A019669 -id:A019669 - 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, Classical continued fractions for some multivariate polynomials generalizing the Genocchi and median Genocchi numbers, arXiv:2212.07232 [math.CO], 2022.
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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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
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P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 144.
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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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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

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

A008441 Number of ways of writing n as the sum of 2 triangular numbers.

Original entry on oeis.org

1, 2, 1, 2, 2, 0, 3, 2, 0, 2, 2, 2, 1, 2, 0, 2, 4, 0, 2, 0, 1, 4, 2, 0, 2, 2, 0, 2, 2, 2, 1, 4, 0, 0, 2, 0, 4, 2, 2, 2, 0, 0, 3, 2, 0, 2, 4, 0, 2, 2, 0, 4, 0, 0, 0, 4, 3, 2, 2, 0, 2, 2, 0, 0, 2, 2, 4, 2, 0, 2, 2, 0, 3, 2, 0, 0, 4, 0, 2, 2, 0, 6, 0, 2, 2, 0, 0, 2, 2, 0, 1, 4, 2, 2, 4, 0, 0, 2, 0, 2, 2, 2, 2, 0, 0

Offset: 0

N. J. A. Sloane

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700). The present sequence gives the expansion coefficients of psi(q)^2.

Also the number of positive odd solutions to equation x^2 + y^2 = 8*n + 2. - Seiichi Manyama, May 28 2017

			G.f. = 1 + 2*x + x^2 + 2*x^3 + 2*x^4 + 3*x^6 + 2*x^7 + 2*x^9 + 2*x^10 + 2*x^11 + ...
G.f. for B(q) = q * A(q^4) = q + 2*q^5 + q^9 + 2*q^13 + 2*q^17 + 3*q^25 + 2*q^29 + 2*q^37 + 2*q^41 + ...

B. C. Berndt, Ramanujan's Notebooks Part III, Springer-Verlag. See p. 139 Example (iv).
J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer-Verlag, p. 102.
R. W. Gosper, Strip Mining in the Abandoned Orefields of Nineteenth Century Mathematics, in Computers in Mathematics (Ed. D. V. Chudnovsky and R. D. Jenks). New York: Dekker, 1990. See p. 279.
R. W. Gosper, Experiments and discoveries in q-trigonometry, in Symbolic Computation, Number Theory, Special Functions, Physics and Combinatorics. Editors: F. G. Garvan and M. E. H. Ismail. Kluwer, Dordrecht, Netherlands, 2001, pp. 79-105. [See Pi_q.]
P. A. MacMahon, Combinatory Analysis, Cambridge Univ. Press, London and New York, Vol. 1, 1915 and Vol. 2, 1916. See vol. 2, p 31, Article 272.
Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, John Wiley and Sons, Inc., NY 1991, p. 165.

T. D. Noe, Table of n, a(n) for n = 0..10000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 575, 16.23.1 and 16.23.2.
R. P. Agarwal, Lambert series and Ramanujan, Prod. Indian Acad. Sci. (Math. Sci.), v. 103, n. 3, 1993, pp. 269-293 (see p. 285).
N. J. Fine, Basic Hypergeometric Series and Applications, Amer. Math. Soc., 1988; p. 72, Eq (31.2); p. 78, Eq. following (32.25).
J. W. L. Glaisher, On the function chi(n), Quarterly Journal of Pure and Applied Mathematics, 20 (1884), 97-167. See p. 108.
R. W. Gosper, Experiments and discoveries in q-trigonometry, Preprint.
R. W. Gosper, q-Trigonometry: Some Prefatory Afterthoughts.
Michael D. Hirschhorn, The number of representations of a number by various forms, Discrete Mathematics 298 (2005), 205-211.
Nadia Heninger, E. M. Rains and N. J. A. Sloane, On the Integrality of n-th Roots of Generating Functions, arXiv:math/0509316 [math.NT], 2005-2006; J. Combinatorial Theory, Series A, 113 (2006), 1732-1745.
Nicco, Conjectured identity of the product of two theta functions, Math Stackexchange, Sep 09 2015.
K. Ono, S. Robins and P. T. Wahl, On the representation of integers as sums of triangular numbers, Aequationes mathematicae, August 1995, Volume 50, Issue 1-2, pp. 73-94.
Hjalmar Rosengren, Sums of triangular numbers from the Frobenius determinant, arXiv:math/0504272 [math.NT], 2005.
James A. Sellers and Roberto Tauraso, Arithmetic properties of MacMahon-type sums of divisors: the odd case, arXiv:2505.12813 [math.NT], 2025.
Michael Somos, Introduction to Ramanujan theta functions.
Eric Weisstein's World of Mathematics, Ramanujan Theta Functions.

