A000367 -id:A000367 - cp's OEIS frontend

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

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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N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
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Bishal Deb and Alan D. Sokal, Continued fractions for cycle-alternating permutations, arXiv:2304.06545 [math.CO], 2023.
K. Dilcher and C. Vignat, Euler and the Strong Law of Small Numbers, Amer. Math. Mnthly, 123 (May 2016), 486-490.
Filippo Disanto and Emanuele Munarini, Local height in weighted Dyck models of random walks and the variability of the number of coalescent histories for caterpillar-shaped gene trees and species trees, SN Applied Sciences (2019), 1:578.
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Michael E. Hoffman, Derivative Polynomials, Euler Polynomials, and Associated Integer Sequences, The Electronic Journal of Combinatorics, vol.6, no.1, #R21, (1999).
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Eric Weisstein's World of Mathematics, Euler Number, Secant Number, Alternating Permutation.
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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

A027641 Numerator of Bernoulli number B_n.

Original entry on oeis.org

1, -1, 1, 0, -1, 0, 1, 0, -1, 0, 5, 0, -691, 0, 7, 0, -3617, 0, 43867, 0, -174611, 0, 854513, 0, -236364091, 0, 8553103, 0, -23749461029, 0, 8615841276005, 0, -7709321041217, 0, 2577687858367, 0, -26315271553053477373, 0, 2929993913841559, 0, -261082718496449122051

Offset: 0

N. J. A. Sloane

a(n)/A027642(n) (Bernoulli numbers) provide the a-sequence for the Sheffer matrix A094816 (coefficients of orthogonal Poisson-Charlier polynomials). See the W. Lang link under A006232 for a- and z-sequences for Sheffer matrices. The corresponding z-sequence is given by the rationals A130189(n)/A130190(n).
Harvey (2008) describes a new algorithm for computing Bernoulli numbers. His method is to compute B(k) modulo p for many small primes p and then reconstruct B(k) via the Chinese Remainder Theorem. The time complexity is O(k^2 log(k)^(2+eps)). The algorithm is especially well-suited to parallelization. - Jonathan Vos Post, Jul 09 2008
Regard the Bernoulli numbers as forming a vector = B_n, and the variant starting (1, 1/2, 1/6, 0, -1/30, ...), (i.e., the first 1/2 has sign +) as forming a vector Bv_n. The relationship between the Pascal triangle matrix, B_n, and Bv_n is as follows: The binomial transform of B_n = Bv_n. B_n is unchanged when multiplied by the Pascal matrix with rows signed (+-+-, ...), i.e., (1; -1,-1; 1,2,1; ...). Bv_n is unchanged when multiplied by the Pascal matrix with columns signed (+-+-, ...), i.e., (1; 1,-1; 1,-2,1; 1,-3,3,-1; ...). - Gary W. Adamson, Jun 29 2012
The sequence of the Bernoulli numbers B_n = a(n)/A027642(n) is the inverse binomial transform of the sequence {A164555(n)/A027642(n)}, illustrated by the fact that they appear as top row and left column in A190339. - Paul Curtz, May 13 2016
Named by de Moivre (1773; "the numbers of Mr. James Bernoulli") after the Swiss mathematician Jacob Bernoulli (1655-1705). - Amiram Eldar, Oct 02 2023

			B_n sequence begins 1, -1/2, 1/6, 0, -1/30, 0, 1/42, 0, -1/30, 0, 5/66, 0, -691/2730, 0, 7/6, 0, -3617/510, ...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 810.
Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 49.
Harold T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 230.
Harold M. Edwards, Riemann's Zeta Function, Academic Press, NY, 1974; see p. 11.
Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.6.1.
Herman H. Goldstine, A History of Numerical Analysis, Springer-Verlag, 1977; Section 2.6.
L. M. Milne-Thompson, Calculus of Finite Differences, 1951, p. 137.
Hans Rademacher, Topics in Analytic Number Theory, Springer, 1973, Chap. 1.

