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

A255882 Expansion of exp( Sum_{n >= 1} A210657(n)*(-x)^n/n ).

Original entry on oeis.org

1, 2, 13, 224, 8170, 522716, 51749722, 7309866728, 1394040714169, 344865267322010, 107361980072755261, 41067497940750566312, 18931745446455458282248, 10350955324610065848650384, 6622526747212249020075069880, 4901565185965701578921602882976
Offset: 0

Views

Author

Peter Bala, Mar 09 2015

Keywords

Comments

A210657(n) = 3^(2*n)*E(2*n,1/3), where E(n,x) is the n-th Euler polynomial. In general it appears that when is k a nonzero integer, the expansion of exp( Sum_{n >= 1} k^(2*n)*E(2*n,1/k)*(-x)^n/n ) has (positive) integer coefficients. See A255881 (k = 2), A255883(k = 4) and A255884 (k = 6).

Crossrefs

Programs

  • Maple
    #A255882
    k := 3:
    exp(add(k^(2*n)*euler(2*n, 1/k)*(-x)^n/n, n = 1 .. 15)): seq(coeftayl(%, x = 0, n), n = 0 .. 15);
  • Mathematica
    A210657[n_]:= 9^n EulerE[2 n, 1/3]; a:= With[{nmax = 80}, CoefficientList[Series[Exp[Sum[A210657[k]*(-x)^(k)/(k), {k, 1, 75}]], {x, 0, nmax}], x]]; Table[a[[n]], {n, 1, 51}] (* G. C. Greubel, Aug 26 2018 *)

Formula

O.g.f.: exp( 2*x + 22*x^2/2 + 602*x^3/3 + 30742*x^4/4 + ... ) = 1 + 2*x + 13*x^2 + 224*x^3 + 8170*x^4 + ....
a(0) = 1 and for n >= 1, n*a(n) = Sum_{k = 1..n} (-1)^k*3^(2*k)*E(2*k,1/3)*a(n-k).
a(n) ~ 2^(2*n + 2) * 3^(2*n + 1/2) * n^(2*n - 1/2) / (exp(2*n) * Pi^(2*n + 1/2)). - Vaclav Kotesovec, Jun 08 2019

A255883 Expansion of exp( Sum_{n >= 1} A000281(n)*x^n/n ).

Original entry on oeis.org

1, 3, 33, 1011, 65985, 7536099, 1329205857, 334169853267, 113370124235649, 49880529542872515, 27614111852126579361, 18782012442066306225843, 15394836674855296870428993, 14965462261283347594195897251, 17023467576167762236198869304545, 22400927665017118737825435362462739
Offset: 0

Views

Author

Peter Bala, Mar 09 2015

Keywords

Comments

A000281(n) =(-1)^n*4^(2*n)*E(2*n,1/4), where E(n,x) denotes the n-th Euler polynomial. In general it appears that when k is a nonzero integer, the expansion of exp( Sum_{n >= 1} k^(2*n)*E(2*n,1/k)*(-x)^n/n ) has (positive) integer coefficients. See A255881 (k = 2), A255882(k = 3) and A255884 (k = 6).

Crossrefs

Programs

  • Maple
    #A255883
    k := 4:
    exp(add(k^(2*n)*euler(2*n, 1/k)*(-x)^n/n, n = 1 .. 15)): seq(coeftayl(%, x = 0, n), n = 0 .. 15);
  • Mathematica
    A000281:= With[{nn = 200}, Take[CoefficientList[Series[Cos[x]/Cos[2 x], {x, 0, nn}], x] Range[0, nn]!, {1, -1, 2}]]; a:= With[{nmax = 80}, CoefficientList[Series[Exp[Sum[A000281[[k + 1]]*x^(k)/(k), {k, 1, 85}]], {x, 0, nmax}], x]]; Table[a[[n]], {n, 1, 50}]  (* G. C. Greubel, Aug 26 2018 *)

Formula

O.g.f.: exp( 3*x + 57*x^2/2 + 2763*x^3/3 + 250737*x^4/4 + ... ) = 1 + 3*x + 33*x^2 + 1011*x^3 + 65985*x^4 + ....
a(0) = 1 and for n >= 1, n*a(n) = Sum_{k = 1..n} (-1)^k*4^(2*k)*E(2*k,1/4)*a(n-k).
a(n) ~ 2^(6*n + 5/2) * n^(2*n - 1/2) / (exp(2*n) * Pi^(2*n + 1/2)). - Vaclav Kotesovec, Jun 08 2019

A255884 Expansion of exp( Sum_{n >= 1} A002438(n)*x^n/n ).

