2)+4*zeta(1/2)^4*zeta'''''(1/2)). (End).
From Peter Bala, Oct 29 2023: (Start)
zeta(3) = (8/45)*Integral_{x >= 1} x^3*log(x)^3*(1 + log(x))*log(1 + 1/x^x) dx =  (2/45)*Integral_{x >= 1} x^4*log(x)^4*(1 + log(x))/(1 + x^x) dx.
zeta(5) = 131/128 + 26*Sum_{n >= 1} (n^2 + 2*n + 40/39)/(n*(n + 1)*(n + 2))^5.
zeta(5) =  5162893/4976640 - 1323520*Sum_{n >= 1} (n^2 + 4*n + 56288/12925)/(n*(n + 1)*(n + 2)*(n + 3)*(n + 4))^5. Taking 10 terms of the series gives a value for zeta(5) correct to 20 decimal places.
Conjecture: for k >= 1, there exist rational numbers A(k), B(k) and c(k) such that zeta(5) = A(k) + B(k)*Sum_{n >= 1} (n^2 + 2*k*n + c(k))/(n*(n + 1)*...*(n + 2*k))^5. A similar conjecture can be made for the constant zeta(3). (End)
zeta(5) = (694/204813)*Pi^5 - Sum_{n >= 1} (6280/3251)*(1/(n^5*(exp(4*Pi*n)-1))) + Sum_{n >= 1} (296/3251)*(1/(n^5*(exp(5*Pi*n)-1))) - Sum_{n >= 1} (1073/6502)*(1/(n^5*(exp(10*Pi*n)-1))) + Sum_{n >= 1} (37/6502)*(1/(n^5*(exp(20*Pi*n)-1))). - Simon Plouffe, Jan 06 2024
From Peter Bala, Apr 27 2025: (Start)
zeta(5) = 1/5! * Integral_{x >= 0} x^5 * exp(x)/(exp(x) - 1)^2 dx = (16/15) * 1/5! * Integral_{x >= 0} x^5 * exp(x)/(exp(x) + 1)^2 dx.
zeta(5) = 1/6! * Integral_{x >= 0} x^6 * exp(x)*(exp(x) + 1)/(exp(x) - 1)^3 dx = 1/(3^3 * 5^2) * Integral_{x >= 0} x^6 * exp(x)*(exp(x) - 1)/(exp(x) + 1)^3 dx. (End)
zeta(5) = Sum_{i, j >= 1} 1/((i^4)*j*binomial(i+j, i)). More generally, zeta(n+1) = Sum_{i, j >= 1} 1/((i^n)*j*binomial(i+j, i)) for n >= 1. - Peter Bala, Aug 07 2025

A003881 Decimal expansion of Pi/4.

Original entry on oeis.org

7, 8, 5, 3, 9, 8, 1, 6, 3, 3, 9, 7, 4, 4, 8, 3, 0, 9, 6, 1, 5, 6, 6, 0, 8, 4, 5, 8, 1, 9, 8, 7, 5, 7, 2, 1, 0, 4, 9, 2, 9, 2, 3, 4, 9, 8, 4, 3, 7, 7, 6, 4, 5, 5, 2, 4, 3, 7, 3, 6, 1, 4, 8, 0, 7, 6, 9, 5, 4, 1, 0, 1, 5, 7, 1, 5, 5, 2, 2, 4, 9, 6, 5, 7, 0, 0, 8, 7, 0, 6, 3, 3, 5, 5, 2, 9, 2, 6, 6, 9, 9, 5, 5, 3, 7

Offset: 0

N. J. A. Sloane, Simon Plouffe

Also the ratio of the area of a circle to the circumscribed square. More generally, the ratio of the area of an ellipse to the circumscribed rectangle. Also the ratio of the volume of a cylinder to the circumscribed cube. - Omar E. Pol, Sep 25 2013
Also the surface area of a quarter-sphere of diameter 1. - Omar E. Pol, Oct 03 2013
Least positive solution to sin(x) = cos(x). - Franklin T. Adams-Watters, Jun 17 2014
Dirichlet L-series of the non-principal character modulo 4 (A101455) at 1. See e.g. Table 22 of arXiv:1008.2547. - R. J. Mathar, May 27 2016
This constant is also equal to the infinite sum of the arctangent functions with nested radicals consisting of square roots of two. Specifically, one of the Viete-like formulas for Pi is given by Pi/4 = Sum_{k = 2..oo} arctan(sqrt(2 - a_{k - 1})/a_k), where the nested radicals are defined by recurrence relations a_k = sqrt(2 + a_{k - 1}) and a_1 = sqrt(2) (see the article [Abrarov and Quine]). - Sanjar Abrarov, Jan 09 2017
Pi/4 is the area enclosed between circumcircle and incircle of a regular polygon of unit side. - Mohammed Yaseen, Nov 29 2023

			0.785398163397448309615660845819875721049292349843776455243736148...
N = 2, m = 6: Pi/4 = 4!*3^4 Sum_{k >= 0} (-1)^k/((2*k - 11)*(2*k - 5)*(2*k + 1)*(2*k + 7)*(2*k + 13)). - _Peter Bala_, Nov 15 2016

Jörg Arndt and Christoph Haenel, Pi: Algorithmen, Computer, Arithmetik, Springer 2000, p. 150.
Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 437.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Sections 6.3 and 8.4, pp. 429 and 492.
Douglas R. Hofstadter, Gödel, Escher, Bach: An Eternal Golden Braid, Basic Books, p. 408.
J. Rivaud, Analyse, Séries, équations différentielles, Mathématiques supérieures et spéciales, Premier cycle universitaire, Vuibert, 1981, Exercice 3, p. 136.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 119.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Sanjar M. Abrarov and Brendan M. Quine, A Viète-like formula for pi based on infinite sum of the arctangent functions with nested radicals, figshare, 4509014, (2017).
Peter Bala, Arctanh(z) and the Legendre polynomials
Jonathan M. Borwein, Peter B. Borwein, and Karl Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Antonio Gracia Llorente, Shifting Constants Through Infinite Product Transformations, OSF Preprint, 2024.
Ronald K. Hoeflin, Titan Test.
Richard J. Mathar, Table of Dirichlet L-Series and Prime Zeta Modulo Functions for Small Moduli, arXiv:1008.2547 [math.NT], 2010-2015.
Literate Programs, Pi with Machin's formula (Haskell).
Michael Penn, A surprising appearance of pie!, YouTube video, 2020.
Michael Penn, Transforming normal identities into "crazy" ones, YouTube video, 2022.
Srinivasa Ramanujan, Question 353, J. Ind. Math. Soc.
Eric Weisstein's World of Mathematics, Prime Products.
Wikipedia, Leibniz formula for Pi.
Index entries for sequences from "Goedel, Escher, Bach".
Index entries for transcendental numbers.

Cf. A000796, A001586, A071904, A019669, A197723, A347152.
Cf. A000182, A000364, A024235, A278080, A278195.
Cf. A006752 (beta(2)=Catalan), A153071 (beta(3)), A175572 (beta(4)), A175571 (beta(5)), A175570 (beta(6)), A258814 (beta(7)), A258815 (beta(8)), A258816 (beta(9)).
Cf. A001622.

-- see link: Literate Programs
import Data.Char (digitToInt)
a003881_list len = map digitToInt $ show $ machin `div` (10 ^ 10) where
   machin = 4 * arccot 5 unity - arccot 239 unity
   unity = 10 ^ (len + 10)
   arccot x unity = arccot' x unity 0 (unity `div` x) 1 1 where
     arccot' x unity summa xpow n sign
    | term == 0 = summa
    | otherwise = arccot'
      x unity (summa + sign * term) (xpow `div` x ^ 2) (n + 2) (- sign)
    where term = xpow `div` n
-- Reinhard Zumkeller, Nov 20 2012

R:= RealField(100); Pi(R)/4; // G. C. Greubel, Mar 08 2018

Maple
```
evalf(Pi/4) ;
```

RealDigits[N[Pi/4,6! ]]  (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)
(* PROGRAM STARTS *)
(* Define the nested radicals a_k by recurrence *)
a[k_] := Nest[Sqrt[2 + #1] & , 0, k]
(* Example of Pi/4 approximation at K = 100 *)
Print["The actual value of Pi/4 is"]
N[Pi/4, 40]
Print["At K = 100 the approximated value of Pi/4 is"]
K := 100;  (* the truncating integer *)
N[Sum[ArcTan[Sqrt[2 - a[k - 1]]/a[k]], {k, 2, K}], 40] (* equation (8) *)
(* Error terms for Pi/4 approximations *)
Print["Error terms for Pi/4"]
k := 1; (* initial value of the index k *)
K := 10; (* initial value of the truncating integer K *)
sqn := {}; (* initiate the sequence *)
AppendTo[sqn, {"Truncating integer K ", " Error term in Pi/4"}];
While[K <= 30,
AppendTo[sqn, {K,
   N[Pi/4 - Sum[ArcTan[Sqrt[2 - a[k - 1]]/a[k]], {k, 2, K}], 1000] //
    N}]; K++]
Print[MatrixForm[sqn]]
(* Sanjar Abrarov, Jan 09 2017 *)