T. D. Noe, Table of n, a(n) for n = 0..200
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
C. M. Bender and K. A. Milton, Continued fraction as a discrete nonlinear transform, arXiv:hep-th/9304052, 1993.
Beáta Bényi and Péter Hajnal, Poly-Bernoulli Numbers and Eulerian Numbers, arXiv:1804.01868 [math.CO], 2018.
H. Bergmann, Eine explizite Darstellung der Bernoullischen Zahlen, Math. Nach. 34 (1967), 377-378. Math Rev 36#4030.
Richard P. Brent and David Harvey, Fast computation of Bernoulli, Tangent and Secant numbers, arXiv preprint arXiv:1108.0286 [math.CO], 2011.
K.-W. Chen, Algorithms for Bernoulli numbers and Euler numbers, J. Integer Sequences, 4 (2001), #01.1.6.
W. Y.C. Chen, J. J. F. Guo and L. X. W. Wang, Log-behavior of the Bernoulli Numbers, arXiv:1208.5213 [math.CO], 2012-2013.
Abraham de Moivre, The Doctrine of Chances, 3rd edition, London, 1733, p. 95.
K. Dilcher, A Bibliography of Bernoulli Numbers (Alphabetically Indexed Authorwise).
Bakir Farhi, Formulas Involving Bernoulli and Stirling Numbers of Both Kinds, Journal of Integer Sequences, Vol. 28 (2025), Article 25.2.6. See p. 16.
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.
H. W. Gould and J. Quaintance, Bernoulli Numbers and a New Binomial Transform Identity, J. Int. Seq. 17 (2014), Article 14.2.2.
M.-P. Grosset and A. P. Veselov, Bernoulli numbers and solitons, arXiv:math/0503175 [math.GM], 2005.
David Harvey, A multimodular algorithm for computing Bernoulli numbers, arXiv:0807.1347 [math.NT], Jul 08 2008.
A. Iványi, Leader election in synchronous networks, Acta Univ. Sapientiae, Mathematica, 5, 2 (2013) 54-82.
Sumit Kumar Jha, A new explicit formula for the Bernoulli numbers in terms of the Stirling numbers of the second kind, Notes on Number Theory and Discrete Mathematics (2020) Vol. 26, No. 2, 148-151.
M. Kaneko, The Akiyama-Tanigawa algorithm for Bernoulli numbers, J. Integer Sequences, 3 (2000), Article 00.2.9.
Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli numbers, arXiv:1707.04451 [math.NT], 2017.
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), Article 14.4.6.
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, The Bernoulli Manifesto. A survey on the occasion of the 300th anniversary of the publication of Jacob Bernoulli's Ars Conjectandi, 1713-2013.
Romeo Meštrović, On a Congruence Modulo n^3 Involving Two Consecutive Sums of Powers, Journal of Integer Sequences, Vol. 17 (2014), Article 14.8.4.
Hisanori Mishima, Bernoulli numbers (n = 2 to 114), (n = 116 to 154) (Factorizations).
Ben Moonen, A remark on the paper of Deninger and Murre, arXiv:2407.05837 [math.AG], 2024. See p. 6.
A. F. Neto, Carlitz's Identity for the Bernoulli Numbers and Zeon Algebra, J. Int. Seq. 18 (2015), Article 15.5.6.
A. F. Neto, A note of a Theorem of Guo, Mezo, and Qi, J. Int. Seq. 19 (2016) Article 16.4.8.
Niels Nielsen, Traite Elementaire des Nombres de Bernoulli, Gauthier-Villars, 1923, pp. 398.
Simon Plouffe, The First 498 Bernoulli numbers. [Project Gutenberg Etext]
Ed Sandifer, How Euler Did It, Bernoulli numbers.
N. J. A. Sloane, "A Handbook of Integer Sequences" Fifty Years Later, arXiv:2301.03149 [math.NT], 2023, p. 5.
J. Singh, On an Arithmetic Convolution, J. Int. Seq. 17 (2014) Article 14.6.7.
J. Sondow and E. Tsukerman, The p-adic order of power sums, the Erdos-Moser equation, and Bernoulli numbers, arXiv:1401.0322 [math.NT], 2014; see section 5.
Eric Weisstein's World of Mathematics, Bernoulli Number.
Eric Weisstein's World of Mathematics, Polygamma Function.
Roman Witula, Damian Slota and Edyta Hetmaniok, Bridges between different known integer sequences, Annales Mathematicae et Informaticae, 41 (2013) pp. 255-263.
Wolfram Research, Generating functions of B_n & B_2n.
Index entries for "core" sequences.
Index entries for sequences related to Bernoulli numbers.