Original entry on oeis.org

1, 5, 115, 7955, 1179715, 304888655, 121350927565, 68751844662605, 52528700295424915, 52031089992310711055, 64835758857480094584265, 99249388572274155967996505, 183075972804988649078529524365, 400493686169423616676960341062705, 1025151296160300228944197705742007715
Offset: 0

Views

Author

Peter Bala, Mar 09 2015

Keywords

Comments

A002438(n+1) =(-1)^n*6^(2*n)*E(2*n,1/6), where E(n,x) denotes the n-th Euler polynomial. In general it appears that when k is a nonzero integer, the expansion of exp( Sum_{n >= 1} k^(2*n)*E(2*n,1/k)*(-x)^n/n ) has (positive) integer coefficients. See A255881 (k = 2), A255882(k = 3) and A255883 (k = 4).

Crossrefs

Programs

  • Maple
    #A255884
    k := 6:
    exp(add(k^(2*n)*euler(2*n, 1/k)*(-x)^n/n, n = 1 .. 14)): seq(coeftayl(%, x = 0, n), n = 0 .. 14);
  • Mathematica
    A000243[n_]:= (1 + 9^(n - 1))*Abs[EulerE[2*(n - 1)]]/2; a:= With[{nmax = 75}, CoefficientList[Series[Exp[Sum[A000243[k + 1]*x^(k)/(k), {k, 1, 85}]], {x, 0, nmax}], x]]; Table[a[[n]], {n, 1, 50}] (* G. C. Greubel, Aug 26 2018 *)

Formula

O.g.f.: exp( 5*x + 205*x^2/2 + 22265*x^3/3 + 4544185 *x^4/4 + ... ) = 1 + 5*x + 115*x^2 + 7955*x^3 + 1179715*x^4 + ....
a(0) = 1 and for n >= 1, n*a(n) = Sum_{k = 1..n} (-1)^k*6^(2*k)*E(2*k,1/6)*a(n-k).
a(n) ~ 2^(4*n + 2) * 3^(2*n) * n^(2*n - 1/2) / (exp(2*n) * Pi^(2*n + 1/2)). - Vaclav Kotesovec, Jun 08 2019

A255926 Expansion of exp( Sum_{n >= 1} A210676(n)*x^n/n ).

Original entry on oeis.org

1, -3, 30, -802, 45414, -4508190, 692197470, -151610017950, 44827810930305, -17193060505570335, 8298004578522898140, -4920774627129981351120, 3516683319021255757053900, -2980761698101283167670391780, 2956463734237276273792194346560, -3392220222832838757465019626175680
Offset: 0

Views

Author

Peter Bala, Mar 11 2015

Keywords

Comments

It appears that this sequence is integer valued.
The o.g.f. A(x) = 1 - 3*x + 30*x^2 - 802*x^3 + ... for this sequence is such that 1 + x*d/dx( log(A(x)) ) is the o.g.f. for A210676.
This sequence is the particular case m = -3 of the following general conjecture.
Let m be an integer and consider the sequence u(n) defined by the recurrence u(n) = m*Sum_{k = 0..n-1} binomial(2*n,2*k)*u(k) with the initial condition u(0) = 1. Then the expansion of exp( Sum_{n >= 1} u(n)*x^n/n ) has integer coefficients.
For cases see A255882(m = -2), A255881(m = -1), A255928(m = 1), A255929(m = 2) and A255930(m = 3).
Note that u(n), as a polynomial in the variable m, is the n-th row polynomial of A241171.