A002654, A008442, A035154, A035181, A035184, A112301, A113406, A113414, A113446, A113652 all satisfy A(4n+1)=a(n).
Cf. A004020, A005883, A104794, A052343, A199015 (partial sums).
Number of ways of writing n as a sum of k triangular numbers, for k=1,...: A010054, A008441, A008443, A008438, A008439, A008440, A226252, A007331, A226253, A226254, A226255, A014787, A014809.
Cf. A002175, A053692, A113407, A121444.
Cf. A000217, A001938, A004018.
Cf. A274621 (reciprocal series).
Cf. A001826, A001842.
Cf. A019669, A079006.

a052343 = (flip div 2) . (+ 1) . a008441
-- Reinhard Zumkeller, Jul 25 2014

A := Basis( ModularForms( Gamma1(8), 1), 420); A[2]; /* Michael Somos, Jan 31 2015 */

sigmamr := proc(n,m,r) local a,d ; a := 0 ; for d in numtheory[divisors](n) do if modp(d,m) = r then a := a+1 ; end if; end do: a; end proc:
A002654 := proc(n) sigmamr(n,4,1)-sigmamr(n,4,3) ; end proc:
A008441 := proc(n) A002654(4*n+1) ; end proc:
seq(A008441(n),n=0..90) ; # R. J. Mathar, Mar 23 2011

Plus@@((-1)^(1/2 (Divisors[4#+1]-1)))& /@ Range[0, 104] (* Ant King, Dec 02 2010 *)
a[ n_] := SeriesCoefficient[ (1/2) EllipticTheta[ 2, 0, q] EllipticTheta[ 3, 0, q], {q, 0, n + 1/4}]; (* Michael Somos, Jun 19 2012 *)
a[ n_] := SeriesCoefficient[ (1/4) EllipticTheta[ 2, 0, q]^2, {q, 0, 2 n + 1/2}]; (* Michael Somos, Jun 19 2012 *)
a[ n_] := If[ n < 0, 0, DivisorSum[ 4 n + 1, (-1)^Quotient[#, 2] &]];  (* Michael Somos, Jun 08 2014 *)
QP = QPochhammer; s = QP[q^2]^4/QP[q]^2 + O[q]^100; CoefficientList[s, q] (* Jean-François Alcover, Nov 27 2015, adapted from PARI *)
TriangleQ[n_] := IntegerQ@Sqrt[8n +1]; Table[Count[FrobeniusSolve[{1, 1}, n], {?TriangleQ}], {n, 0, 104}] (* Robert G. Wilson v, Apr 15 2017 *)

{a(n) = if( n<1, n==0, polcoeff( sum(k=0, (sqrtint(8*n + 1) - 1)\2, x^(k * (k+1)/2), x * O(x^n))^2, n) )};

{a(n) = if( n<0, 0, n = 4*n + 1; sumdiv(n, d, (-1)^(d\2)))}; /* Michael Somos, Sep 02 2005 */

{a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x^2 + A)^4 / eta(x + A)^2, n))};

{a(n) = if( n<0, 0, n = 4*n + 1; sumdiv( n, d, (d%4==1) - (d%4==3)))}; /* Michael Somos, Sep 14 2005 */