Pi/4 \\ Charles R Greathouse IV, Jul 07 2014

# Leibniz/Cohen/Villegas/Zagier/Arndt/Haenel
def FastLeibniz(n):
    b = 2^(2*n-1); c = b; s = 0
    for k in range(n-1,-1,-1):
        t = 2*k+1
        s = s + c/t if is_even(k) else s - c/t
        b *= (t*(k+1))/(2*(n-k)*(n+k))
        c += b
    return s/c
A003881 = RealField(3333)(FastLeibniz(1330))
print(A003881)  # Peter Luschny, Nov 20 2012

Equals Integral_{x=0..oo} sin(2x)/(2x) dx.
Equals lim_{n->oo} n*A001586(n-1)/A001586(n) (conjecture). - Mats Granvik, Feb 23 2011
Equals Integral_{x=0..1} 1/(1+x^2) dx. - Gary W. Adamson, Jun 22 2003
Equals Integral_{x=0..Pi/2} sin(x)^2 dx, or Integral_{x=0..Pi/2} cos(x)^2 dx. - Jean-François Alcover, Mar 26 2013
Equals (Sum_{x=0..oo} sin(x)*cos(x)/x) - 1/2. - Bruno Berselli, May 13 2013
Equals (-digamma(1/4) + digamma(3/4))/4. - Jean-François Alcover, May 31 2013
Equals Sum_{n>=0} (-1)^n/(2*n+1). - Geoffrey Critzer, Nov 03 2013
Equals Integral_{x=0..1} Product_{k>=1} (1-x^(8*k))^3 dx [cf. A258414]. - Vaclav Kotesovec, May 30 2015
Equals Product_{k in A071904} (if k mod 4 = 1 then (k-1)/(k+1)) else (if k mod 4 = 3 then (k+1)/(k-1)). - Dimitris Valianatos, Oct 05 2016
From Peter Bala, Nov 15 2016: (Start)
For N even: 2*(Pi/4 - Sum_{k = 1..N/2} (-1)^(k-1)/(2*k - 1)) ~ (-1)^(N/2)*(1/N - 1/N^3 + 5/N^5 - 61/N^7 + 1385/N^9 - ...), where the sequence of unsigned coefficients [1, 1, 5, 61, 1385, ...] is A000364. See Borwein et al., Theorem 1 (a).
For N odd: 2*(Pi/4 - Sum_{k = 1..(N-1)/2} (-1)^(k-1)/(2*k - 1)) ~ (-1)^((N-1)/2)*(1/N - 1/N^2 + 2/N^4 - 16/N^6 + 272/N^8 - ...), where the sequence of unsigned coefficients [1, 1, 2, 16, 272, ...] is A000182 with an extra initial term of 1.
For N = 0,1,2,... and m = 1,3,5,... there holds Pi/4 = (2*N)! * m^(2*N) * Sum_{k >= 0} ( (-1)^(N+k) * 1/Product_{j = -N..N} (2*k + 1 + 2*m*j) ); when N = 0 we get the Madhava-Gregory-Leibniz series for Pi/4.
For examples of asymptotic expansions for the tails of these series representations for Pi/4 see A024235 (case N = 1, m = 1), A278080 (case N = 2, m = 1) and A278195 (case N = 3, m = 1).
For N = 0,1,2,..., Pi/4 = 4^(N-1)*N!/(2*N)! * Sum_{k >= 0} 2^(k+1)*(k + N)!* (k + 2*N)!/(2*k + 2*N + 1)!, follows by applying Euler's series transformation to the above series representation for Pi/4 in the case m = 1. (End)
From Peter Bala, Nov 05 2019: (Start)
For k = 0,1,2,..., Pi/4 = k!*Sum_{n = -oo..oo} 1/((4*n+1)*(4*n+3)* ...*(4*n+2*k+1)), where Sum_{n = -oo..oo} f(n) is understood as lim_{j -> oo} Sum_{n = -j..j} f(n).
Equals Integral_{x = 0..oo} sin(x)^4/x^2 dx = Sum_{n >= 1} sin(n)^4/n^2, by the Abel-Plana formula.
Equals Integral_{x = 0..oo} sin(x)^3/x dx = Sum_{n >= 1} sin(n)^3/n, by the Abel-Plana formula. (End)
From Amiram Eldar, Aug 19 2020: (Start)
Equals arcsin(1/sqrt(2)).
Equals Product_{k>=1} (1 - 1/(2*k+1)^2).
Equals Integral_{x=0..oo} x/(x^4 + 1) dx.
Equals Integral_{x=0..oo} 1/(x^2 + 4) dx. (End)
With offset 1, equals 5 * Pi / 2. - Sean A. Irvine, Aug 19 2021
Equals (1/2)!^2 = Gamma(3/2)^2. - Gary W. Adamson, Aug 23 2021
Equals Integral_{x = 0..oo} exp(-x)*sin(x)/x dx (see Rivaud reference). - Bernard Schott, Jan 28 2022
From Amiram Eldar, Nov 06 2023: (Start)
Equals beta(1), where beta is the Dirichlet beta function.
Equals Product_{p prime >= 3} (1 - (-1)^((p-1)/2)/p)^(-1). (End)
Equals arctan( F(1)/F(4) ) + arctan( F(2)/F(3) ), where F(1), F(2), F(3), and F(4) are any four consecutive Fibonacci numbers. - Gary W. Adamson, Mar 03 2024
Pi/4 = Sum_{n >= 1} i/(n*P(n, i)*P(n-1, i)) = (1/2)*Sum_{n >= 1} (-1)^(n+1)*4^n/(n*A006139(n)*A006139(n-1)), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial. The n-th summand of the series is O( 1/(3 + 2*sqrt(3))^n ). - Peter Bala, Mar 16 2024
Equals arctan( phi^(-3) ) + arctan(phi^(-1) ). - Gary W. Adamson, Mar 27 2024
Equals Sum_{n>=1} eta(n)/2^n, where eta(n) is the Dirichlet eta function. - Antonio Graciá Llorente, Oct 04 2024
Equals Product_{k>=2} ((k + 1)^(k*(2*k + 1))*(k - 1)^(k*(2*k - 1)))/k^(4*k^2). - Antonio Graciá Llorente, Apr 12 2025
Equals Integral_{x=sqrt(2)..oo} dx/(x*sqrt(x^2 - 1)). - Kritsada Moomuang, May 29 2025

a(98) and a(99) corrected by Reinhard Zumkeller, Nov 20 2012

A122045 Euler (or secant) numbers E(n).

Original entry on oeis.org

1, 0, -1, 0, 5, 0, -61, 0, 1385, 0, -50521, 0, 2702765, 0, -199360981, 0, 19391512145, 0, -2404879675441, 0, 370371188237525, 0, -69348874393137901, 0, 15514534163557086905, 0, -4087072509293123892361, 0, 1252259641403629865468285, 0, -441543893249023104553682821

Offset: 0

Roger L. Bagula, Sep 13 2006

sign

The convention in the OEIS is that the alternate zeros are normally omitted in such sequences. See A000364 for the official version of this sequence.

Odd primes p such that p | E(p-1) are primes p == 1 (mod 4), A002144. Conjecture: odd composites m such that m | E(m-1) are Carmichael numbers m such that p == 1 (mod 4) for every prime p|m, A265237. - Thomas Ordowski, Feb 06 2020

			G.f. = 1 - x^2 + 5*x^4 - 61*x^6 + 1385*x^8 - 50521*x^10 + 2702765*x^12 + ...

Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 5, page 41.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Fernando Barbero G., Juan Margalef-Bentabol, and Eduardo J. S. Villaseñor, A two-sided Faulhaber-like formula involving Bernoulli polynomials, arXiv:2002.00550 [math.NT], 2020.
Marilena Barnabei, Flavio Bonetti, Niccolò Castronuovo, and Matteo Silimbani, Ascending runs in permutations and valued Dyck paths, Ars Mathematica Contemporanea (2019) Vol. 16, No. 2, 445-463.
Hacène Belbachir and Yassine Otmani, Quadrinomial-Like Versions for Wolstenholme, Morley and Glaisher Congruences, Integers (2023) Vol. 23.
Fatima Zohra Bensaci, Rachid Boumahdi, and Laala Khaldi, Finite Sums Involving Fibonacci and Lucas Numbers, J. Int. Seq. (2024). See p. 2.
Michael D. Hirschhorn, Binomial Identities and Congruences for Euler Numbers, Fibonacci Quart. 53 (2015), no. 4, 319-322.
Guodong Liu, Generating functions and generalized Euler numbers, Proc. Japan Acad. Ser. A Math. Sci., Volume 84, Number 2 (2008), 29-34. See p 32.
F. Luca, A. Pizarro-Madariaga, and C. Pomerance, On the counting function of irregular primes, 2014.
Emanuele Munarini, Two-Parameter Identities for q-Appell Polynomials, Journal of Integer Sequences, Vol. 26 (2023), Article 23.3.1.
A. Niedermaier and J. Remmel, Analogues of up-down permutations for colored permutations, J. Int. Seq. 13 (2010) 10.5.6.
Bruce E. Sagan, Generalized Euler numbers and ordered set partitions, arXiv:2501.07692 [math.NT], 2025. See p. 2.
T. J. Stieltjes, Sur la réduction en fraction continue d'une série procédant suivant les puissances descendantes d'une variable, Ann. Fac. Sc. Toulouse 3 (1889) 1-17.

Cf. A000364, A002144, A028296, A265237, A331978, A350972.

m:=35; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!( 1/Cosh(x) )); [Factorial(n-1)*b[n]: n in [1..m-1]]; // G. C. Greubel, Feb 13 2020

seq(euler(n) , n=0..31); # Zerinvary Lajos, Mar 15 2009
P := proc(n,x) option remember; if n = 0 then 1 else
   (n*x+(1/2)*(1-x))*P(n-1,x)+x*(1-x)*diff(P(n-1,x),x); expand(%) fi end:
A122045 := n -> (-1)^n*subs(x=-1, P(n,x)):
seq(A122045(n), n=0..30);  # Peter Luschny, Mar 07 2014
ptan := proc(n) option remember; if irem(n, 2) = 1 then 0 else
    -add(`if`(k=0, 1, binomial(n, k)*ptan(n - k)), k = 0..n-1,2) fi end:
A122045 := n -> ifelse(n = 0, 1, ptan(n)):
seq(A122045(n), n = 0..30); # Peter Luschny, Jun 06 2022

Table[EulerE[n], {n, 0, 30}]
Range[0, 30]! CoefficientList[ Series[ Sech[x], {x, 0, 30}], x] (* Robert G. Wilson v, Aug 08 2018 *)
a[0] := 1; a[n_?OddQ] := 0; a[n_?EvenQ] := a[n] = -Sum[a[k] Binomial[n, k], {k, 0, n - 1, 2}]; Map[a, Range[0, 30]] (* Oliver Seipel, May 20 2024 *)

a[n]:=if n<2 then 1-n else sum(-a[n-2*k]*binomial(n,2*k),k,1,floor(n/2));
makelist(a[n],n,0,50); /* Tani Akinari, Sep 15 2023 */

x='x+O('x^66); Vec(serlaplace(1/cosh(x))) \\ Joerg Arndt, Mar 10 2014

a(n) = 2^n*2^(n+1)*(subst(bernpol(n+1,x), x, 3/4) - subst(bernpol(n+1,x), x, 1/4))/(n+1); \\ Michel Marcus, May 20 2017

from sympy import bernoulli as B
def a(n): return int(2**n*2**(n + 1)*(B(n + 1, 3/4) - B(n + 1, 1/4))/(n + 1))
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 24 2017, after PARI code by Michel Marcus

from functools import cache
from math import comb as binomial
@cache
def ptan(n):   # see also A331978 and A350972.
    return (0 if n % 2 == 1 else
    -sum(binomial(n,k) * ptan(n-k) if k > 0 else 1 for k in range(0, n-1, 2)))
def A122045(n): return 1 if n == 0 else ptan(n)
print([A122045(n) for n in range(31)])  # Peter Luschny, Jun 06 2022

[euler_number(i) for i in range(31)] # Zerinvary Lajos, Mar 15 2009

E.g.f.: sech(x). - Michael Somos, Mar 11 2014
a(n) = (Sum_{k>=0} (-1)^k*(2*k+1)^n)*2. - Gottfried Helms, Mar 09 2012
From Sergei N. Gladkovskii, Oct 14 2012 - Oct 13 2013: (Start)
Continued fractions:
G.f.: 1/U(0) where U(k) = 1 - x + x*(k+1)/(1 - x*(k+1)/U(k+1)).
G.f.: 1/U(0) where U(k) = 1 - x^2 + x^2*(2*k+1)*(2*k+2)/(1 + x^2*(2*k+1)*(2*k+2)/ U(k+1)).
E.g.f.: (1-x)/U(0) where U(k) = 1 - x/(1 - x/(x - (2*k+1)*(2*k+2)/U(k+1))).
E.g.f.: 1 - x^2/U(0) where U(k) = (2*k+1)*(2*k+2) + x^2 - x^2*(2*k+1)*(2*k+2)/U(k+1).
E.g.f.: 1/U(0) where U(k) = 1 + x^2/(2*(2*k+1)*(4*k+1) - 2*x^2*(2*k+1)*(4*k+1)/(x^2 + 4*(4*k+3)*(k+1)/U(k+1))).
E.g.f.: (2 + x^4/(U(0)*(x^2-2) - 2))/(2-x^2) where U(k) = 4*k + 4 + 1/(1 + x^2/(2 - x^2 + (2*k+3)*(2*k+4)/U(k+1))).
G.f.: 1/G(0) where G(k) = 1 + x^2*(2*k+1)^2/(1 + x^2*(2*k+2)^2/G(k+1)) (due to T. J. Stieltjes).
G.f.: 1/S(0) where S(k) = 1 + x^2*(k+1)^2/S(k+1) (due to T. J. Stieltjes).
G.f.: 1 - x/(1+x) + x/(1+x)/Q(0) where Q(k) = 1 - x + x*(k+2)/(1 - x*(k+1)/Q(k+1)).
G.f.: -(1/x)/Q(0) where Q(k) = -1/x + (k+1)^2/Q(k+1) (due to T. J. Stieltjes).
G.f.: (1/(1-x))/Q(0) + 1/(1-x) where Q(k) = 1 - 1/x + (k+1)*(k+2)/Q(k+1).
G.f.: (x/(x-1))/Q(0) + 1/(1-x) where Q(k) = 1 - x + x^2*(k+1)*(k+2)/Q(k+1).
G.f.: 1 - x/(1+x) + (x/(1+x))/Q(0) where Q(k) = 1 + x + (k+1)*(k+2)*x^2/Q(k+1).
E.g.f.: 1 - T(0)*x^2/(2+x^2) where T(k) = 1 - x^2*(2*k+1)*(2*k+2)/(x^2*(2*k+1)*(2*k+2) - ((2*k+1)*(2*k+2) + x^2)*((2*k+3)*(2*k+4) + x^2)/T(k+1)).
G.f.: T(0) where T(k) = 1 - x^2*(k+1)^2/(x^2*(k+1)^2 + 1/T(k+1)). (End)
a(n) = 2^(2*n+1)*(zeta(-n,1/4) - zeta(-n,3/4)), where zeta(a, z) is the generalized Riemann zeta function. - Peter Luschny, Mar 11 2015
a(n) = 2^n*(2^(n+1)/(n+1))*(B(n+1, 3/4) - B(n+1, 1/4)) where B(n,x) is the n-th Bernoulli polynomial. See Liu link. - Michel Marcus, May 20 2017 [This is the same as: a(n) = -4^(n+1)*B(n+1, 1/4)*((n+1) mod 2)/(n+1). Peter Luschny, Oct 30 2020]
a(n) = 2*Im(PolyLog(-n, I)). - Peter Luschny, Sep 29 2020
a(4n) == 5 (mod 60) and a(4n+2) == -1 (mod 60). See Hirschhorn. - Michel Marcus, Jan 11 2022
For n > 1, a(n) = -Sum_{k=1..floor(n/2)} a(n-2*k)*binomial(n,2*k). - Tani Akinari, Sep 15 2023

Edited by N. J. A. Sloane, Sep 17 2006

A002105 Reduced tangent numbers: 2^n(2^{2n} - 1)|B_{2n}|/n, where B_n = Bernoulli numbers.

