cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

1, 1, 5, 61, 1385, 50521, 2702765, 199360981, 19391512145, 2404879675441, 370371188237525, 69348874393137901, 15514534163557086905, 4087072509293123892361, 1252259641403629865468285, 441543893249023104553682821, 177519391579539289436664789665
Offset: 0

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Comments

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

Examples

			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)
		

References

  • 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; gives a version with signs: E_{2n} = (-1)^n*a(n) (this is A028296).
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
  • J. M. Borwein and D. M. Bailey, Mathematics by Experiment, Peters, Boston, 2004; p. 49
  • J. M. Borwein, D. H. Bailey, and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 141.
  • Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 420.
  • G. Chrystal, Algebra, Vol. II, p. 342.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 49.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 110.
  • H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 69.
  • L. Euler, Inst. Calc. Diff., Section 224.
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 444.
  • L. Seidel, Über eine einfache Entstehungsweise der Bernoulli'schen Zahlen und einiger verwandten Reihen, Sitzungsberichte der mathematisch-physikalischen Classe der königlich bayerischen Akademie der Wissenschaften zu München, volume 7 (1877), 157-187.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapters 5 and 33, pages 41, 314.
  • J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 269.

Crossrefs

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).

Programs

  • Maple
    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
  • Mathematica
    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 *)
  • Maxima
    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 */
    
  • Maxima
    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 */
    
  • PARI
    {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
    
  • PARI
    {a(n) = if( n<0, 0, (2*n)! * polcoeff( 1 / cos(x + O(x^(2*n + 1))), 2*n))}; /* Michael Somos, Jun 18 2002 */
    
  • PARI
    {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 */
    
  • PARI
    {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
    
  • PARI
    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
    
  • PARI
    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
    
  • PARI
    a(n)=abs(eulerfrac(2*n)) \\ Charles R Greathouse IV, Mar 23 2022
    
  • PARI
    \\ 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
    
  • Python
    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
    
  • Python
    # 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
  • Sage
    # 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
    

Formula

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

Extensions

Typo in name corrected by Anders Claesson, Dec 01 2015

A002436 E.g.f.: Sum_{n >= 0} a(n)*x^(2*n)/(2*n)! = sec(2*x).

Original entry on oeis.org

1, 4, 80, 3904, 354560, 51733504, 11070525440, 3266330312704, 1270842139934720, 630424777638805504, 388362339077351014400, 290870261262635870715904, 260290690801376575335956480, 274278793184290987427604987904, 336150887870579862992197737512960
Offset: 0

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Author

Keywords

Examples

			G.f. = 1 + 4*x + 80*x^2 + 3904*x^3 + 354560*x^4 + 51733504*x^5 + 11070525440*x^6 + ...
		

References

  • 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. 75.
  • J. W. L. Glaisher, On the last two figures in certain coefficients analogous to the Eulerian numbers, Quart. J. Pure Appl. Math., 44 (1913), 105-112.
  • 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).

Crossrefs

(-1)^n*a(n) give the alternating row sums of A060187(2*n), n >= 0. The alternating sums for odd numbered rows vanish. - Wolfdieter Lang, Jul 12 2017

Programs

  • Magma
    m:=35; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!( (1+Tan(x))/(1-Tan(x)) )); [Factorial(n-1)*b[n]: n in [1..m by 2]]; // Vincenzo Librandi, May 30 2019
  • Maple
    A := n -> (-4)^n*euler(2*n); # (Then A(n) = a(n+1) for n >= 0.) # Peter Luschny, Jan 27 2009
  • Mathematica
    Rest@ Union[ Range[0, 24]! CoefficientList[ Series[ Sec[ 2x], {x, 0, 24}], x]] (* Robert G. Wilson v, Apr 16 2011 *)
    a[ n_] :=  If[ n < 0, 0, 2 (-16)^n LerchPhi[ -1, -2 n, 1/2]]; (* Michael Somos, Oct 14 2014 *)
    With[{nn=30},Take[CoefficientList[Series[Sec[2x],{x,0,nn}],x] Range[0,nn]!,{1,-1,2}]] (* Harvey P. Dale, May 06 2018 *)
  • PARI
    {a(n) = local(m); if( n<0, 0, m = 2*n; m! * polcoeff( 1 / cos( 2*x + x * O(x^m)), m))}; /* Michael Somos, Apr 16 2011 */
    
  • Sage
    @CachedFunction
    def sp(n,x) :
        if n == 0 : return 1
        return -add(2^(n-k)*sp(k,1/2)*binomial(n,k) for k in range(n)[::2])
    A002436 = lambda n : abs(sp(2*(n-1),x))
    [A002436(n) for n in (1..15)]   # Peter Luschny, Jul 30 2012
    