Crossrefs

Cf. A210676, A241171, A255882(m = -2), A255881(m = -1), A255928(m = 1), A255929(m = 2), A255930(m = 3).

Programs

  • Maple
    A210676 := proc (n) option remember; if n = 0 then 1 else -3*add(binomial(2*n, 2*k)*A210676(k), k = 0 .. n-1) end if; end proc:
    A255926 := proc (n) option remember; if n = 0 then 1 else add(A210676(n-k)*A255926(k), k = 0 .. n-1)/n end if; end proc:
    seq(A255926(n), n = 0 .. 16);

Formula

O.g.f.: exp(-3*x + 51*x^2/2 - 2163*x^3/3 + 171231*x^4/4 + ...) = 1 - 3*x + 30*x^2 - 802*x^3 + 45414*x^4 - ....
a(0) = 1 and a(n) = (1/n)*Sum_{k = 0..n-1} A210676(n-k)*a(k) for n >= 1.

A255928 Expansion of exp( Sum_{n >= 1} A094088(n)*x^n/n ).

Original entry on oeis.org

1, 1, 4, 44, 1025, 41693, 2617128, 234091692, 28251572652, 4421489003700, 870650503128708, 210629395976568828, 61405707768736724472, 21231253444779700476672, 8589776776743377081599500, 4020181599664131540547091076, 2155088041310451318611119556661
Offset: 0

Views

Author

Peter Bala, Mar 11 2015

Keywords

Comments

It appears that this sequence is integer valued.
The o.g.f. A(x) = 1 + x + 4*x^2 + 44*x^3 + ... for this sequence is such that 1 + x*d/dx( log(A(x)) ) is the o.g.f. for A094088.
This sequence is the particular case m = 1 of the following general conjecture.
Let m be an integer and consider the sequence u(n) defined by the recurrence u(n) = m*Sum_{k = 0..n-1} binomial(2*n,2*k)*u(k) with the initial condition u(0) = 1. Then the expansion of exp( Sum_{n >= 1} u(n)*x^n/n ) has integer coefficients.
For cases see A255926(m = -3), A255882(m = -2), A255881(m = -1), A255929(m = 2) and A255930(m = 3).
Note that u(n), as a polynomial in the variable m, is the n-th row generating polynomial of A241171.

Crossrefs

Cf. A094088, A241171, A255926(m = -3), A255882(m = -2), A255881(m = -1), A255929(m = 2), A255930(m = 3).

Programs

  • Maple
    A094088 := proc (n) option remember; if n = 0 then 1 else add(binomial(2*n, 2*k)*A094088(k), k = 0 .. n-1) end if; end proc:
    A255928 := proc (n) option remember; if n = 0 then 1 else add(A094088(n-k)*A255928(k), k = 0 .. n-1)/n end if; end proc:
    seq(A255928(n), n = 0 .. 16);

Formula

O.g.f.: exp(x + 7*x^2/2 + 121*x^3/3 + 3907*x^4/4 + ...) = 1 + x + 4*x^2 + 44*x^3 + 1025*x^4 + ....
a(0) = 1 and a(n) = (1/n)*Sum_{k = 0..n-1} A094088(n-k)*a(k) for n >= 1.

A255929 Expansion of exp( Sum_{n >= 1} A210672(n)*x^n/n ).

Original entry on oeis.org

1, 2, 15, 308, 13399, 1019106, 119698377, 20039968920, 4527610159068, 1326616296092984, 489092182592254708, 221537815033845709776, 120928125204565597029220, 78286897353506845258973144, 59305342759674536454338570652, 51970719684035315747385128783808
Offset: 0

Views

Author

Peter Bala, Mar 11 2015

Keywords

Comments

It appears that this sequence is integer valued.
The o.g.f. A(x) = 1 + 2*x + 15*x^2 + 308*x^3 + ... for this sequence is such that 1 + x*d/dx( log(A(x)) ) is the o.g.f. for A210672.
This sequence is the particular case m = 2 of the following general conjecture.
Let m be an integer and consider the sequence u(n) defined by the recurrence u(n) = m*Sum_{k = 0..n-1} binomial(2*n,2*k)*u(k) with the initial condition u(0) = 1. Then the expansion of exp( Sum_{n >= 1} u(n)*x^n/n ) has integer coefficients.
For cases see A255926(m = -3), A255882(m = -2), A255881(m = -1), A255928(m = 1) and A255930(m = 3).
Note that u(n), as a polynomial in the variable m, is the n-th row generating polynomial of A241171.