Original entry on oeis.org

1, 1, 4, 34, 496, 11056, 349504, 14873104, 819786496, 56814228736, 4835447317504, 495812444583424, 60283564499562496, 8575634961418940416, 1411083019275488149504, 265929039218907754399744, 56906245479134057176170496, 13722623393637762299131396096, 3704005473270641755597685653504

Offset: 1

N. J. A. Sloane

Comments from R. L. Graham, Apr 25 2006 and Jun 08 2006: "This sequence also gives the number of ways of arranging 2n tokens in a row, with 2 copies of each token from 1 through n, such that the first token is a 1 and between every pair of tokens labeled i (i=1..n-1) there is exactly one token labeled i+1.
"For example, for n=3, there are 4 possibilities: 123123, 121323, 132312 and 132132 and indeed a(3) = 4. This is the work of my Ph. D. student Nan Zang. See also A117513, A117514, A117515.
"Develin and Sullivant give another occurrence of this sequence and show that their numbers have the same generating function, although they were unable to find a 1-1-mapping between their problem and Poupard's."
The sequence 1,0,1,0,4,0,34,0,496,0,11056, ... counts increasing complete binary trees with e.g.f. sec^2(x/sqrt 2). - Wenjin Woan, Oct 03 2007
a(n) = number of increasing full binary trees on vertex set [2n-1] with the left-largest property: the largest descendant of each non-leaf vertex occurs in its left subtree (Poupard). The first Mathematica recurrence below counts these trees by number 2k-1 of vertices in the left subtree of the root: the root is necessarily labeled 1 and n necessarily occurs in the left subtree and so there are Binomial[2n-3,2k-2] ways to choose the remaining labels for the left subtree. - David Callan, Nov 29 2007
Number of bilabeled unordered increasing trees with 2n labels. - Markus Kuba, Nov 18 2014
Conjecture: taking the sequence modulo an integer k gives an eventually purely periodic sequence with period dividing phi(k). For example, the sequence taken modulo 10 begins [1, 1, 4, 4, 6, 6, 4, 4, 6, 6, 4, 4, 6, 6, ...] with an apparent period [4, 4, 6, 6] of length 4 = phi(10) beginning at a(3). - Peter Bala, May 08 2023
Let c(1), c(2), c(3), ... be a geometric progression and s = (2*c(1)/c(2))^(1/2). Then c(1)*s*tan(x/s) = Sum_{n>0} a(n) * c(n) * x^(2*n-1) / (2*n-1)!. - Michael Somos, Jan 15 2025

			G.f. = x + x^2 + 4*x^3 + 34*x^4 + 496*x^5 + 11056*x^6 + 349504*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 = 1..100
G. E. Andrews, J. Jimenez-Urroz, and K. Ono, q-series identities and values of certain L-functions, Duke Math J. Volume 108, Number 3 (2001), 395-419.
R. C. Archibald, Review of Terrill-Terrill paper, Math. Comp., 1 (1945), pp. 385-386.
Peter Bala, A continued fraction expansion related to A002105
Paul Barry, Three Études on a sequence transformation pipeline, arXiv:1803.06408 [math.CO], 2018.
R. B. Brent, Generalizing Tuenter's Binomial Sums, J. Int. Seq. 18 (2015) # 15.3.2.
M. P. Develin and S. P. Sullivant, Markov Bases of Binary Graph Models, Annals of Combinatorics 7 (2003) 441-466.
Dominique Foata and Guo-Niu Han, The doubloon polynomial triangle, Ram. J. 23 (2010), 107-126
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, 2012. - From _N. J. A. Sloane_, Feb 02 2013
R. L. Graham and Nan Zang, Enumerating split-pair arrangements, J. Combin. Theory Ser. A 115 (2008), no. 2, 293-303.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
Markus Kuba and Alois Panholzer, Combinatorial families of multilabelled increasing trees and hook-length formulas, arXiv:1411.4587 [math.CO], 2014.
B. A. Kupershmidt, On skew-symmetric and general deformations of Lax pseudodifferential operators, J. Math. Sci. 151 (4) (2008) 3139, eq. (1.13).
Zhicong Lin, Shi-Mei Ma, David G. L. Wang, and Liuquan Wang, Positivity and divisibility of alternating descent polynomials, arXiv:2011.02685 [math.CO], 2020.
E. Norton, Symplectic Reflection Algebras in Positive Characteristic as Ore Extensions, arXiv preprint arXiv:1302.5411 [math.RA], 2013.
Ronald Orozco López, Solution of the Differential Equation y^(k)= e^(a*y), Special Values of Bell Polynomials and (k,a)-Autonomous Coefficients, Universidad de los Andes (Colombia 2021).
C. Poupard, Deux proprietes des arbres binaires ordonnes stricts, European J. Combin., 10 (1989), 369-374.
M. Safaryan, Problem 11725, Amer. Math. Monthly, 120 (2013), 661.
H. M. Terrill and E. M. Terrill, Tables of numbers related to the tangent coefficients, J. Franklin Inst., 239 (1945), 66-67.
H. M. Terrill and E. M. Terrill, Tables of numbers related to the tangent coefficients, J. Franklin Inst., 239 (1945), 64-67. [Annotated scanned copy]
A. Vieru, Agoh's conjecture: its proof, its generalizations, its analogues, arXiv preprint arXiv:1107.2938 [math.NT], 2011.
Index entries for sequences related to Bernoulli numbers.

Row sums of A008301.
Left edge of triangle A210108.
Cf. A000111, A000217, A000364, A000464, A002439, A008281, A079144, A158690.

A002105:= func< n | (-1)^(n+1)*2^n*(4^n - 1)*Bernoulli(2*n)/n >;
[A002105(n): n in [1..30]]; // G. C. Greubel, Sep 20 2024

S := proc(n, k) option remember; if k=0 then `if`(n=0, 1, 0) else S(n, k-1) + S(n-1, n-k) fi end: A002105 := n -> S(2*n-1, 2*n-1)/2^(n-1): seq(A002105(i),i=1..16); # Peter Luschny, Jul 08 2012
# The above function written as a formula: a(n) = A008281(2*n-1, 2*n-1)/2^(n-1).
# Alternatively, based on the triangular numbers A000217:
T := proc(n, k) option remember; if k = 0 then 1 else if k = n then T(n, k-1) else
A000217(n - k + 1) * T(n, k - 1) + T(n - 1, k) fi fi end:
a := n -> T(n, n): seq(a(n), n = 0..18);  # Peter Luschny, Sep 30 2023

u[1] = 1; u[n_]/;n>=2 := u[n] = Sum[Binomial[2n-3,2k-2]u[k]u[n-k],{k,n-1}]; Table[u[n],{n,8}] (* Poupard and also Develin and Sullivant, give a different recurrence that involves a symmetric sum: v[1] = 1; v[n_]/;n>=2 := v[n] = 1/2 Sum[Binomial[2n-2,2k-1]v[k]v[n-k],{k,n-1}] *) (*David Callan, Nov 29 2007 *)
a[n_] := (-1)^n 2^(n+1) PolyLog[1-2n, -1]; Array[a, 10] (* Vladimir Reshetnikov, Jan 23 2011 *)
Table[(-1)^(n+1)*2^n*(2^(2n)-1)*BernoulliB[2n]/n,{n,1,20}] (* Vaclav Kotesovec, Nov 03 2014 *)
eulerCF[f_, len_] := Module[{g}, g[len-1]=1; g[k_]:=g[k]=1-f[k]/(f[k]-1/g[k+1]); CoefficientList[g[0] + O[x]^len, x]]; A002105List[len_] := eulerCF[(1/2) x (#+1) (#+2)&, len]; A002105List[19] (* Peter Luschny, Aug 08 2015 after Sergei N. Gladkovskii *)
Table[PolyGamma[2n-1, 1/2] 2^(2-n)/Pi^(2n), {n, 1, 10}] (* Vladimir Reshetnikov, Oct 18 2015 *)
Table[EulerE[2n-1, 0] (-2)^n, {n, 1, 10}] (* Vladimir Reshetnikov, Oct 21 2015 *)

{a(n) = if( n<1, 0, ((-2)^n - (-8)^n) * bernfrac(2*n) / n)}; /* Michael Somos, Jun 22 2002 */

{a(n) = if( n<0, 0, (2*n)! * polcoeff( -2 * log( cos(x / quadgen(8) + O(x^(2*n + 1)))), 2*n))}; /* Michael Somos, Jul 17 2003 */

{a(n) = if( n<0, 0, -(-2)^(n+1) * sum(i=1, 2*n, 2^-i * sum(j=1, i, (-1)^j * binomial( i-1, j-1) * j^(2*n - 1))))}; /* Michael Somos, Sep 07 2013 */