Formula

a(n) = A000831(2*n) = 4^n * A000364(n). a(n) = 2 * A000816(n) except n=0. - Michael Somos, Apr 26 2011
E.g.f.: sec(2*x) = 1 + 2*(x^2)/G(0); G(k) = (k+1)*(2*k+1) - 2*(x^2) + (x^2)*(2*k+1)*(2*k+2)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 01 2011
E.g.f.: sec(2*x) = 1/cos(2*x) = 1/(cos(x)^2 - sin(x)^2). - Arkadiusz Wesolowski, Jul 25 2012
From Sergei N. Gladkovskii, Oct 23 2012 (Start)
G.f.: 1/U(0) where U(k) = 1 - 2*(4*k+1)*(4*k+2)*x/(1 - 2*(4*k+3)*(4*k+4)*x/U(k+1)); (continued fraction, 2-step).
E.g.f.: 1/S(0) where S(k) = 1 - 2*x^2/((4*k+1)*(2*k+1) - x^2*(4*k+1)*(2*k+1)/(x^2 - (4*k+3)*(k+1)/S(k+1))); (continued fraction, 3rd kind, 3-step). (End)
G.f.: 1/U(0) where U(k) = 1 - (4*k+2)*(4*k+2)*x^2/(1 - (4*k+4)*(4*k+4)*x^2/U(k+1)); (continued fraction, 2-step). - Sergei N. Gladkovskii, Nov 06 2012
G.f.: 1/G(0) where G(k) = 1 - 4*x*(k+1)^2/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 12 2013
a(n+1) = | 2*16^n*lerchphi(-1, -2*n, 1/2) |, n>=0. - Peter Luschny, Apr 27 2013
G.f.: Q(0), where Q(k) = 1 - x*(2*k+2)^2/( x*(2*k+2)^2 - 1/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 10 2013
E.g.f.: sec(2*x) = 1/cos(2*x) = 1 + 2*x^2/(1-2*x^2)*T(0), where T(k) = 1 - x^2*(2*k+1)*(2*k+2)/( x^2*(2*k+1)*(2*k+2) + ((k+1)*(2*k+1) - 2*x^2)*((k+2)*(2*k+3) - 2*x^2)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 25 2013
a(n) = (-1)^n*2^(6*n+1)*(Zeta(-2*n,1/4) - Zeta(-2*n, 3/4)), where Zeta(a, z) is the generalized Riemann zeta function. - Peter Luschny, Mar 11 2015

Extensions

More terms from Michael Somos, Jun 21 2002

A326483 a(n) = 2^n*E2_{n}(1/2) with E2_{n} the polynomials defined in A326480.

Original entry on oeis.org

1, -2, -4, 40, 80, -1952, -3904, 177280, 354560, -25866752, -51733504, 5535262720, 11070525440, -1633165156352, -3266330312704, 635421069967360, 1270842139934720, -315212388819402752, -630424777638805504, 194181169538675507200
Offset: 0

Views

Author

Peter Luschny, Jul 12 2019

Keywords

Comments

For comments see A326480.

Crossrefs

Bisections (up to signs): A002436 (even), A000816 (odd).

Programs

  • Maple
    # The function E2(n) is defined in A326480.
    seq(subs(x=1/2, 2^n*E2(n)), n=0..22);

Formula

From Emanuele Munarini, Aug 22 2022: (Start)
E.g.f. for the sequence of the absolute values: (1+tan(2*t))/cos(2*t).
|a(2*n)| = 2^(2*n) |E(2*n)|.
|a(2*n+1)| = 2^(2*n+1) Sum_{k=0..n} binomial(2*n+1,2*k) |E(2*k)| T(n-k+1), where the E(n) are the Euler numbers (A122045) and the T(n) are the tangent numbers (A000182). (End)

A000822 Expansion of (sin^2 x + sin x) /cos 2x.

Original entry on oeis.org

0, 1, 2, 11, 40, 361, 1952, 24611, 177280, 2873041, 25866752, 512343611, 5535262720, 129570724921, 1633165156352, 44110959165011, 635421069967360, 19450718635716001, 315212388819402752, 10784052561125704811
Offset: 0

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Author

Keywords

Crossrefs

Bisections are A000464 and A000816.

A173226 Partial sums of A000364.

Original entry on oeis.org

1, 2, 7, 68, 1453, 51974, 2754739, 202115720, 19593627865, 2424473303306, 372795661540831, 69721670054678732, 15584255833611765637, 4102656765126735657998, 1256362298168756601126283, 442800255547191861154809104, 177962191835086481297819598769
Offset: 0

Views

Author

Jonathan Vos Post, Feb 13 2010

Keywords

Comments

Partial sums of Euler numbers. Partial sums of secant or "Zig" numbers. The subsequence of prime partial sum of Euler numbers begins 2, 7, 1453, no more through a(17). What is the next such prime?