{ my(q='q+O('q^166)); Vec(eta(q^2)^4 / eta(q)^2) } \\ Joerg Arndt, Apr 16 2017

ModularForms( Gamma1(8), 1, prec=420).1; # Michael Somos, Jun 08 2014

This sequence is the quadrisection of many sequences. Here are two examples:
a(n) = A002654(4n+1), the difference between the number of divisors of 4*n+1 of form 4*k+1 and the number of form 4*k-1. - David Broadhurst, Oct 20 2002
a(n) = b(4*n + 1), where b(n) is multiplicative and b(2^e) = 0^e, b(p^e) = (1 + (-1)^e) / 2 if p == 3 (mod 4), b(p^e) = e+1 if p == 1 (mod 4). - Michael Somos, Sep 14 2005
G.f.: (Sum_{k>=0} x^((k^2 + k)/2))^2 = (Sum_{k>=0} x^(k^2 + k)) * (Sum_{k in Z} x^(k^2)).
Expansion of Jacobi theta (theta_2(0, sqrt(q)))^2 / (4 * q^(1/4)).
Sum[d|(4n+1), (-1)^((d-1)/2) ].
Given g.f. A(x), then B(q) = q * A(q^4) satisfies 0 = f(B(q), B(q^2), B(q^4)) where f(u, v, w) = v^3 + 4 * v * w^2 - u^2 * w. - Michael Somos, Sep 14 2005
Given g.f. A(x), then B(q) = q * A(q^4) satisfies 0 = f(B(q), B(q^2), B(q^3), B(q^6)) where f(u1, u2, u3, u6) = u1 * u3 - (u2 - u6) * (u2 + 3*u6). - Michael Somos, Sep 14 2005
Expansion of Jacobi k/(4*q^(1/2)) * (2/Pi)* K(k) in powers of q^2. - Michael Somos, Sep 14 2005. Convolution of A001938 and A004018. This appears in the denominator of the Jacobi sn and cn formula given in the Abramowitz-Stegun reference, p. 575, 16.23.1 and 16.23.2, where m=k^2. - Wolfdieter Lang, Jul 05 2016
G.f.: Sum_{k>=0} a(k) * x^(2*k) = Sum_{k>=0} x^k / (1 + x^(2*k + 1)).
G.f.: Sum_{k in Z} x^k / (1 - x^(4*k + 1)). - Michael Somos, Nov 03 2005
Expansion of psi(x)^2 = phi(x) * psi(x^2) in powers of x where phi(), psi() are Ramanujan theta functions.
Moebius transform is period 8 sequence [ 1, -1, -1, 0, 1, 1, -1, 0, ...]. - Michael Somos, Jan 25 2008
G.f. is a period 1 Fourier series which satisfies f(-1 / (8 t)) = 1/2 (t/i) g(t) where q = exp(2 Pi i t) and g() is the g.f. for A104794.
Euler transform of period 2 sequence [ 2, -2, ...].
G.f.: q^(-1/4) * eta(q^2)^4 / eta(q)^2. See also the Fine reference.
a(n) = Sum_{k=0..n} A010054(k)*A010054(n-k). - Reinhard Zumkeller, Nov 03 2009
A004020(n) = 2 * a(n). A005883(n) = 4 * a(n).
Convolution square of A010054.
G.f.: Product_{k>0} (1 - x^(2*k))^2 / (1 - x^(2*k-1))^2.
a(2*n) = A113407(n). a(2*n + 1) = A053692(n). a(3*n) = A002175(n). a(3*n + 1) = 2 * A121444(n). a(9*n + 2) = a(n). a(9*n + 5) = a(9*n + 8) = 0. - Michael Somos, Jun 08 2014
G.f.: exp( Sum_{n>=1} 2*(x^n/n) / (1 + x^n) ). - Paul D. Hanna, Mar 01 2016
a(n) = A001826(2+8*n) - A001842(2+8*n), the difference between the number of divisors 1 (mod 4) and 3 (mod 4) of 2+8*n. See the Ono et al. link, Corollary 1, or directly the Niven et al. reference, p. 165, Corollary (3.23). - Wolfdieter Lang, Jan 11 2017
Expansion of continued fraction 1 / (1 - x^1 + x^1*(1 - x^1)^2 / (1 - x^3 + x^2*(1 - x^2)^2 / (1 - x^5 + x^3*(1 - x^3)^2 / ...))) in powers of x^2. - Michael Somos, Apr 20 2017
Given g.f. A(x), and B(x) is the g.f. for A079006, then B(x) = A(x^2) / A(x) and B(x) * B(x^2) * B(x^4) * ... = 1 / A(x). - Michael Somos, Apr 20 2017
a(0) = 1, a(n) = (2/n)*Sum_{k=1..n} A002129(k)*a(n-k) for n > 0. - Seiichi Manyama, May 06 2017
From Paul D. Hanna, Aug 10 2019: (Start)
G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} binomial(n,k) * (x^(2*n+1) - x^(2*k))^(n-k) = Sum_{n>=0} a(n)*x^(2*n).
G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} binomial(n,k) * (x^(2*n+1) + x^(2*k))^(n-k) * (-1)^k = Sum_{n>=0} a(n)*x^(2*n). (End)
From Peter Bala, Jan 05 2021: (Start)
G.f.: Sum_{n = -oo..oo} x^(4*n^2+2*n) * (1 + x^(4*n+1))/(1 - x^(4*n+1)). See Agarwal, p. 285, equation 6.20 with i = j = 1 and mu = 4.
For prime p of the form 4*k + 3, a(n*p^2 + (p^2 - 1)/4) = a(n).
If n > 0 and p are coprime then a(n*p + (p^2 - 1)/4) = 0. The proofs are similar to those given for the corresponding results for A115110. Cf. A000729.
For prime p of the form 4*k + 1 and for n not congruent to (p - 1)/4 (mod p) we have a(n*p^2 + (p^2 - 1)/4) = 3*a(n) (since b(n), where b(4*n+1) = a(n), is multiplicative). (End)
From Peter Bala, Mar 22 2021: (Start)
G.f. A(q) satisfies:
A(q^2) = Sum_{n = -oo..oo} q^n/(1 - q^(4*n+2)) (set z = q, alpha = q^2, mu = 4 in Agarwal, equation 6.15).
A(q^2) = Sum_{n = -oo..oo} q^(2*n)/(1 - q^(4*n+1)) (set z = q^2, alpha = q, mu = 4 in Agarwal, equation 6.15).
A(q^2) = Sum_{n = -oo..oo} q^n/(1 + q^(2*n+1))^2 = Sum_{n = -oo..oo} q^(3*n+1)/(1 + q^(2*n+1))^2. (End)
G.f.: Sum_{k>=0} a(k) * q^k = Sum_{k>=0} (-1)^k * q^(k*(k+1)) + 2 * Sum_{n>=1, k>=0} (-1)^k * q^(k*(k+2*n+1)+n). - Mamuka Jibladze, May 17 2021
G.f.: Sum_{k>=0} a(k) * q^k = Sum_{k>=0} (-1)^k * q^(k*(k+1)) * (1 + q^(2*k+1))/(1 - q^(2*k+1)). - Mamuka Jibladze, Jun 06 2021
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Pi/2 (A019669). - Amiram Eldar, Oct 15 2022

More terms and information from Michael Somos, Mar 23 2003

A000466 a(n) = 4*n^2 - 1.