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

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

from sympy import bernoulli
def A002105(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 A002105_list(n) :
    D = [0]*(n+2); D[1] = 1
    R = []; z = 1/2; b = True
    for i in(0..2*n-1) :
        h = i//2 + 1
        if b :
            for k in range(h-1, 0, -1) : D[k] += D[k+1]
            z *= 2
        else :
            for k in range(1, h+1, 1) :  D[k] += D[k-1]
        b = not b
        if b : R.append(D[h]*z/h)
    return R
A002105_list(16) # Peter Luschny, Jun 29 2012

def A002105(n): return (-1)^(n+1)*2^n*(4^n -1)*bernoulli(2*n)/n
[A002105(n) for n in range(1,31)] # G. C. Greubel, Sep 20 2024

E.g.f.: 2*log(sec(x / sqrt(2))) = Sum_{n>0} a(n) * x^(2*n) / (2*n)!. - Michael Somos, Jun 22 2002
A000182(n) = 2^(n-1) * a(n). - Michael Somos, Jun 22 2002
a(n) = 2^(n-1)/n * A110501(n). - Don Knuth, Jan 16 2007
a(n+1) = Sum_{k = 0..n} A094665(n, k). - Philippe Deléham, Jun 11 2004
O.g.f.: A(x) = x/(1-x/(1-3*x/(1-6*x/(1-10*x/(1-15*x/(... -n*(n+1)/2*x/(1 - ...))))))) (continued fraction). - Paul D. Hanna, Oct 07 2005
sqrt(2) tan( x/sqrt(2)) = Sum_(n>=0) (x^(2n+1)/(2n+1)!) a_n. - Dominique Foata and Guo-Niu Han, Oct 24 2008
Basic hypergeometric generating function: Sum_{n>=0} Product {k = 1..n} (1-exp(-2*k*t))/Product {k = 1..n} (1+exp(-2*k*t)) = 1 + t + 4*t^2/2! + 34*t^3/3! + 496*t^4/4! + ... [Andrews et al., Theorem 4]. For other sequences with generating functions of a similar type see A000364, A000464, A002439, A079144 and A158690. - Peter Bala, Mar 24 2009
E.g.f.: Sum_{n>=0} Product_{k=1..n} tanh(k*x) = Sum_{n>=0} a(n)*x^n/n!. - Paul D. Hanna, May 11 2010
a(n) = (-1)^(n+1)*sum(j!*stirling2(2*n+1,j)*2^(n+1-j)*(-1)^(j),j,1,2*n+1), n>=0. - Vladimir Kruchinin, Aug 23 2010
From Gary W. Adamson, Jul 14 2011: (Start)
a(n) = upper left term in M^n, a(n+1) = sum of top row terms in M^n; where M = the infinite square production matrix:
   1, 3, 0,  0,  0, 0, 0, ...
   1, 3, 6,  0,  0, 0, 0, ...
   1, 3, 6, 10,  0, 0, 0, ...
   1, 3, 6, 10, 15, 0, 0, ... (End)
E.g.f. A(x) satisfies differential equation A''(x)=exp(A(x)). - Vladimir Kruchinin, Nov 18 2011
E.g.f.: For E(x)=sqrt(2)* tan( x/sqrt(2))=x/G(0); G(k)= 4*k + 1 - x^2/(8*k + 6 - x^2/G(k+1)); (from continued fraction Lambert's, 2-step). - Sergei N. Gladkovskii, Jan 14 2012
a(n) = (-1)^n*2^(n+1)*Li_{1-2*n}(-1). (See also the Mathematica prog. by Vladimir Reshetnikov.) - Peter Luschny, Jun 28 2012
G.f.: 1/G(0) where G(k) = 1 - x*( 4*k^2 + 4*k + 1 ) - x^2*(k+1)^2*( 4*k^2 + 8*k + 3)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 14 2013
G.f.: 1/Q(0), where Q(k) = 1 - (k+1)*(k+2)/2*x/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 03 2013
G.f.: (1/G(0))/sqrt(x) - 1/sqrt(x), where G(k) = 1 - sqrt(x)*(2*k+1)/(1 + sqrt(x)*(2*k+1)/(1 + sqrt(x)*(k+1)/(1 - sqrt(x)*(k+1)/G(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Jul 07 2013
log(2) - 1/1 + 1/2 - 1/3 + ... + (-1)^n / n = (-1)^n / 2 * (1/n - 1 / (2*n^2) + 1 / (2*n^2)^2 - 4 / (2*n^2)^3 + ... + (-1)^k * a(k) / (2*n^2)^k + ...) asymptotic expansion. - Michael Somos, Sep 07 2013
G.f.: T(0), where T(k) = 1-x*(k+1)*(k+2)/(x*(k+1)*(k+2)-2/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 24 2013
a(n) ~ 2^(3*n+2) * n^(2*n-1/2) / (exp(2*n) * Pi^(2*n-1/2)). - Vaclav Kotesovec, Nov 03 2014
From Peter Bala, Sep 10 2015: (Start)
The e.g.f. A(x) = sqrt(2)*tan(x/sqrt(2)) satisfies A''(x) = A(x)*A'(x), hence the recurrence a(0) = 0, a(1) = 1, else a(n) = Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i) for the aerated sequence [0,1,0,1,0,4,0,34,0,496,...].
Note that the same recurrence, but with the initial conditions a(0) = 1 and a(1) = 1, produces the sequence [1,1,1,2,5,16,61,272,...] = A000111. (End)
a(n) = polygamma(2*n-1, 1/2)*2^(2-n)/Pi^(2*n). - Vladimir Reshetnikov, Oct 18 2015
E.g.f.: sqrt(2)*tan(x/sqrt(2)) = Sum_{n>0} a(n) * x^(2*n-1) / (2*n-1)!. - Michael Somos, Mar 05 2017
From Peter Bala, May 05 2017: (Start)
Let B(x) = A(x)/x = 1 + x + 4*x^2 + 34*x^3 + ... denote the shifted o.g.f. Then B(x) = 1/(1 + 2*x - 3*x/(1 - x/(1 + 2*x - 10*x/(1 - 6*x/(1 + 2*x - 21*x/(1 - 15*x/(1 + 2*x - 36*x/(1 - 28*x/(1 + 2*x - ...))))))))), where the coefficient sequence [3, 1, 10, 6, 21, 15, 36, 28, ...] in the partial numerators of the continued fraction is obtained by swapping adjacent triangular numbers. Cf. A079144.
It follows (by means of an equivalence transformation) that the second binomial transform of B(x), with g.f. equal to 1/(1 - 2*x)*B(x/(1 - 2*x)), has the S-fraction representation 1/(1 - 3*x/(1 - x/(1 - 10*x/(1 - 6*x/(1 - 21*x/(1 - 15*x/(1 - 36*x/(1 - 28*x/(1 - ...))))))))). Compare with the S-fraction representation of the g.f. A(x) given above by Hanna, dated Oct 07 2005. (End)
The computation can be based on the triangular numbers, a(n) = T(n, n) where T(n, k) = A000217(n - k + 1) * T(n, k - 1) + T(n - 1, k) for 0 < k < n, and T(n, 0) = 1, T(n, n) = T(n, k-1) if k > 0. This is equivalent to Paul D. Hanna's continued fraction 2005. - Peter Luschny, Sep 30 2023

Additional comments from Michael Somos, Jun 25 2002

A028296 Expansion of e.g.f. Gudermannian(x) = 2*arctan(exp(x)) - Pi/2.

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

The Euler numbers A000364 with alternating signs.
The first column of the inverse to the matrix with entries C(2*i,2*j), i,j >=0. The full matrix is lower triangular with the i-th subdiagonal having entries a(i)*C(2*j,2*i) j>=i. - Nolan Wallach (nwallach(AT)ucsd.edu), Dec 26 2005
This sequence is also EulerE(2*n). - Paul Abbott (paul(AT)physics.uwa.edu.au), Apr 14 2006
Consider the sequence defined by a(0)=1; thereafter a(n) = c*Sum_{k=1..n} binomial(2n,2k)*a(n-k). For c = -3, -2, -1, 1, 2, 3, 4 this is A210676, A210657, A028296, A094088, A210672, A210674, A249939.
To avoid possible confusion: these are the odd e.g.f. coefficients of Gudermannian(x) with the offset shifted by -1 (even coefficients are zero). They are identical to the even e.g.f. coefficients for 1/cosh(x) = -Gudermannian'(x) (see the Example). Since the complex root of cosh(z) with the smallest absolute value is z0 = i*Pi/2, the radius of convergence for the Taylor series of all these functions is Pi/2 = A019669. - Stanislav Sykora, Oct 07 2016

			Gudermannian(x) = x - (1/6)*x^3 + (1/24)*x^5 - (61/5040)*x^7 + (277/72576)*x^9 + ....
Gudermannian'(x) = 1/cosh(x) = (1/1!)*x^0 - (1/2!)*x^2 + (5/4!)*x^4 - (61/6!)*x^6 + (1385/8!)*x^8 + .... - _Stanislav Sykora_, Oct 07 2016

Gradshteyn and Ryzhik, Tables, 5th ed., Section 1.490, pp. 51-52.
R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section B45.