This is the main entry for the Bernoulli numbers and has all the references, links and formulas. Sequences A027642 (the denominators of B_n) and A000367/A002445 = B_{2n} are also important!
A refinement is A194587.
Cf. A000146, A002882, A003245, A127187, A127188, A074909, A164555, A176327, A190339, A131689.
Cf. A094816, A130189, A130190, A155585.

[Numerator(Bernoulli(n)): n in [0..40]]; // Vincenzo Librandi, Mar 17 2014

B := n -> add((-1)^m*m!*Stirling2(n, m)/(m+1), m=0..n);
B := n -> bernoulli(n);
seq(numer(bernoulli(n)), n=0..40); # Zerinvary Lajos, Apr 08 2009

Table[ Numerator[ BernoulliB[ n]], {n, 0, 40}] (* Robert G. Wilson v, Oct 11 2004 *)
Numerator[ Range[0, 40]! CoefficientList[ Series[x/(E^x - 1), {x, 0, 40}], x]]
Numerator[CoefficientList[Series[PolyGamma[1, 1/x]/x - x, {x, 0, 40}, Assumptions -> x > 0], x]] (* Vladimir Reshetnikov, Apr 24 2013 *)

B(n):=(-1)^((n))*sum((stirling1(n,k)*stirling2(n+k,n))/binomial(n+k,k),k,0,n);
makelist(num(B(n)),n,0,20); /* Vladimir Kruchinin, Mar 16 2013 */

PARI
```
a(n)=numerator(bernfrac(n))
```

from sympy import bernoulli
from fractions import Fraction
[bernoulli(i).as_numer_denom()[0] for i in range(51)]  # Indranil Ghosh, Mar 18 2017

from sympy import bernoulli
def A027641(n): return bernoulli(n).p
print([A027641(n) for n in range(80)])  # M. F. Hasler, Jun 11 2019

[bernoulli(n).numerator() for n in range(41)]  # Peter Luschny, Feb 19 2016

# Alternatively:
def A027641_list(len):
    f, R, C = 1, [1], [1]+[0]*(len-1)
    for n in (1..len-1):
        f *= n
        for k in range(n, 0, -1):
            C[k] = C[k-1] / (k+1)
        C[0] = -sum(C[k] for k in (1..n))
        R.append((C[0]*f).numerator())
    return R
A027641_list(41)  # Peter Luschny, Feb 20 2016