Original entry on oeis.org

-1, 3, 15, 35, 63, 99, 143, 195, 255, 323, 399, 483, 575, 675, 783, 899, 1023, 1155, 1295, 1443, 1599, 1763, 1935, 2115, 2303, 2499, 2703, 2915, 3135, 3363, 3599, 3843, 4095, 4355, 4623, 4899, 5183, 5475, 5775, 6083, 6399, 6723, 7055, 7395

Offset: 0

Chan Siu Kee (skchan5(AT)hkein.ie.cuhk.hk)

Sum_{n>=1} (-1)^n*a(n)/n! = 1 - 1/e = A068996. - Gerald McGarvey, Nov 06 2007
Sequence arises from reading the line from -1, in the direction -1, 15, ... and the same line from 3, in the direction 3, 35, ..., in the square spiral whose nonnegative vertices are the squares A000290. - Omar E. Pol, May 24 2008
a(n) is the product of the consecutive odd integers 2n-1 and 2n+1 (cf. A005408). - Doug Bell, Mar 08 2009
For n>0: a(n) = A176271(2*n,n); cf. A016754, A053755. - Reinhard Zumkeller, Apr 13 2010
a(n+1) gives the curvature c(n) of the n-th circle touching the two equal semicircles of the symmetric arbelos (1/2, 1/2) and the (n-1)-st circle, with input c(0) = 3 =  A059100(1) (referring to the second circle of the Pappus chain), for n >= 0. - Wolfdieter Lang and Kival Ngaokrajang, Jul 03 2015
After 3, a(n) is pseudoprime to base 2n. For example: (2*2)^(a(2)-1) == 1 (mod a(2)), in fact 4^14 = 15*17895697+1. - Bruno Berselli, Sep 24 2015
Numbers m such that m+1 and (m+1)/4 are squares. - Bruno Berselli, Mar 03 2016
After -1, the least common multiple of 2*m+1 and 2*m-1. - Colin Barker, Feb 11 2017
This sequence contains all products of the twin prime pairs (see A037074). - Charles Kusniec, Oct 03 2019

T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.
L. B. W. Jolley, Summation of Series, Dover, 2nd ed., 1961.
Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968), pp. 980-981.
A. Languasco and A. Zaccagnini, Manuale di Crittografia, Ulrico Hoepli Editore (2015), p. 259.

Vincenzo Librandi, Table of n, a(n) for n = 0..900
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, and Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.
Milan Janjic and Boris Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Kival Ngaokrajang, Illustration of the Pappus chain (downwards direction).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A001539, A016286, A016742.
Factor of A160466. Superset of A037074.
Cf. A059100 (curvatures for a Pappus chain).

[4*n^2-1: n in [0..50]]; // Vincenzo Librandi, Apr 26 2011

A000466:=n->4*n^2-1; seq(A000466(n), n=0..100); # Wesley Ivan Hurt, Nov 19 2013

4 Range[0,50]^2-1 (* Harvey P. Dale, Jan 23 2011 *)

makelist(4*n^2-1, n, 0, 50); /* Martin Ettl, Nov 12 2012 */

a(n)=4*n^2-1 \\ Charles R Greathouse IV, Oct 27 2011

[4*n^2-1 for n in (0..50)] # Bruno Berselli, Sep 24 2015

O.g.f.: ( 1-6*x-3*x^2 ) / (x-1)^3 . - R. J. Mathar, Mar 24 2011
E.g.f.: (-1 + 4*x + 4*x^2)*exp(x). - Ilya Gutkovskiy, May 26 2016
Sum_{n>=1} 1/a(n) = 1/2 [Jolley eq. 233]. - Benoit Cloitre, Apr 05 2002
Sum_{n>=1} 2/a(n) = 1 = 2/3 + 2/15 + 2/35 + 2/63 + 2/99 + 2/143, ..., with partial sums: 2/3, 4/5, 6/7, 8/9, 10/11, 12/13, 14/15, ... - Gary W. Adamson, Jun 16 2003
1/3 + Sum_{n>=2} 4/a(n) = 1 = 1/3 + 4/15 + 4/35 + 4/63, ..., with partial sums: 1/3, 3/5, 5/7, 7/9, 9/11, ..., (2n+1)/(2n+3). - Gary W. Adamson, Jun 18 2003
Sum_{n>=0} 2/a(2*n+1) = Pi/4 = 2/3 + 2/35 + 2/99, ... = (1 - 1/3) + (1/5 - 2/7) + (1/9 - 1/11) + ... = Sum_{n>=0} (-1)^n/(2*n+1). - Gary W. Adamson, Jun 22 2003
Product(n>=1, (a(n)+1)/a(n)) = Pi/2 (Wallis formula). - Mohammed Bouayoun (mohammed.bouayoun(AT)sanef.com), Mar 03 2004
a(n)+2 = A053755(n). - Zak Seidov, Jan 16 2007
a(n)^2 + A008586(n)^2 = A053755(n)^2 (Pythagorean triple). - Zak Seidov, Jan 16 2007
a(n) = a(n-1) + 8*n - 4 for n > 0, a(0)=-1. - Vincenzo Librandi, Dec 17 2010
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/4 - 1/2 = (A019669-1)/2. [Jolley eq (366)]. - R. J. Mathar, Mar 24 2011
For n>0, a(n) = 2/(Integral_{x=0..Pi/2} (sin(x))^3*(cos(x))^(2*n-2)). - Francesco Daddi, Aug 02 2011
Nonlinear recurrence for c(n) = a(n+1) (see the arbelos comment above) from Descartes' three circle theorem (see the links under A259555): c(n) = 4 + c(n-1) + 4*sqrt(c(n-1) + 1), with input c(0) = 3  =  A059100(1), for n >= 0. The appropriate solution of this recurrence is c(n-1) + 1 = 4*n^2. - Wolfdieter Lang, Jul 03 2015
a(n) = 3*Pochhammer(5/2,n-1)/Pochhammer(1/2,n-1). Hence, the e.g.f. for a(n+1), i.e., dropping the first term, is 3* 1F1(5/2;1/2;x), with 1F1 being the confluent hypergeometric function (also known as Kummer's). - Stanislav Sykora, May 26 2016
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/sqrt(2))/sqrt(2). - Amiram Eldar, Feb 04 2021