T. D. Noe, Table of n, a(n) for n = 0..100
Beáta Bényi and Toshiki Matsusaka, Extensions of the combinatorics of poly-Bernoulli numbers, arXiv:2106.05585 [math.CO], 2021.
Filippo Callegaro and Giovanni Gaiffi, On models of the braid arrangement and their hidden symmetries, arXiv preprint arXiv:1406.1304 [math.AT], 2014.
Karl Dilcher and Christophe Vignat, Euler and the Strong Law of Small Numbers, Amer. Math. Mnthly, 123 (May 2016), 486-490.
Allan L. Edmonds and Steven Klee, The combinatorics of hyperbolized manifolds, arXiv preprint arXiv:1210.7396 [math.CO], 2012. - From _N. J. A. Sloane_, Jan 02 2013
Guodong Liu, On congruences of Euler numbers modulo powers of two, Journal of Number Theory, Volume 128, Issue 12, December 2008, Pages 3063-3071.
Emanuele Munarini, Two-Parameter Identities for q-Appell Polynomials, Journal of Integer Sequences, Vol. 26 (2023), Article 23.3.1.
Niels Erik Nørlund, Vorlesungen über Differenzenrechnung, Springer 1924, p. 25.
Joe Santmyer, Derivative Polynomials for Trigonometric and Hyperbolic Functions, Arhimede Math. J. (2023) Vol. 10, No. 2, 152-158.
Jan W. H. Swanepoel, A Short Simple Probabilistic Proof of a Well Known Identity and the Derivation of Related New Identities Involving the Bernoulli Numbers and the Euler Numbers, Integers (2025) Vol. 25, Art. No. A50. See p. 4.
Zhi-Hong Sun, On the further properties of {U_n}, arXiv:1203.5977v1 [math.NT], Mar 27 2012.

Absolute values are the Euler numbers A000364.

Cf. A019669.

A028296 := proc(n) a :=0 ; for k from 1 to 2*n+1 by 2 do a := a+(-1)^((k-1)/2)/2^k/k *add( (-1)^i *(k-2*i)^(2*n+1) *binomial(k,i), i=0..k) ; end do: a ; end proc:
seq(A028296(n),n=0..10) ; # R. J. Mathar, Apr 20 2011

Table[EulerE[2*n], {n, 0, 30}] (* Paul Abbott, Apr 14 2006 *)
Table[(CoefficientList[Series[1/Cosh[x],{x,0,40}],x]*Range[0,40]!)[[2*n+1]],{n,0,20}] (* Vaclav Kotesovec, Aug 04 2014*)
With[{nn=40},Take[CoefficientList[Series[Gudermannian[x],{x,0,nn}],x] Range[ 0,nn-1]!,{2,-1,2}]] (* Harvey P. Dale, Feb 24 2018 *)
{1, Table[2*(-I)*PolyLog[-2*n, I], {n, 1, 12}]} // Flatten (* Peter Luschny, Aug 12 2021 *)
a[0] := 1; a[n_] := a[n] = -Sum[Binomial[2 n, 2 k] a[k], {k, 0, n - 1}]; Map[a, Range[0, 16]] (* Oliver Seipel, May 19 2024 *)

a(n):=sum((-1+(-1)^(k))*(-1)^((k+1)/2)/(2^(k+1)*k)*sum((-1)^i*(k-2*i)^n*binomial(k,i),i,0,k),k,1,n); /* with interpolated zeros, Vladimir Kruchinin, Apr 20 2011 */

a(n) = 2*imag(polylog(-2*n, I)); \\ Michel Marcus, May 30 2018

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

from sympy import euler
def A028296(n): return euler(n<<1) # Chai Wah Wu, Apr 16 2023

def A028296_list(len):
    f = lambda k: x*(k+1)^2
    g = 1
    for k in range(len-2,-1,-1):
        g = (1-f(k)/(f(k)+1/g)).simplify_rational()
    return taylor(g, x, 0, len-1).list()
print(A028296_list(17))

def A028296(n):
    shapes = ([x*2 for x in p] for p in Partitions(n))
    return sum((-1)^len(s)*factorial(len(s))*SetPartitions(sum(s), s).cardinality() for s in shapes)
print([A028296(n) for n in (0..16)]) # Peter Luschny, Aug 10 2015

E.g.f.: sech(x) = 1/cosh(x) (even terms), or Gudermannian(x) (odd terms).
Recurrence: a(n) = -Sum_{i=0..n-1} a(i)*binomial(2*n, 2*i). - Ralf Stephan, Feb 24 2005
a(n) = Sum_{k=1,3,5,..,2n+1} ((-1)^((k-1)/2) /(2^k*k)) * Sum_{i=0..k} (-1)^i*(k-2*i)^(2n+1) * binomial(k,i). - Vladimir Kruchinin, Apr 20 2011
a(n) = 2^(4*n+1)*(zeta(-2*n,1/4) - zeta(-2*n,3/4)). - Gerry Martens, May 27 2011
From Sergei N. Gladkovskii, Dec 15 2011 - Oct 09 2013: (Start)
Continued fractions:
G.f.: A(x) = 1 - x/(S(0)+x), S(k) = euler(2*k) + x*euler(2*k+2) - x*euler(2*k)* euler(2*k+4)/S(k+1).
E.g.f.: E(x) = 1 - x/(S(0)+x); S(k) = (k+1)*euler(2*k) + x*euler(2*k+2) - x*(k+1)* euler(2*k)*euler(2*k+4)/S(k+1).
2*arctan(exp(z)) - Pi/2 = z*(1 - z^2/(G(0) + z^2)), G(k) = 2*(k+1)*(2*k+3)*euler(2*k) + z^2*euler(2*k+2) - 2*z^2*(k+1)*(2*k+3)*euler(2*k)*euler(2*k+4)/G(k+1).
G.f.: A(x) = 1/S(0) where S(k) = 1 + x*(k+1)^2/S(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^4/( (x^2+2)*Q(0) + 2))/(2+x^2) where Q(k) = 4*k + 4 + 1/( 1 - x^2/( 2 + x^2 + (2*k+3)*(2*k+4)/Q(k+1))).
E.g.f.: 1/cosh(x) = 8*(1-x^2)/(8 - 4*x^2 - 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));
G.f.: 1/Q(0), where Q(k) = 1 - sqrt(x) + sqrt(x)*(k+1)/(1-sqrt(x)*(k+1)/Q(k+1));
G.f.: (1/Q(0) + 1)/(1-sqrt(x)), where Q(k) = 1 - 1/sqrt(x) + (k+1)*(k+2)/Q(k+1);
G.f.: Q(0), where Q(k) = 1 - x*(k+1)^2/( x*(k+1)^2 + 1/Q(k+1)). (End)
a(n) ~ (-1)^n * (2*n)! * 2^(2*n+2) / Pi^(2*n+1). - Vaclav Kotesovec, Aug 04 2014
a(n) = 2*Im(Li_{-2n}(i)), where Li_n(x) is polylogarithm, i=sqrt(-1). - Vladimir Reshetnikov, Oct 22 2015

A086645 Triangle read by rows: T(n, k) = binomial(2n, 2k).