Crossrefs

A210672, A241171, A255926(m = -3), A255882(m = -2), A255881(m = -1), A255928(m = 1), A255930(m = 3).

Programs

  • Maple
    #A255929
    A210672 := proc (n) option remember; if n = 0 then 1 else 2*add(binomial(2*n, 2*k)*A210672(k), k = 0 .. n-1) end if; end proc:
    A255929 := proc (n) option remember; if n = 0 then 1 else add(A210672(n-k)*A255929(k), k = 0 .. n-1)/n end if; end proc:
    seq(A255929(n), n = 0 .. 15);

Formula

O.g.f.: exp(2*x + 26*x^2/2 + 842*x^3/3 + 50906*x^4/4 + ...) = 1 + 2*x + 15*x^2 + 308*x^3 + 13399*x^4 + ....
a(0) = 1 and a(n) = 1/n*Sum_{k = 0..n-1} A210672(n-k)*a(k) for n >= 1.

A255930 Expansion of exp( Sum_{n >= 1} A210674(n)*x^n/n ).

Original entry on oeis.org

1, 3, 33, 991, 63060, 7018860, 1206748720, 295775068680, 97835325011235, 41970842737399345, 22655642596496388759, 15025240474194493147857, 12008582230377080862401692, 11382727559611560650861409564, 12625404970864692720119281536900, 16199644066580777034289339157904220
Offset: 0

Views

Author

Peter Bala, Mar 11 2015

Keywords

Comments

It appears that this sequence is integer valued.
The o.g.f. A(x) = 1 + 3*x + 33*x^2 + 991*x^3 + ... for this sequence is such that 1 + x*d/dx( log(A(x)) ) is the o.g.f. for A210674.
This sequence is the particular case m = 3 of the following general conjecture.
Let m be an integer and consider the sequence u(n) defined by the recurrence u(n) = m*Sum_{k = 0..n-1} binomial(2*n,2*k)*u(k) with the initial condition u(0) = 1. Then the expansion of exp( Sum_{n >= 1} u(n)*x^n/n ) has integer coefficients.
For cases see A255926(m = -3), A255882(m = -2), A255881(m = -1), A255928 (m = 1) and A255929(m = 2).
Note that u(n), as a polynomial in the variable m, is the n-th row generating polynomial of A241171.

Crossrefs

Cf. A210674, A241171, A255926(m = -3), A255882(m = -2), A255881(m = -1), A255928(m = 1), A255929(m = 2).

Programs

  • Maple
    #A255930
    A210674 := proc (n) option remember; if n = 0 then 1 else 3*add(binomial(2*n, 2*k)*A210674(k), k = 0 .. n-1) end if; end proc:
    A255930 := proc (n) option remember; if n = 0 then 1 else add(A210674(n-k)*A255930(k), k = 0 .. n-1)/n end if; end proc:
    seq(A255930(n), n = 0 .. 15);

Formula

O.g.f.: exp(3*x + 57*x^2/2 + 2703*x^3/3 + 239277*x^4/4 + ...) = 1 + 3*x + 33*x^2 + 991*x^3 + 63060*x^4 + ....
a(0) = 1 and a(n) = 1/n*Sum_{k = 0..n-1} A210674(n-k)*a(k) for n >= 1.

A280442 Numerators of coefficients in the Taylor series expansion of Sum_{n>=0} exp((-1)^n*euler(2*n)*x^n/(2*n)).