A197476 Decimal expansion of least x>0 having cos(x) = cos(2*x)^2.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling, Oct 15 2011

The Mathematica program includes a graph.  Guide for least x>0 satisfying cos(b*x) = cos(c*x)^2, for selected b and c:
b.....c......x
....2.......A197476
....3.......A197477
....4.......A197478
....1.......A000796, Pi
....3.......A197479
....4.......A197480
....1.......A019669, Pi/2
....2.......A197482
....4.......A197483
....1.......A168229, arctan(sqrt(7))
....2.......A019669, Pi/2
....3.......A019679, Pi/12
....6.......A197485
....8.......A197486
....2.......A003881, Pi/4
....3.......A019670, Pi/3, tangency point
....4.......A197488
....8.......A197489
....4*Pi....A197334
....3*Pi....A197335
....2*Pi....A197490
....3*Pi/2..A197491
....Pi......A197492
....Pi/2....A197493
....Pi/3....A197494
....Pi/4....A197495
....2*Pi/3..A197506
....3*Pi....A197507
....3*Pi/2..A197508
....2*Pi....A197509
....Pi......A197510
....Pi/2....A197511
....Pi/3....A197512
....Pi/4....A197513
....Pi/6....A197514
Pi....1.......A197515
Pi....2.......A197516
Pi....1/2.....A197517
2*Pi..1.......A197518
2*Pi..2.......A197519
2*Pi..3.......A197520
Pi/2..Pi/3....A197521
Pi/2..Pi/6....3
Pi/3..1.......A197582
Pi/3..2.......A197583
Pi/3..1/3.....A197584
See A197133 for a guide for least x>0 satisfying sin(b*x) = sin(c*x)^2 for selected b and c.

			1.137743932905455557789449860055008349584...

Index entries for transcendental numbers.

Cf. A197133.

b = 1; c = 2; f[x_] := Cos[x]
t = x /. FindRoot[f[b*x] == f[c*x]^2, {x, 1.1, 1.3}, WorkingPrecision -> 200]
RealDigits[t] (* A197476 *)
Plot[{f[b*x], f[c*x]^2}, {x, 0, 2}]
(* or *)
RealDigits[ ArcCos[ ((19 - 3*Sqrt[33])^(1/3) + (19 + 3*Sqrt[33])^(1/3) - 2)/6], 10, 99] // First (* Jean-François Alcover, Feb 19 2013 *)

Edited by Georg Fischer, Jul 28 2021

A019670 Decimal expansion of Pi/3.

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane, Dec 11 1996

With an offset of zero, also the decimal expansion of Pi/30 ~ 0.104719... which is the average arithmetic area  of the 0-winding sectors enclosed by a closed Brownian planar path, of a given length t, according to Desbois, p. 1. - Jonathan Vos Post, Jan 23 2011
Polar angle (or apex angle) of the cone that subtends exactly one quarter of the full solid angle. See comments in A238238. - Stanislav Sykora, Jun 07 2014
60 degrees in radians. - M. F. Hasler, Jul 08 2016
Volume of a quarter sphere of radius 1. - Omar E. Pol, Aug 17 2019
Also smallest positive zero of Sum_{k>=1} cos(k*x)/k = -log(2*|sin(x/2)|). Proof of this identity: Sum_{k>=1} cos(k*x)/k = Re(Sum_{k>=1} exp(k*x*i)/k) = Re(-log(1-exp(x*i))) = -log(2*|sin(x/2)|), x != 2*m*Pi, where i = sqrt(-1). - Jianing Song, Nov 09 2019
The area of a circle circumscribing a unit-area regular dodecagon. - Amiram Eldar, Nov 05 2020

			Pi/3 = 1.04719755119659774615421446109316762806572313312503527365831486...
From _Peter Bala_, Nov 16 2016: (Start)
Case n = 1. Pi/3 = 18 * Sum_{k >= 0} (-1)^(k+1)( 1/((6*k - 5)*(6*k + 1)*(6*k + 7)) + 1/((6*k - 1)*(6*k + 5)*(6*k + 11)) ).
Using the methods of Borwein et al. we can find the following asymptotic expansion for the tails of this series: for N divisible by 6 there holds Sum_{k >= N/6} (-1)^(k+1)( 1/((6*k - 5)*(6*k + 1)*(6*k + 7)) + 1/((6*k - 1)*(6*k + 5)*(6*k + 11)) ) ~ 1/N^3 + 6/N^5 + 1671/N ^7 - 241604/N^9 + ..., where the sequence [1, 0, 6, 0, 1671, 0, -241604, 0, ...] is the sequence of coefficients in the expansion of ((1/18)*cosh(2*x)/cosh(3*x)) * sinh(3*x)^2 = x^2/2! + 6*x^4/4! + 1671*x^6/6! - 241604*x^8/8! + .... Cf. A024235, A278080 and A278195. (End)

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 8.3, p. 489.