E.g.f: x/(exp(x) - 1); take numerators.
Recurrence: B^n = (1+B)^n, n >= 2 (interpreting B^j as B_j).
B_{2n}/(2n)! = 2*(-1)^(n-1)*(2*Pi)^(-2n) Sum_{k>=1} 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).
Sum_{i=1..n-1} i^k = ((n+B)^(k+1)-B^(k+1))/(k+1) (interpreting B^j as B_j).
B_{n-1} = - Sum_{r=1..n} (-1)^r binomial(n, r) r^(-1) Sum_{k=1..r} k^(n-1). More concisely, B_n = 1 - (1-C)^(n+1), where C^r is replaced by the arithmetic mean of the first r n-th powers of natural numbers in the expansion of the right-hand side. [Bergmann]
Sum_{i>=1} 1/i^(2k) = zeta(2k) = (2*Pi)^(2k)*|B_{2k}|/(2*(2k)!).
B_{2n} = (-1)^(m-1)/2^(2m+1) * Integral{-inf..inf, [d^(m-1)/dx^(m-1) sech(x)^2 ]^2 dx} (see Grosset/Veselov).
Let B(s,z) = -2^(1-s)(i/Pi)^s s! PolyLog(s,exp(-2*i*Pi/z)). Then B(2n,1) = B_{2n} for n >= 1. Similarly the numbers B(2n+1,1), which might be called Co-Bernoulli numbers, can be considered, and it is remarkable that Leonhard Euler in 1755 already calculated B(3,1) and B(5,1) (Opera Omnia, Ser. 1, Vol. 10, p. 351). (Cf. the Luschny reference for a discussion.) - Peter Luschny, May 02 2009
The B_n sequence is the left column of the inverse of triangle A074909, the "beheaded" Pascal's triangle. - Gary W. Adamson, Mar 05 2012
From Sergei N. Gladkovskii, Dec 04 2012: (Start)
E.g.f. E(x)= 2 - x/(tan(x) + sec(x) - 1)= Sum_{n>=0} a(n)*x^n/n!, a(n)=|B(n)|, where B(n) is Bernoulli number B_n.
E(x)= 2 + x - B(0), where B(k)=  4*k+1 + x/(2 + x/(4*k+3 - x/(2 - x/B(k+1)))); (continued fraction, 4-step). (End)
E.g.f.: x/(exp(x)-1)= U(0); U(k)= 2*k+1 - x(2*k+1)/(x + (2*k+2)/(1 + x/U(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 05 2012
E.g.f.: 2*(x-1)/(x*Q(0)-2) where Q(k) = 1 + 2*x*(k+1)/((2*k+1)*(2*k+3) - x*(2*k+1)*(2*k+3)^2/(x*(2*k+3) + 4*(k+1)*(k+2)/Q(k+1))); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 26 2013
a(n) = numerator(B(n)), B(n) = (-1)^n*Sum_{k=0..n} Stirling1(n,k) * Stirling2(n+k,n) / binomial(n+k,k). - Vladimir Kruchinin, Mar 16 2013
E.g.f.: x/(exp(x)-1) = E(0) where E(k) = 2*k+1 - x/(2 + x/E(k+1)); (continued fraction). - Sergei N. Gladkovskii, Mar 16 2013
G.f. for Bernoulli(n) = a(n)/A027642(n): psi_1(1/x)/x - x, where psi_n(z) is the polygamma function, psi_n(z) = (d/dz)^(n+1) log(Gamma(z)). - Vladimir Reshetnikov, Apr 24 2013
E.g.f.: 2*E(0) - 2*x, where E(k)= x + (k+1)/(1 + 1/(1 - x/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 10 2013
B_n = Sum_{m=0..n} (-1)^m *A131689(n, m)/(m + 1), n >= 0. See one of the Maple programs. - Wolfdieter Lang, May 05 2017
a(n) = numerator((-1)^n*A155585(n-1)*n/(4^n-2^n)), for n>=1. - Mats Granvik, Nov 26 2017
From Artur Jasinski, Dec 30 2020: (Start)
a(n) = numerator(-2*cos(Pi*n/2)*Gamma(n+1)*zeta(n)/(2*Pi)^n), for n=0 and n>1.
a(n) = numerator(-n*zeta(1-n)), for n=0 and n>1. (End)
a(n) = numerator(Sum_{k=0..n-1} (-1)^(k-1)*k!*Stirling2(n-1,k) / ((k+1)*(k+2))), for n>0 (see Jha link). - Bill McEachen, Jul 17 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).
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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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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]
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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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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. - _N. J. A. Sloane_, Dec 28 2012 [broken link]
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Eric Weisstein's World of Mathematics, Tangent Number
Eric Weisstein's World of Mathematics, Alternating Permutation
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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

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

A017329 a(n) = 10*n + 5.

Original entry on oeis.org

5, 15, 25, 35, 45, 55, 65, 75, 85, 95, 105, 115, 125, 135, 145, 155, 165, 175, 185, 195, 205, 215, 225, 235, 245, 255, 265, 275, 285, 295, 305, 315, 325, 335, 345, 355, 365, 375, 385, 395, 405, 415, 425, 435, 445, 455, 465, 475, 485, 495, 505, 515, 525, 535

Offset: 0

N. J. A. Sloane

Continued fraction expansion of tanh(1/5). - Benoit Cloitre, Dec 17 2002
n such that 5 divides the numerator of B(2n) where B(2n) = the 2n-th Bernoulli number. - Benoit Cloitre, Jan 01 2004
5 times odd numbers. - Omar E. Pol, May 02 2008
5th transversal numbers (or 5-transversal numbers): Numbers of the 5th column of positive numbers in the square array of nonnegative and polygonal numbers A139600. Also, numbers of the 5th column in the square array A057145. - Omar E. Pol, May 02 2008
Successive sums: 5, 20, 45, 80, 125, ... (see A033429). - Philippe Deléham, Dec 08 2011
3^a(n) + 1 is divisible by 61. - Vincenzo Librandi, Feb 05 2013
If the initial 5 is changed to 1, giving 1,15,25,35,45,..., these are values of m such that A323288(m)/m reaches a new record high value. - N. J. A. Sloane, Jan 23 2019