Crossrefs

Programs

  • Mathematica
    Accumulate[Table[Abs[EulerE[2n]],{n,0,20}]] (* Harvey P. Dale, Aug 10 2024 *)
  • Python
    from sympy import euler
    def A173226(n): return sum(abs(euler(i)) for i in range(0,(n<<1)+1,2)) # Chai Wah Wu, Apr 16 2023

Formula

a(n) = Sum_{i=0..n} A000364(i).
G.f.: 1/U(0)/(1-x) where U(k)=1 + x - x*(2*k+1)*(2*k+2)/(1 - x*(2*k+1)*(2*k+2)/U(k+1)) ; (continued fraction, 2-step). - Sergei N. Gladkovskii, Oct 15 2012
G.f.: 1/(1-x)/Q(0), where Q(k)= 1 - x*(2*k+1)^2/(1 - x*(2*k+2)^2/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 27 2013
G.f.: Q(0)/(1-x), where Q(k) = 1 - x*(k+1)^2/( x*(k+1)^2 - 1/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 22 2013

A000819 E.g.f.: cos(x)^2 / cos(2x) = Sum_{n >= 0} a(n) * x^(2n) / (2n)!.

Original entry on oeis.org

1, 2, 40, 1952, 177280, 25866752, 5535262720, 1633165156352, 635421069967360, 315212388819402752, 194181169538675507200, 145435130631317935357952, 130145345400688287667978240, 137139396592145493713802493952
Offset: 0

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Author

Keywords

Examples

			G.f. = 1 + 2*x + 40*x^2 + 1952*x^3 + 177280*x^4 + 25866752*x^5 + ... - _Michael Somos_, Apr 04 2017
		

Crossrefs

Essentially the same as A000816.
Second column of array A103905.

Programs

  • Mathematica
    With[{nn=30},Take[CoefficientList[Series[Cos[x]^2/Cos[2x],{x,0,nn}],x] Range[ 0,nn]!,{1,-1,2}]] (* Harvey P. Dale, Jul 06 2014 *)
    a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, m! SeriesCoefficient[ 1 / (1 - Tan[x]^2), {x, 0, m}]]]; (* Michael Somos, Apr 04 2017 *)
  • PARI
    {a(n) = my(m); if( n<0, 0, m = 2*n; m! * polcoeff( 1 / (1 - tan(x + x * O(x^m))^2), m))}; /* Michael Somos, Apr 04 2017 */

Formula

E.g.f.: cos(x)^2/cos(2x)=1/Q(0)+1/2; Q(k)=1+1/(1-2*(x^2)/(2*(x^2)-(k+1)*(2k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 18 2011
From Michael Somos, Apr 04 2017: (Start)
E.g.f.: cos(x)^2 / cos(2*x) = (1 + sec(2*x)) / 2 = tan(2*x) / (2 * tan(x)) = 1 / (1 - tan(x)^2).
a(n) = A000816(n) unless n=0.
a(n) = 1/2 * A002436(n) unless n=0.
a(n) = 2^(2*n - 1) * A000364(n). (End)

A352602 a(n) = 4^n*(2^(2*n+1)-1)*(2*n)!.

Original entry on oeis.org

1, 56, 11904, 5852160, 5274501120, 7606429286400, 16070664624537600, 46802060374022553600, 179724025424120905728000, 879933863508054097526784000, 5350005543376937290448240640000, 39547255119844566012586402775040000, 349281388446657765223160470894018560000
Offset: 0

Views

Author

Artur Jasinski, Mar 22 2022

Keywords

Comments

For n>0, PolyGamma(2*n,1/4) = -a(n)*Zeta(2*n+1) - A000816(n)*Pi^(2n+1) = -2^(2*n-1)*(A331839(n)*Zeta(2*n+1) + A000364(n)*Pi^(2n+1)).

Examples

			PolyGamma(2,1/4) = -56*zeta(3) - 2*Pi^3
PolyGamma(4,1/4) = -11904*zeta(5) - 40*Pi^5
PolyGamma(6,1/4) = -5852160*zeta(7) - 1952*Pi^7
		

Crossrefs

Programs

  • Maple
    A352602 := proc(n)
        4^n*(2^(2*n+1)-1)*(2*n)! ;
    end proc:
    seq(A352602(n),n=0..30) ; # R. J. Mathar, Aug 19 2022
  • Mathematica
    Table[4^n*(2^(2*n + 1) - 1)*(2*n)!, {n, 0, 12}]
  • PARI
    a(n) = n<<=1; my(f=n!<Kevin Ryde, Mar 23 2022

Formula

a(n) = (-Pi^(2*n+1)*A000816(n) - PolyGamma(2*n,1/4))/zeta(2*n+1).
a(n) = 2^(2*n-1)*A331839(n).
D-finite with recurrence a(n) -40*n*(2*n-1)*a(n-1) +256*n*(n-1)*(2*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Aug 19 2022
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