Ivan Panchenko, Table of n, a(n) for n = 1..1000
Kunle Adegoke, Infinite arctangent sums involving Fibonacci and Lucas numbers, arXiv:1603.08097 [math.NT], 2016.
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Jean Desbois and Stéphane Ouvry, Algebraic and arithmetic area for m planar Brownian paths, arXiv:1101.4135 [math-ph], Jan 21, 2011.
Index entries for transcendental numbers.

Cf. A000796 (Pi), A019669 (Pi/2), A019692 (2*Pi), A122952 (3*Pi), A003881(Pi/4), A137914, A238238, A002388, A091925, A093954, A016910, A019694, A136017, A352324.

Integral_{x=0..oo} 1/(1+x^m) dx: A013661 (m=2), A248897 (m=3), A093954 (m=4), A352324 (m=5), this sequence (m=6), A352125 (m=8), A094888 (m=10).

RealDigits[N[Pi/3,6! ]] (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)

Pi/3 \\ Charles R Greathouse IV, Sep 08 2011

N=150; default(realprecision, 3+N); digits(Pi*10^N\3) \\ M. F. Hasler, Jul 08 2016

A third of A000796, a sixth of A019692, the square root of A100044.
Sum_{k >= 0} (-1)^k/(6k+1) + (-1)^k/(6k+5). - Charles R Greathouse IV, Sep 08 2011
Product_{k >= 1}(1-(6k)^(-2))^(-1). - Fred Daniel Kline, May 30 2013
From Peter Bala, Feb 05 2015: (Start)
Pi/3 = Sum {k >= 0} binomial(2*k,k)*1/(2*k + 1)*(1/16)^k = 2F1(1/2,1/2;3/2;1/4). Similar series expansions hold for Pi^2 (A002388), Pi^3 (A091925) and Pi/(2*sqrt(2)) (A093954.)
The integer sequences A(n) := 4^n*(2*n + 1)! and B(n) := A(n)*( Sum {k = 0..n} binomial(2*k,k)*1/(2*k + 1)*(1/16)^k ) both satisfy the second-order recurrence equation u(n) = (20*n^2 + 4*n + 1)*u(n-1) - 8*(n - 1)*(2*n - 1)^3*u(n-2). From this observation we can obtain the continued fraction expansion Pi/3 = 1 + 1/(24 - 8*3^3/(89 - 8*2*5^3/(193 - 8*3*7^3/(337 - ... - 8*(n - 1)*(2*n - 1)^3/((20*n^2 + 4*n + 1) - ... ))))). Cf. A002388 and A093954. (End)
Equals Sum_{k >= 1} arctan(sqrt(3)*L(2k)/L(4k)) where L=A000032. See also A005248 and A056854. - Michel Marcus, Mar 29 2016
Equals Product_{n >= 1} A016910(n) / A136017(n). - Fred Daniel Kline, Jun 09 2016
Equals Integral_{x=-oo..oo} sech(x)/3 dx. - Ilya Gutkovskiy, Jun 09 2016
From Peter Bala, Nov 16 2016: (Start)
Euler's series transformation applied to the series representation Pi/3 = Sum_{k >= 0} (-1)^k/(6*k + 1) + (-1)^k/(6*k + 5) given above by Greathouse produces the faster converging series Pi/3 = (1/2) * Sum_{n >= 0} 3^n*n!*( 1/(Product_{k = 0..n} (6*k + 1)) + 1/(Product_{k = 0..n} (6*k + 5)) ).
The series given above by Greathouse is the case n = 0 of the more general result Pi/3 = 9^n*(2*n)! * Sum_{k >= 0} (-1)^(k+n)*( 1/(Product_{j = -n..n} (6*k + 1 + 6*j)) + 1/(Product_{j = -n..n} (6*k + 5 + 6*j)) ) for n = 0,1,2,.... Cf. A003881. See the example section for notes on the case n = 1.(End)
Equals Product_{p>=5, p prime} p/sqrt(p^2-1). - Dimitris Valianatos, May 13 2017
Equals A019699/4 or A019693/2. - Omar E. Pol, Aug 17 2019
Equals Integral_{x >= 0} (sin(x)/x)^4 = 1/2 + Sum_{n >= 0} (sin(n)/n)^4, by the Abel-Plana formula. - Peter Bala, Nov 05 2019
Equals Integral_{x=0..oo} 1/(1 + x^6) dx. - Bernard Schott, Mar 12 2022
Pi/3 = -Sum_{n >= 1} i/(n*P(n, 1/sqrt(-3))*P(n-1, 1/sqrt(-3))), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial. The first twenty terms of the series gives the approximation Pi/3 = 1.04719755(06...) correct to 8 decimal places. - Peter Bala, Mar 16 2024
Equals Integral_{x >= 0} (2*x^2 + 1)/((x^2 + 1)*(4*x^2 + 1)) dx. - Peter Bala, Feb 12 2025

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

A093954 Decimal expansion of Pi/(2*sqrt(2)).