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189. - From N. J. A. Sloane, Dec 01 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences.
Index to sequences related to polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A000367, A002193, A016957, A057145, A094884, A139600, A139606, A182007.
Subsequence of A034709, together with A017281, A017293, A139222, A139245, A139249, A139264, A139279 and A139280.
Cf. A005408, A008587, A033429, A267541, A323288.

a017329 = (+ 5) . (* 10)
a017329_list = [5, 15 ..]  -- Reinhard Zumkeller, Apr 10 2015

[10*n + 5: n in [0..60]]; // Vincenzo Librandi, Feb 05 2013

Range[5, 1000, 10] (* Vladimir Joseph Stephan Orlovsky, May 28 2011 *)
LinearRecurrence[{2,-1},{5,15},60] (* Harvey P. Dale, Nov 16 2019 *)

a(n)=10*n+5 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 5*A005408(n). - Omar E. Pol, Oct 19 2008
a(n) = 20*n - a(n-1) (with a(0)=5). - Vincenzo Librandi, Nov 19 2010
G.f.: 5*(x+1)/(x-1)^2. - Colin Barker, Nov 14 2012
a(n) = A057145(n+2,5). - R. J. Mathar, Jul 28 2016
E.g.f.: 5*exp(x)*(1 + 2*x). - Stefano Spezia, Feb 14 2020
Sum_{n>=0} (-1)^n/a(n) = Pi/20. - Amiram Eldar, Dec 12 2021
From Amiram Eldar, Nov 23 2024: (Start)
Product_{n>=0} (1 - (-1)^n/a(n)) = sqrt(5-sqrt(5))/2 = sqrt(2)*sin(Pi/5) = A182007/A002193.
Product_{n>=0} (1 + (-1)^n/a(n)) = phi/sqrt(2) (A094884). (End)
a(n) = (n+3)^2 - (n-2)^2. - Alexander Yutkin, Mar 16 2025
From Elmo R. Oliveira, Apr 12 2025: (Start)
a(n) = 2*a(n-1) - a(n-2).
a(n) = A008587(2*n+1). (End)

A001067 Numerator of Bernoulli(2n)/(2n).

Original entry on oeis.org

1, -1, 1, -1, 1, -691, 1, -3617, 43867, -174611, 77683, -236364091, 657931, -3392780147, 1723168255201, -7709321041217, 151628697551, -26315271553053477373, 154210205991661, -261082718496449122051, 1520097643918070802691, -2530297234481911294093

Offset: 1

N. J. A. Sloane, Richard E. Borcherds (reb(AT)math.berkeley.edu)

It was incorrectly claimed that a(n) is "also numerator of "modified Bernoulli number" b(2n) = Bernoulli(2*n)/(2*n*n!)"; actually, the numerators of these fractions and the numerators of "modified Bernoulli numbers" (see A057868 for details) differ from each other and from this sequence. - Andrey Zabolotskiy, Dec 03 2022
Ramanujan incorrectly conjectured that the sequence contains only primes (and 1). - Jud McCranie. See A112548, A119766.
a(n) = A046968(n) if n < 574; a(574) = 37 * A046968(574). - Michael Somos, Feb 01 2004
Absolute values give denominators of constant terms of Fourier series of meromorphic modular forms E_k/Delta, where E_k is the normalized k th Eisenstein series [cf. Gunning or Serre references] and Delta is the normalized unique weight-twelve cusp form for the full modular group (the generating function of Ramanujan's tau function.) - Barry Brent (barrybrent(AT)iphouse.com), Jun 01 2009
|a(n)| is a product of powers of irregular primes (A000928), with the exception of n = 1,2,3,4,5,7. - Peter Luschny, Jul 28 2009
Conjecture: If there is a prime p such that 2*n+1 < p and p divides a(n), then p^2 does not divide a(n). This conjecture is true for p < 12 million. - Seiichi Manyama, Jan 21 2017