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 15, 15, 1, 1, 28, 70, 28, 1, 1, 45, 210, 210, 45, 1, 1, 66, 495, 924, 495, 66, 1, 1, 91, 1001, 3003, 3003, 1001, 91, 1, 1, 120, 1820, 8008, 12870, 8008, 1820, 120, 1, 1, 153, 3060, 18564, 43758, 43758, 18564, 3060, 153, 1, 1, 190, 4845, 38760

Offset: 0

Philippe Deléham, Jul 26 2003

Terms have the same parity as those of Pascal's triangle.
Coefficients of polynomials (1/2)*((1 + x^(1/2))^(2n) + (1 - x^(1/2))^(2n)).
Number of compositions of 2n having k parts greater than 1; example: T(3, 2) = 15 because we have 4+2, 2+4, 3+2+1, 3+1+2, 2+3+1, 2+1+3, 1+3+2, 1+2+3, 2+2+1+1, 2+1+2+1, 2+1+1+2, 1+2+2+1, 1+2+1+2, 1+1+2+2, 3+3. - Philippe Deléham, May 18 2005
Number of binary words of length 2n - 1 having k runs of consecutive 1's; example: T(3,2) = 15 because we have 00101, 01001, 01010, 01011, 01101, 10001, 10010, 10011, 10100, 10110, 10111, 11001, 11010, 11011, 11101. - Philippe Deléham, May 18 2005
Let M_n be the n X n matrix M_n(i, j) = T(i, j-1); then for n > 0, det(M_n) = A000364(n), Euler numbers; example: det([1, 1, 0, 0; 1, 6, 1, 0; 1, 15, 15, 1; 1, 28, 70, 28 ]) = 1385 = A000364(4). - Philippe Deléham, Sep 04 2005
Equals ConvOffsStoT transform of the hexagonal numbers, A000384: (1, 6, 15, 28, 45, ...); e.g., ConvOffs transform of (1, 6, 15, 28) = (1, 28, 70, 28, 1). - Gary W. Adamson, Apr 22 2008
From Peter Bala, Oct 23 2008: (Start)
Let C_n be the root lattice generated as a monoid by {+-2*e_i: 1 <= i <= n; +-e_i +- e_j: 1 <= i not equal to j <= n}. Let P(C_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the h-vectors of a unimodular triangulation of P(C_n) [Ardila et al.]. See A127674 for (a signed version of) the corresponding array of f-vectors for these type C_n polytopes. See A008459 for the array of h-vectors for type A_n polytopes and A108558 for the array of h-vectors associated with type D_n polytopes.
The Hilbert transform of this triangle is A142992 (see A145905 for the definition of this term).
(End)
Diagonal sums: A108479. - Philippe Deléham, Sep 08 2009
Coefficients of Product_{k=1..n} (cot(k*Pi/(2n+1))^2 - x) = Sum_{k=0..n} (-1)^k*binomial(2n,2k)*x^k/(2n+1-2k). - David Ingerman (daviddavifree(AT)gmail.com), Mar 30 2010
Generalized Narayana triangle for 4^n (or cosh(2x)). - Paul Barry, Sep 28 2010
Coefficients of the matrix inverse appear to be T^(-1)(n,k) = (-1)^(n+k)*A086646(n,k). - R. J. Mathar, Mar 12 2013
Let E(y) = Sum_{n>=0} y^n/(2*n)! = cosh(sqrt(y)). Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence (2*n)! as defined in Wang and Wang. Cf. A103327. - Peter Bala, Aug 06 2013
Row 6, (1,66,495,924,495,66,1), plays a role in expansions of powers of the Dedekind eta function. See the Chan link, p. 534, and A034839. - Tom Copeland, Dec 12 2016

			From _Peter Bala_, Oct 23 2008: (Start)
The triangle begins
n\k|..0.....1.....2.....3.....4.....5.....6
===========================================
0..|..1
1..|..1.....1
2..|..1.....6.....1
3..|..1....15....15.....1
4..|..1....28....70....28.....1
5..|..1....45...210...210....45.....1
6..|..1....66...495...924...495....66.....1
...
(End)
From _Peter Bala_, Aug 06 2013: (Start)
Viewed as the generalized Riordan array (cosh(sqrt(y)), y) with respect to the sequence (2*n)! the column generating functions begin
1st col: cosh(sqrt(y)) = 1 + y/2! + y^2/4! + y^3/6! + y^4/8! + ....
2nd col: 1/2!*y*cosh(sqrt(y)) = y/2! + 6*y^2/4! + 15*y^3/6! + 28*y^4/8! + ....
3rd col: 1/4!*y^2*cosh(sqrt(y)) = y^2/4! + 15*y^3/6! + 70*y^4/8! + 210*y^5/10! + .... (End)

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 224.

Indranil Ghosh, Rows 0.. 120 of triangle, flattened
F. Ardila, M. Beck, S. Hosten, J. Pfeifle and K. Seashore, Root polytopes and growth series of root lattices arXiv:0809.5123 [math.CO], 2008.
H. Chan, S. Cooper and P. Toh, The 26th power of Dedekind's eta function Advances in Mathematics, 207 (2006) 532-543.
T. Copeland, Infinitesimal Generators, the Pascal Pyramid, and the Witt and Virasoro Algebras, 2012.
T. Copeland, Skipping over Dimensions, Juggling Zeros in the Matrix, 2020.
E. Lucas, Théorie des fonctions numériques simplement périodiques, Baltimore, 1878. See page 145 equation (153).
W. Wang and T. Wang, Generalized Riordan array, Discrete Mathematics, Vol. 308, No. 24, 6466-6500.

Cf. A098158, A000384, A081294.
Cf. A008459, A108558, A127674, A142992. - Peter Bala, Oct 23 2008
Cf. A103327 (binomial(2n+1, 2k+1)), A103328 (binomial(2n, 2k+1)), A091042 (binomial(2n+1, 2k)). -Wolfdieter Lang, Jan 06 2013
Cf. A086646 (unsigned matrix inverse), A103327.
Cf. A034839.

/* As triangle: */ [[Binomial(2*n, 2*k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Dec 14 2016

A086645:=(n,k)->binomial(2*n,2*k): seq(seq(A086645(n,k),k=0..n),n=0..12);

Table[Binomial[2 n, 2 k], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 13 2016 *)

create_list(binomial(2*n,2*k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */

PARI
```
{T(n, k) = binomial(2*n, 2*k)};
```

{T(n, k) = sum( i=0, min(k, n-k), 4^i * binomial(n, 2*i) * binomial(n - 2*i, k-i))}; /* Michael Somos, May 26 2005 */

T(n, k) = (2*n)!/((2*(n-k))!*(2*k)!) row sums = A081294. COLUMNS: A000012, A000384
Sum_{k>=0} T(n, k)*A000364(k) = A000795(n) = (2^n)*A005647(n).
Sum_{k>=0} T(n, k)*2^k = A001541(n). Sum_{k>=0} T(n, k)*3^k = 2^n*A001075(n). Sum_{k>=0} T(n, k)*4^k = A083884(n). - Philippe Deléham, Feb 29 2004
O.g.f.: (1 - z*(1+x))/(x^2*z^2 - 2*x*z*(1+z) + (1-z)^2) = 1 + (1 + x)*z +(1 + 6*x + x^2)*z^2 + ... . - Peter Bala, Oct 23 2008
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A081294(n), A001541(n), A090965(n), A083884(n), A099140(n), A099141(n), A099142(n), A165224(n), A026244(n) for x = 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Sep 08 2009
Product_{k=1..n} (cot(k*Pi/(2n+1))^2 - x) = Sum_{k=0..n} (-1)^k*binomial(2n,2k)*x^k/(2n+1-2k). - David Ingerman (daviddavifree(AT)gmail.com), Mar 30 2010
From Paul Barry, Sep 28 2010: (Start)
G.f.: 1/(1-x-x*y-4*x^2*y/(1-x-x*y)) = (1-x*(1+y))/(1-2*x*(1+y)+x^2*(1-y)^2);
E.g.f.: exp((1+y)*x)*cosh(2*sqrt(y)*x);
T(n,k) = Sum_{j=0..n} C(n,j)*C(n-j,2*(k-j))*4^(k-j). (End)
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) + 2*T(n-2,k-1) - T(n-2,k) - T(n-2,k-2), with T(0,0)=T(1,0)=T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 26 2013
From Peter Bala, Sep 22 2021: (Start)
n-th row polynomial R(n,x) = (1-x)^n*T(n,(1+x)/(1-x)), where T(n,x) is the n-th Chebyshev polynomial of the first kind. Cf. A008459.
R(n,x) = Sum_{k = 0..n} binomial(n,2*k)*(4*x)^k*(1 + x)^(n-2*k).
R(n,x) = n*Sum_{k = 0..n} (n+k-1)!/((n-k)!*(2*k)!)*(4*x)^k*(1-x)^(n-k) for n >= 1. (End)

A133853 a(n) = (64^n - 1)/63.

Original entry on oeis.org

0, 1, 65, 4161, 266305, 17043521, 1090785345, 69810262081, 4467856773185, 285942833483841, 18300341342965825, 1171221845949812801, 74958198140788019265, 4797324681010433232961, 307028779584667726909505, 19649841893418734522208321, 1257589881178799009421332545

Offset: 0

Paul Curtz, Jan 07 2008

Partial sums of powers of 64 (A089357), a.k.a. q-numbers for q=64.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Quynh Nguyen, Jean Pedersen, and Hien T. Vu, New Integer Sequences Arising From 3-Period Folding Numbers, Vol. 19 (2016), Article 16.3.1. See Table 1.
Index entries related to partial sums
Index entries related to q-numbers
Index entries for linear recurrences with constant coefficients, signature (65,-64).