Original entry on oeis.org

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

Offset: 1

Eric W. Weisstein, Apr 19 2004

The value is the length Pi*sqrt(2)/4 of the diagonal in the square with side length Pi/4 = Sum_{n>=0} (-1)^n/(2n+1) = A003881. The area of the circumcircle of this square is Pi*(Pi*sqrt(2)/8)^2 = Pi^3/32 = A153071. - Eric Desbiaux, Jan 18 2009
This is the value of the Dirichlet L-function of modulus m=8 at argument s=1 for the non-principal character (1,0,1,0,-1,0,-1,0). See arXiv:1008.2547. - R. J. Mathar, Mar 22 2011
Archimedes's-like scheme: set p(0) = sqrt(2), q(0) = 1; p(n+1) = 2*p(n)*q(n)/(p(n)+q(n)) (harmonic mean, i.e., 1/p(n+1) = (1/p(n) + 1/q(n))/2), q(n+1) = sqrt(p(n+1)*q(n)) (geometric mean, i.e., log(q(n+1)) = (log(p(n+1)) + log(q(n)))/2), for n >= 0. The error of p(n) and q(n) decreases by a factor of approximately 4 each iteration, i.e., approximately 2 bits are gained by each iteration. Set r(n) = (2*q(n) + p(n))/3, the error decreases by a factor of approximately 16 for each iteration, i.e., approximately 4 bits are gained by each iteration. For a similar scheme see also A244644. - A.H.M. Smeets, Jul 12 2018
The area of a circle circumscribing a unit-area regular octagon. - Amiram Eldar, Nov 05 2020

			1.11072073453959156175397...
From _Peter Bala_, Mar 03 2015: (Start)
Asymptotic expansion at n = 5000.
The truncated series Sum_{k = 0..5000 - 1} (-1)^floor(k/2)/(2*k + 1) = 1.110(6)207345(42)591561(18)3970(5238)1.... The bracketed digits show where this decimal expansion differs from that of Pi/(2*sqrt(2)). The numbers 1, -3, 57, -2763 must be added to the bracketed numbers to give the correct decimal expansion to 30 digits: Pi/(2*sqrt(2)) = 1.110(7)207345(39)591561(75)3970 (2475)1.... (End)
From _Peter Bala_, Nov 24 2016: (Start)
Case m = 1, n = 1:
Pi/(2*sqrt(2)) = 4*Sum_{k >= 0} (-1)^(1 + floor(k/2))/((2*k - 1)*(2*k + 1)*(2*k + 3)).
We appear to have the following asymptotic expansion for the tails of this series: for N divisible by 4, Sum_{k >= N/2} (-1)^floor(k/2)/((2*k - 1)*(2*k + 1)*(2*k + 3)) ~ 1/N^3 - 14/N^5 + 691/N^7 - 62684/N^9 - ..., where the coefficient sequence [1, 0, -14, 0, 691, 0, -62684, ...] appears to come from the e.g.f. (1/2!)*cosh(x)/cosh(2*x)*sinh(x)^2 = x^2/2! - 14*x^4/4! + 691*x^6/6! - 62684*x^8/8! + .... Cf. A019670.
For example, take N = 10^5. The truncated series Sum_{k = 0..N/2 -1} (-1)^(1+floor(k/2))/((2*k - 1)*(2*k + 1)*(2*k + 3)) = 0.27768018363489(8)89043849(11)61878(80026)6163(351171)58.... The bracketed digits show where this decimal expansion differs from that of (1/4)*Pi/(2*sqrt(2)). The numbers -1, 14, -691, 62684 must be added to the bracketed numbers to give the correct decimal expansion: (1/4)*Pi/(2*sqrt(2)) = 0.27768018363489(7) 89043849(25)61878(79335)6163(413855)58... (End)

J. M. Arnaudiès, P. Delezoide et H. Fraysse, Exercices résolus d'Analyse du cours de mathématiques - 2, Dunod, 1993, Exercice 5, p. 240.
George Boros and Victor H. Moll, Irresistible integrals, Cambridge University Press, 2006, p. 149.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 1.4.1, p. 20.
L. B. W. Jolley, Summation of Series, Dover, 1961, eq. 76, page 16.
Joel L. Schiff, The Laplace Transform: Theory and Applications, Springer-Verlag New York, Inc. (1999). See p. 149.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 53.

Harry J. Smith, Table of n, a(n) for n = 1..20000
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
R. J. Mathar, Table of Dirichlet L-series and prime zeta modulo functions for small moduli, arXiv:1008.2547 [math.NT], 2010-2015, table 7 and section 2.2, value of L(m=8,r=4,s=1).
Michael Penn, Newton's sum (2023), YouTube video.
Michael I. Shamos, A catalog of the real numbers, (2007). See p. 149.
Eric Weisstein's World of Mathematics, Bifoliate.
Index entries for transcendental numbers.

Cf. A161684 (continued fraction).
Cf. A002388, A019670.
Cf. A000281, A063448, A247719, A193887, A244976, A181048, A181049.
Cf. A003881, A251809, A188510.
Cf. A352324, A248897, A019669.

simplify( sum((cos((1/2)*k*Pi)+sin((1/2)*k*Pi))/(2*k+1), k = 0 .. infinity) );  # Peter Bala, Mar 09 2015

RealDigits[Pi/Sqrt@8, 10, 111][[1]] (* Michael De Vlieger, Sep 23 2016 and slightly modified by Robert G. Wilson v, Jul 23 2018 *)

default(realprecision, 20080); x=Pi*sqrt(2)/4; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b093954.txt", n, " ", d)); \\ Harry J. Smith, Jun 17 2009