			The sequence Bernoulli(2*n)/(2*n) (n >= 1) begins 1/12, -1/120, 1/252, -1/240, 1/132, -691/32760, 1/12, -3617/8160, ...
The sequence of modified Bernoulli numbers begins 1/48, -1/5760, 1/362880, -1/19353600, 1/958003200, -691/31384184832000, ...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 259, (6.3.18) and (6.3.19); also p. 810.
L. V. Ahlfors, Complex Analysis, McGraw-Hill, 1979, p. 205
R. C. Gunning, Lectures on Modular Forms. Princeton Univ. Press, Princeton, NJ, 1962, p. 53.
R. Kanigel, The Man Who Knew Infinity, pp. 91-92.
J. W. Milnor and J. D. Stasheff, Characteristic Classes, Princeton, 1974, p. 285.
J.-P. Serre, A Course in Arithmetic, Springer-Verlag, 1973, p. 93.

Seiichi Manyama, Table of n, a(n) for n = 1..314 (first 100 terms from T. D. Noe)
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math.Series 55, Tenth Printing, 1972, p. 259, (6.3.18) and (6.3.19).
D. Bar-Natan, T. T. Q. Le and D. P. Thurston, Two applications of elementary knot theory to Lie algebras and Vassiliev invariants, arXiv:math/0204311 [math.QA], 2002-2003; Geometry and Topology 7-1 (2003) 1-31.
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.
E. Z. Goren, Tables of values of Riemann zeta functions
Milan Janjic, Hessenberg Matrices and Integer Sequences, J. Int. Seq. 13 (2010) # 10.7.8, section 3.
J. Sondow and E. W. Weisstein, MathWorld: Riemann Zeta Function
Eric Weisstein's World of Mathematics, Eisenstein Series.
Eric Weisstein's World of Mathematics, Bernoulli Number.
Wikipedia, Kummer-Vandiver conjecture
Index entries for sequences related to Bernoulli numbers

Similar to but different from A046968. See A090495, A090496.
Denominators given by A006953.
Cf. A000367, A006863, A033563, A046968.
Cf. A141590, A255505.

List([1..25], n-> NumeratorRat(Bernoulli(2*n)/(2*n)));  # G. C. Greubel, Sep 19 2019

[Numerator(Bernoulli(2*n)/(2*n)):n in [1..40]]; // Vincenzo Librandi, Sep 17 2015

A001067_list := proc(n) 1/(1-1/exp(z)); series(%,z,2*n+4);
seq(numer((2*i+1)!*coeff(%,z,2*i+1)),i=0..n) end:
A001067_list(21); # Peter Luschny, Jul 12 2012

Table[ Numerator[ BernoulliB[2n]/(2n)], {n, 1, 22}] (* Robert G. Wilson v, Feb 03 2004 *)

{a(n) = if( n<1, 0, numerator( bernfrac(2*n) / (2*n)))}; /* Michael Somos, Feb 01 2004 */

@CachedFunction
def S(n, k) :
    if k == 0 :
        if n == 0 : return 1
        else: return 0
    return S(n, k-1) + S(n-1, n-k)
def BernoulliDivN(n) :
    if n == 0 : return 1
    return (-1)^n*S(2*n-1,2*n-1)/(4^n-16^n)
[BernoulliDivN(n).numerator() for n in (1..22)]
# Peter Luschny, Jul 08 2012

[numerator(bernoulli(2*n)/(2*n)) for n in (1..25)] # G. C. Greubel, Sep 19 2019

Zeta(1-2*n) = - Bernoulli(2*n)/(2*n).
G.f.: numerators of coefficients of z^(2*n) in z/(exp(z)-1). - Benoit Cloitre, Jun 02 2003
For 2 <= k <= 1000 and k != 7, the 2-order of the full constant term of E_k/Delta = 3 + ord_2(k - 7). - Barry Brent (barrybrent(AT)iphouse.com), Jun 01 2009
G.f. for Bernoulli(2*n)/(2*n) = a(n)/A006953(n): (-1)^n/((2*Pi)^(2*n)*(2*n))*integral(log(1-1/t)^(2*n) dt,t=0,1). - Gerry Martens, May 18 2011
E.g.f.: a(n) = numerator((2*n+1)!*[x^(2*n+1)](1/(1-1/exp(x)))). - Peter Luschny, Jul 12 2012
|a(n)| = numerator of Integral_{r=0..1} HurwitzZeta(1-n, r)^2 dr. More general: |Bernoulli(2*n)| = binomial(2*n,n)*n^2*I(n) for n >= 1 where I(n) denotes the integral. - Peter Luschny, May 24 2015

A249134 Numbers k such that Bernoulli number B_k has denominator 2730.