Cf. A000364.

Cf. similar sequences of the form (k^n-1)/(k-1) listed in A269025.

[(64^n-1)/63: n in [0..20]]; // Vincenzo Librandi, Aug 10 2011

LinearRecurrence[{65,-64},{0,1},20] (* Harvey P. Dale, Aug 20 2017 *)

makelist((64^n-1)/63, n, 0, 20); /* Martin Ettl, Nov 12 2012 */

PARI
```
A133853(n)=64^n\63
```

From Wolfdieter Lang, Apr 08 2022: (Start)
a(n) = Sum_{j=0..n-1} 2^(6*j). See the comment.
G.f.: x/((1 - 64*x)*(1 - x)).
E.g.f.: exp(x)*(exp(63*x) - 1)/63. (End)

a(6)-a(15) from Vincenzo Librandi, Aug 10 2011

A034947 Jacobi (or Kronecker) symbol (-1/n).

Original entry on oeis.org

1, 1, -1, 1, 1, -1, -1, 1, 1, 1, -1, -1, 1, -1, -1, 1, 1, 1, -1, 1, 1, -1, -1, -1, 1, 1, -1, -1, 1, -1, -1, 1, 1, 1, -1, 1, 1, -1, -1, 1, 1, 1, -1, -1, 1, -1, -1, -1, 1, 1, -1, 1, 1, -1, -1, -1, 1, 1, -1, -1, 1, -1, -1, 1, 1, 1, -1, 1, 1, -1, -1, 1, 1, 1, -1, -1, 1, -1, -1, 1, 1

Offset: 1

N. J. A. Sloane

Also the regular paper-folding sequence.
For a proof that a(n) equals the paper-folding sequence, see Allouche and Sondow, arXiv v4. - Jean-Paul Allouche and Jonathan Sondow, May 19 2015
It appears that, replacing +1 with 0 and -1 with 1, we obtain A038189. Alternatively, replacing -1 with 0 we obtain (allowing for offset) A014577. - Jeremy Gardiner, Nov 08 2004
Partial sums = A005811 starting (1, 2, 1, 2, 3, 2, 1, 2, 3, ...). - Gary W. Adamson, Jul 23 2008
The congruence in {-1,1} of the odd part of n modulo 4 (Cf. A099545). - Peter Munn, Jul 09 2022

			G.f. = x + x^2 - x^3 + x^4 + x^5 - x^6 - x^7 + x^8 + x^9 + x^10 - x^11 - x^12 + ...

J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, pp. 155, 182.
H. Cohen, Course in Computational Number Theory, p. 28.

N. J. A. Sloane, Table of n, a(n) for n = 1..10000
J.-P. Allouche, G.-N. Han, and Jeffrey Shallit, On some conjectures of P. Barry, arXiv:2006.08909 [math.NT], 2020.
J.-P. Allouche and Jonathan Sondow, Summation of rational series twisted by strongly B-multiplicative coefficients, Electron. J. Combin., 22 #1 (2015) P1.59; see p. 8.
J.-P. Allouche and Jonathan Sondow, Summation of rational series twisted by strongly B-multiplicative coefficients, arXiv:1408.5770 [math.NT] v4, 2015; see p. 9.
Jean-Paul Allouche and Leo Goldmakher, Mock characters and the Kronecker symbol, arXiv:1608.03957 [math.NT], 2016.
L. Almodovar, V. H. Moll, H. Quand, Infinite products arising in paperfolding, JIS 19 (2016) # 16.5.1 eq. (1)
Joerg Arndt, Matters Computational (The Fxtbook), section 38.8.4 Differences of the sum of Gray code digits, coefficients of polynomials L.
Danielle Cox and K. McLellan, A problem on generation sets containing Fibonacci numbers, Fib. Quart., 55 (No. 2, 2017), 105-113.
Chandler Davis and Donald E. Knuth, Number Representations and Dragon Curves -- I and II, Journal of Recreational Mathematics, volume 3, number 2, April 1970, pages 66-81, and number 3, July 1970, pages 133-149. Reprinted with addendum in Donald E. Knuth, Selected Papers on Fun and Games, CSLI Publications, 2010, pages 571-614. a(n) = d(n) at equation 3.1.
Chandler Davis and Donald E. Knuth, Number Representations and Dragon Curves, Journal of Recreational Mathematics, volume 3, number 2, April 1970, pages 66-81, and number 3, July 1970, pages 133-149. [Cached copy, with permission]
A. Iványi, Leader election in synchronous networks, Acta Univ. Sapientiae, Mathematica, 5, 2 (2013) 54-82.
Jeffrey Shallit, Cloitre's Self-Generating Sequence, arXiv:2501.00784 [math.CO], 2025. See page 4.
Eric Weisstein's World of Mathematics, Kronecker Symbol.
Index entries for sequences obtained by enumerating foldings.

Cf. A000265, A005811, A000364, A008836, A065339, A097706, A099545, A209615.
Moebius transform of A035184.
Cf. A091072 (indices of 1), A091067 (indices of -1), A371594 (indices of run starts).
The following are all essentially the same sequence: A014577, A014707, A014709, A014710, A034947, A038189, A082410. - N. J. A. Sloane, Jul 27 2012

[KroneckerSymbol(-1,n): n in [1..100]]; // Vincenzo Librandi, Aug 16 2016

with(numtheory): A034947 := n->jacobi(-1,n);

Table[KroneckerSymbol[ -1, n], {n, 0, 100}] (* Corrected by Jean-François Alcover, Dec 04 2013 *)

PARI
```
{a(n) = kronecker(-1, n)};
```

for(n=1, 81, f=factor(n); print1((-1)^sum(s=1, omega(n), f[s, 2]*(Mod(f[s, 1], 4)==3)), ", ")); \\ Arkadiusz Wesolowski, Nov 05 2013

a(n)=direuler(p=1,n,if(p==2,1/(1-kronecker(-4, p)*X)/(1-X),1/(1-kronecker(-4, p)*X))) /* Ralf Stephan, Mar 27 2015 */

a(n) = if(n%2==0, a(n/2), (n+2)%4-2) \\ Peter Munn, Jul 09 2022

def A034947(n):
    s = bin(n)[2:]
    m = len(s)
    i = s[::-1].find('1')
    return 1-2*int(s[m-i-2]) if m-i-2 >= 0 else 1 # Chai Wah Wu, Apr 08 2021

def A034947(n): return -1 if n>>(-n&n).bit_length()&1 else 1 # Chai Wah Wu, Feb 26 2025

Multiplicative with a(2^e) = 1, a(p^e) = (-1)^(e*(p-1)/2) if p>2.
a(2*n) = a(n), a(4*n+1) = 1, a(4*n+3) = -1, a(-n) = -a(n). a(n) = 2*A014577(n-1)-1.
a(prime(n)) = A070750(n) for n > 1. - T. D. Noe, Nov 08 2004
This sequence can be constructed by starting with w = "empty string", and repeatedly applying the map w -> w 1 reverse(-w) [See Allouche and Shallit p. 182]. - N. J. A. Sloane, Jul 27 2012
a(n) = (-1)^A065339(n) = lambda(A097706(n)), where A065339(n) is number of primes of the form 4*m + 3 dividing n (counted with multiplicity) and lambda is Liouville's function, A008836. - Arkadiusz Wesolowski, Nov 05 2013 and Peter Munn, Jun 22 2022
Sum_{n>=1} a(n)/n = Pi/2, due to F. von Haeseler; more generally, Sum_{n>=1} a(n)/n^(2*d+1) = Pi^(2*d+1)*A000364(d)/(2^(2*d+2)-2)(2*d)! for d >= 0; see Allouche and Sondow, 2015. - Jean-Paul Allouche and Jonathan Sondow, Mar 20 2015
Dirichlet g.f.: beta(s)/(1-2^(-s)) = L(chi_2(4),s)/(1-2^(-s)). - Ralf Stephan, Mar 27 2015
a(n) = A209615(n) * (-1)^(v2(n)), where v2(n) = A007814(n) is the 2-adic valuation of n. - Jianing Song, Apr 24 2021
a(n) = 2 - A099545(n) == A000265(n) (mod 4). - Peter Munn, Jun 22 2022 and Jul 09 2022

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A013663 Decimal expansion of zeta(5).

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A003881 Decimal expansion of Pi/4.

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A002105 Reduced tangent numbers: 2^n(2^{2n} - 1)|B_{2n}|/n, where B_n = Bernoulli numbers.