Equals 1/A112628.
Equals Integral_{x=0..oo} 1/(x^4+1) dx. - Jean-François Alcover, Apr 29 2013
From Peter Bala, Feb 05 2015: (Start)
Pi/(2*sqrt(2)) = Sum_{k >= 0} binomial(2*k,k)*1/(2*k + 1)*(1/8)^k.
The integer sequences A(n) := 2^n*(2*n + 1)! and B(n) := A(n)*( Sum {k = 0..n} binomial(2*k,k)*1/(2*k + 1)*(1/8)^k ) both satisfy the second order recurrence equation u(n) = (12*n^2 + 1)*u(n-1) - 4*(n - 1)*(2*n - 1)^3*u(n-2). From this observation we can obtain the continued fraction expansion Pi/(2*sqrt(2)) = 1 + 1/(12 - 4*3^3/(49 - 4*2*5^3/(109 - 4*3*7^3/(193 - ... - 4*(n - 1)*(2*n - 1)^3/((12*n^2 + 1) - ... ))))). Cf. A002388 and A019670. (End)
From Peter Bala, Mar 03 2015: (Start)
Pi/(2*sqrt(2)) = Sum_{k >= 0} (-1)^floor(k/2)/(2*k + 1) = limit (n -> infinity) Sum_{k = -n .. n - 1} (-1)^k/(4*k + 1). See Wells.
We conjecture the asymptotic expansion Pi/(2*sqrt(2)) - Sum {k = 0..n - 1} (-1)^floor(k/2)/(2*k + 1) ~ 1/(2*n) - 3/(2*n)^3 + 57/(2*n)^5 - 2763/(2*n)^7 + ..., where n is a multiple of 4 and the sequence of unsigned coefficients [1, 3, 57, 2763, ...] is A000281. An example with n = 5000 is given below. (End)
From Peter Bala, Sep 21 2016: (Start)
c = 2 * Sum_{k >= 0} (-1)^k * (4*k + 2)/((4*k + 1)*(4*k + 3)) = A181048 + A181049. The asymptotic expansion conjectured above follows from the asymptotic expansions given in A181048 and A181049.
c = 1/2 * Integral_{x = 0..Pi/2} sqrt(tan(x)) dx. (End)
From Peter Bala, Nov 24 2016: (Start)
Let m be an odd integer and n a nonnegative integer. Then Pi/(2*sqrt(2)) = 2^n*m^(2*n)*(2*n)!*Sum_{k >= 0} (-1)^(n+floor(k/2)) * 1/Product_{j = -n..n} (2*k + 1 + 2*m*j). Cf. A003881.
In the particular case m = 1 the result has the equivalent form: for n a nonnegative integer, Pi/(2*sqrt(2)) = 2^n*(2*n)!*Sum_{k >= 0} (-1)^(n+k)*(8*k + 4)* 1/Product_{j = -n..n+1} (4*k + 2*j + 1). The case m = 1, n = 1 is considered in the Example section below.
Let m be an odd integer and n a nonnegative integer. Then Pi/(2*sqrt(2)) = 4^n*m^(2*n)*(2*n)!*Sum_{k >= 0} (-1)^(n+floor(k/2)) * 1/Product_{j = -n..n} (2*k + 1 + 4*m*j). (End)
Equals Integral_{x = 0..oo} cosh(x)/cosh(2*x) dx. - Peter Bala, Nov 01 2019
Equals Sum_{k>=1} A188510(k)/k = Sum_{k>=1} Kronecker(-8,k)/k = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 - 1/13 - 1/15 + ... - Jianing Song, Nov 16 2019
From Amiram Eldar, Jul 16 2020: (Start)
Equals Product_{k>=1} (1 - (-1)^k/(2*k+1)).
Equals Integral_{x=0..oo} dx/(x^2 + 2).
Equals Integral_{x=0..Pi/2} dx/(sin(x)^2 + 1). (End)
Equals Integral_{x=0..oo} x^2/(x^4 + 1) dx (Arnaudiès). - Bernard Schott, May 19 2022
Equals Integral_{x = 0..1} 1/(2*x^2 + (1 - x)^2) dx. - Peter Bala, Jul 22 2022
Equals Integral_{x = 0..1} 1/(1 - x^4)^(1/4) dx. - Terry D. Grant, Mar 17 2023
Equals 1/Product_{p prime} (1 - Kronecker(-8,p)/p), where Kronecker(-8,p) = 0 if p = 2, 1 if p == 1 or 3 (mod 8) or -1 if p == 5 or 7 (mod 8). - Amiram Eldar, Dec 17 2023
Equals A068465*A068467. - R. J. Mathar, Jun 27 2024
From Stefano Spezia, Jun 05 2025: (Start)
Equals Sum_{k>=1} (-1)^(k+1)(1/(4*k - 3) + 1/(4*k - 1)).
Equals Product_{k=0..oo} (1 + (-1)^k/(2*k + 3)).
Equals Integral_{x=0..oo} 1/(2*x^2 + 1).
Equals Integral_{x=0..1} 1/((1 + x^2)*sqrt(1 - x^2)). (End)

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A387194 Decimal expansion of J_0(Pi/2), Bessel Function of the first kind of index 0 at A019669.

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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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A003881 Decimal expansion of Pi/4.

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A008441 Number of ways of writing n as the sum of 2 triangular numbers.

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A000466 a(n) = 4*n^2 - 1.

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