Original entry on oeis.org

1, 1, 11, 173, 22931, 1319183, 233526463, 29412432709, 39959591850371, 8797116290975003, 4872532317019728133, 1657631603843299234219, 2718086236621937756966743, 1321397724505770800453750299, 1503342018433974345747514544039
Offset: 0

Views

Author

Johannes W. Meijer and Joseph Abate, Jan 03 2017

Keywords

Comments

This sequence is related in a peculiar way to A223067, a sequence related to the period T of a simple gravity pendulum for arbitrary amplitudes. See A280443 for more information.

Crossrefs

Cf. A046161 (denominators).
Cf. A000364 (Euler numbers), A223067, A255881, A280443.

Programs

  • Maple
    nmax:=14: f := series(exp(add((-1)^n*euler(2*n) * x^n/(2*n), n=1..nmax+1)), x=0, nmax+1): for n from 0 to nmax do a(n) := numer(coeff(f, x, n)) od: seq(a(n), n=0..nmax);
  • Sage
    def A280442_list(prec):
        P. = PowerSeriesRing(QQ, default_prec=2*prec)
        def g(x): return exp(sum((-1)^k*euler_number(2*k)*x^k/(2*k) for k in (1..prec+1)))
        R = P(g(x)).coefficients()
        d = lambda n: 2^(2*n - sum(n.digits(2)))
        return [d(n)*R[n] for n in (0..prec)]
    print(A280442_list(14)) # Peter Luschny, Jan 05 2017

Formula

a(n) = numerators of coefficients in the Taylor series expansion of Sum_{n>=0} exp((-1)^n * euler(2*n)*x^n/(2*n)).
Let S = Sum_{n>=0} (-1)^n*euler(2*n)*x^n/(2*n) and w(n) = A005187(n) then a(n) = 2^w(n) * [x^n] exp(S). - Peter Luschny, Jan 05 2017

A262145 O.g.f.: exp( Sum_{n >= 1} A000182(n+1)*x^n/n ), where A000182 is the sequence of tangent numbers.

Original entry on oeis.org

1, 2, 10, 108, 2214, 75708, 3895236, 280356120, 26824493574, 3287849716332, 501916845156012, 93337607623037544, 20766799390944491100, 5446109742113077482456, 1662395457873577922274888
Offset: 0

Views

Author

Peter Bala, Sep 13 2015

Keywords

Comments

It appears that the sequence has integer entries. Calculation suggests the following conjecture: the expansion of exp( Sum_{n >= 1} A000182(n + m)*x^n/n ) has integer coefficients for m = 1, 2, 3, .... This is the case m = 1. Cf. A255881 and A255895.
First row of square array A262144.

Crossrefs

Cf. A000182, A255881, A255895, A262144 (first row).

Programs

  • Maple
    #A262145
    #define tangent numbers A000182
    A000182 := n -> (1/2) * 2^(2*n) * (2^(2*n) - 1) * abs(bernoulli(2*n))/n:
    a := proc (n) option remember;
    if n = 0 then 1 else
      add(A000182(k+1)*a(n-k), k = 1 .. n)/n
    end if;
    end proc:
    seq(a(n), n = 0 .. 15);
  • Mathematica
    max = 15; CoefficientList[E^Sum[(-1)^n*2^(2*n+1)*(4^(n+1)-1)*BernoulliB[2*(n+1)]*x^n / (n*(n+1)), {n, 1, max}] + O[x]^max, x] (* Jean-François Alcover, Sep 18 2015 *)
  • Sage
    def a_list(n):
        T = [0]*(n+2); T[1] = 1
        for k in range(2, n+1): T[k] = (k-1)*T[k-1]
        for k in range(2, n+1):
            for j in range(k, n+1): T[j] = (j-k)*T[j-1]+(j-k+2)*T[j]
        @cached_function
        def a(n): return sum(T[k+1]*a(n-k) for k in (1..n))//n if n> 0 else 1
        return [a(k) for k in range(n)]
    a_list(15) # Peter Luschny, Sep 18 2015

Formula

Recurrence: a(n) = 1/n * Sum_{k = 1..n} A000182(k+1)*a(n-k).
Showing 1-10 of 17 results. Next