Original entry on oeis.org

12, 24, 1308, 1884, 2004, 2364, 2532, 2724, 3804, 4008, 4044, 4188, 4236, 4668, 5052, 5064, 5268, 5388, 5484, 6252, 6492, 6564, 6756, 6852, 7044, 7188, 7356, 7404, 7608, 7764, 8124, 8412, 8472, 8796, 9084, 9228, 9852, 9876, 9924

Offset: 1

Jean-François Alcover and Paul Curtz, Oct 22 2014

nonn

2730 = 2 * 3 * 5 * 7 * 13.

			BernoulliB(12) is -691/2730, hence 12 is in the sequence.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A000367, A002445, A006954, A027642, A027760, A027762, A051222, A051225-A051230, A245056, A271634, A271635, A272138, A272139, A272140, A272183, A272184, A272185, A272186.

Reap[For[n = 0, n <= 10^4, n = n+12, If[Denominator[BernoulliB[n]] == 2730, Print[n]; Sow[n]]]][[2, 1]]
Select[Table[n, {n, 2, 10000}], Denominator@BernoulliB[#]==2730 &] (* Vincenzo Librandi, Apr 02 2015 *)

is(n)=denominator(bernfrac(n))==2730 \\ Charles R Greathouse IV, Oct 22 2014

is(n)=if(n%12 || n%16==0 || n%9==0, return(0)); forprime(p=5,107, if(n%p==0, return(0))); fordiv(n,d, if(isprime(d+1) && d>13, return(0))); 1 \\ Charles R Greathouse IV, Oct 22 2014

A002432 Denominators of zeta(2n)/Pi^(2n).

Original entry on oeis.org

2, 6, 90, 945, 9450, 93555, 638512875, 18243225, 325641566250, 38979295480125, 1531329465290625, 13447856940643125, 201919571963756521875, 11094481976030578125, 564653660170076273671875, 5660878804669082674070015625, 62490220571022341207266406250

Offset: 0

N. J. A. Sloane

Also denominators in expansion of Psi(x).
For n >= 1 a(n) is always divisible by 3 (by the von Staudt-Clausen theorem, see A002445).
A comment due to G. Campbell: The original approach taken by Euler was to derive the infinite product for sin(Pi*x)/(Pi*x) equal to (1 - x^2/1^2) (1 - x^2/2^2)(1 - x^2/3^2) ... treating sin(Pi*x)/(Pi*x) as if it were a polynomial. Differentiating the logarithm of both sides and equating coefficients gives all of the zeta function values for 2, 4, 6, 8, .... - M. F. Hasler, Mar 29 2015
Note that 2n+1 divides a(n) for every n. If 2n+1 > 9 is composite, then (2n+1)^2 divides a(n). If 2n+1 is prime, then (2n+1)^2 does not divide a(n). My theorem: for n > 4, (2n+1)^2 divides a(n) if and only if the number 2n+1 is composite. - Thomas Ordowski, Nov 07 2022

			(zeta(2n)/Pi^(2n), n = 0, 1, 2, 3, ...) = (-1/2, 1/6, 1/90, 1/945, 1/9450, 1/93555, 691/638512875, 2/18243225, 3617/325641566250, ...), i.e.: zeta(0) = -1/2, zeta(2) = Pi^2/6, zeta(4) = Pi^4/90, zeta(6) = Pi^6/945, zeta(8) = Pi^8/9450, zeta(10) = Pi^10/93555, zeta(12) = Pi^12*691/638512875, ...
In Maple, series(Psi(x),x,20) gives -1*x^(-1) + (-gamma) + 1/6*Pi^2*x + (-Zeta(3))*x^2 + 1/90*Pi^4*x^3 + (-Zeta(5))*x^4 + 1/945*Pi^6*x^5 + (-Zeta(7))*x^6 + 1/9450*Pi^8*x^7 + (-Zeta(9))*x^8 + 1/93555*Pi^10*x^9 + ...
a(5) = 93555 = 10!/(2^9 * B(10)) = 3628800/(512*5/66). - _Frank Ellermann_, Apr 03 2020

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 88.
A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 84.
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 and J.P. Martin-Flatin, Table of n, a(n) for n = 0..250 (first 100 terms were computed by T. D. Noe).
G. Campbell, Some series like ζ(3), ζ(5), ζ(7). Number Theory group on LinkedIn.com, March 2015.
N. D. Elkies, On the sums Sum((4k+1)^(-n),k,-inf,+inf), arXiv:math/0101168 [math.CA], 2001-2003.
N. D. Elkies, On the sums Sum_{k = -infinity .. infinity} (4k+1)^(-n), Amer. Math. Monthly, 110 (No. 7, 2003), 561-573.
Masato Kobayashi and Shunji Sasaki, Values of zeta-one functions at positive even integers, arXiv:2202.11835 [math.NT], 2022. See p. 4.
J. Sondow and E. W. Weisstein, MathWorld: Riemann Zeta Function
I. Song, A recursive formula for even order harmonic series, J. Computational and Appl. Math., 21 (1988), 251-256.
I. Song, A recursive formula for even order harmonic series, J. Computational and Appl. Math., 21 (1988), 251-256. [Annotated scanned copy]
Index entries for zeta function.

Cf. A046988 (numerators), A006003.

seq(denom(Zeta(2*n)/Pi^(2*n)),n=0..24); # Martin Renner, Sep 07 2016
A002432List := proc(len) series(-x*cot(x)/2, x, 2*len+1):
seq(denom(coeff(%, x, n)), n=0..2*len-1, 2) end:
A002432List(17); # Peter Luschny, Jun 07 2020

Table[Denominator[Zeta[2 n]/Pi^(2 n)], {n, 0, 30}] (* Artur Jasinski, Mar 11 2010 *)
Denominator[Zeta[2*Range[0, 20]]] (* Harvey P. Dale, Sep 05 2013 *)

a(n)=numerator(bestappr(Pi^(2*n)/zeta(2*n))) \\ Requires sufficient realprecision. The standard value of 38 digits yields erroneous values for n>9. \p99 is more than enough to get the 3 lines of displayed data. - M. F. Hasler, Mar 29 2015

a002432(n) = denominator(polcoeff((1-x*cotan(x))/2,n*2))
default(seriesprecision, 33); for(i=0,16,print1(a002432(i),",")) \\ Chris Boyd, Dec 21 2015

Sum_{n>=1} 2/(n^2 + 1) = Pi*coth(Pi)-1. 2*Sum_{k>=1} (-1)^(k + 1)/n^(2*k) = 2/(n^2+1). - Shmuel Spiegel (shmualm(AT)hotmail.com), Aug 13 2001
zeta(2n)/(2i * ( log(1-i)-log(1+i) ))^(2n) = zeta(2n)/(-i*log(-1))^(2n). - Eric Desbiaux, Dec 12 2008
zeta(2n) = Sum_{k >= 1} k^(-2n) = (-1)^(n-1)*B_{2n}*2^(2n-1)*Pi^(2n)/(2n)!.
a(n) = -A046988(n)*A010050(n)*A002445(n)/(A009117(n)*A000367(n))
a(n) = sqrt(denominator(Sum_{i>=1} A000005(i)/i^2n)). - Enrique Pérez Herrero, Jan 19 2012
Sum_{k >= 1} zeta(2k)*x^(2k) = (1-Pi*x*cot(Pi*x))/2. - Chris Boyd, Dec 21 2015
a(n) = denominator([x^(2*n)] -x*cot(x)/2). - Peter Luschny, Jun 07 2020

Formula and link from Henry Bottomley, Mar 10 2000
Formula corrected by Bjoern Boettcher, May 15 2003
Corrected and edited by M. F. Hasler, Mar 29 2015
a(0) = 2 prepended by Peter Luschny, Jun 07 2020

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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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A027641 Numerator of Bernoulli number B_n.

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

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A002445 Denominators of Bernoulli numbers B_{2n}.

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A001067 Numerator of Bernoulli(2n)/(2n).

A002432 Denominators of zeta(2n)/Pi^(2n).