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.

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

Original entry on oeis.org

1, 2, 16, 272, 7936, 353792, 22368256, 1903757312, 209865342976, 29088885112832, 4951498053124096, 1015423886506852352, 246921480190207983616, 70251601603943959887872, 23119184187809597841473536, 8713962757125169296170811392, 3729407703720529571097509625856
Offset: 1

Views

Author

Keywords

Comments

Number of Joyce trees with 2n-1 nodes. Number of tremolo permutations of {0,1,...,2n}. - Ralf Stephan, Mar 28 2003
The Hankel transform of this sequence is A000178(n) for n odd = 1, 12, 34560, ...; example: det([1, 2, 16; 2, 16, 272, 16, 272, 7936]) = 34560. - Philippe Deléham, Mar 07 2004
a(n) is the number of increasing labeled full binary trees with 2n-1 vertices. Full binary means every non-leaf vertex has two children, distinguished as left and right; labeled means the vertices are labeled 1,2,...,2n-1; increasing means every child has a label greater than its parent. - David Callan, Nov 29 2007
From Micha Hofri (hofri(AT)wpi.edu), May 27 2009: (Start)
a(n) was found to be the number of permutations of [2n] which when inserted in order, to form a binary search tree, yield the maximally full possible tree (with only one single-child node).
The e.g.f. is sec^2(x)=1+tan^2(x), and the same coefficients can be manufactured from the tan(x) itself, which is the e.g.f. for the number of trees as above for odd number of nodes. (End)
a(n) is the number of increasing strict binary trees with 2n-1 nodes. For more information about increasing strict binary trees with an associated permutation, see A245894. - Manda Riehl, Aug 07 2014
For relations to alternating permutations, Euler and Bernoulli polynomials, zigzag numbers, trigonometric functions, Fourier transform of a square wave, quantum algebras, and integrals over and in n-dimensional hypercubes and over Green functions, see Hodges and Sukumar. For further discussion on the quantum algebra, see the later Hodges and Sukumar reference and the paper by Hetyei presenting connections to the general combinatorial theory of Viennot on orthogonal polynomials, inverse polynomials, tridiagonal matrices, and lattice paths (thereby related to continued fractions and cumulants). - Tom Copeland, Nov 30 2014
The Zigzag Hankel transform is A000178. That is, A000178(2*n - k) = det( [a(i+j - k)]{i,j = 1..n} ) for n>0 and k=0,1. - _Michael Somos, Mar 12 2015
a(n) is the number of standard Young tableaux of skew shape (n,n,n-1,n-2,...,3,2)/(n-1,n-2,n-3,...,2,1). - Ran Pan, Apr 10 2015
For relations to the Sheffer Appell operator calculus and a Riccati differential equation for generating the Meixner-Pollaczek and Krawtchouk orthogonal polynomials, see page 45 of the Feinsilver link and Rzadkowski. - Tom Copeland, Sep 28 2015
For relations to an elliptic curve, a Weierstrass elliptic function, the Lorentz formal group law, a Lie infinitesimal generator, and the Eulerian numbers A008292, see A155585. - Tom Copeland, Sep 30 2015
Absolute values of the alternating sums of the odd-numbered rows (where the single 1 at the apex of the triangle is counted as row #1) of the Eulerian triangle, A008292. The actual alternating sums alternate in sign, e.g., 1, -2, 16, -272, etc. (Even-numbered rows have alternating sums always 0.) - Gregory Gerard Wojnar, Sep 28 2018
The sequence is periodic modulo any odd prime p. The minimal period is (p-1)/2 if p == 1 mod 4 and p-1 if p == 3 mod 4 [Knuth & Buckholtz, 1967, Theorem 1]. - Allen Stenger, Aug 03 2020
From Peter Bala, Dec 24 2021: (Start)
Conjectures:
1) The sequence taken modulo any integer k eventually becomes periodic with period dividing phi(k).
2) The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k, except when p = 2, n = 1 and k = 1 or 2.
3) For i >= 1 define a_i(n) = a(n+i). The Gauss congruences a_i(n*p^k) == a_i(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k. If true, then for each i >= 1 the expansion of exp(Sum_{n >= 1} a_i(n)*x^n/n) has integer coefficients. For an example, see A262145.(End)

Examples

			tan(x) = x + 2*x^3/3! + 16*x^5/5! + 272*x^7/7! + ... = x + 1/3*x^3 + 2/15*x^5 + 17/315*x^7 + 62/2835*x^9 + O(x^11).
tanh(x) = x - 1/3*x^3 + 2/15*x^5 - 17/315*x^7 + 62/2835*x^9 - 1382/155925*x^11 + ...
(sec x)^2 = 1 + x^2 + 2/3*x^4 + 17/45*x^6 + ...
a(3)=16 because we have: {1, 3, 2, 5, 4}, {1, 4, 2, 5, 3}, {1, 4, 3, 5, 2},
  {1, 5, 2, 4, 3}, {1, 5, 3, 4, 2}, {2, 3, 1, 5, 4}, {2, 4, 1, 5, 3},
  {2, 4, 3, 5, 1}, {2, 5, 1, 4, 3}, {2, 5, 3, 4, 1}, {3, 4, 1, 5, 2},
  {3, 4, 2, 5, 1}, {3, 5, 1, 4, 2}, {3, 5, 2, 4, 1}, {4, 5, 1, 3, 2},
  {4, 5, 2, 3, 1}. - _Geoffrey Critzer_, May 19 2013
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 88.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 111.
  • H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 69.
  • L. M. Milne-Thompson, Calculus of Finite Differences, 1951, p. 148 (the numbers |C^{2n-1}|).
  • J. W. Milnor and J. D. Stasheff, Characteristic Classes, Princeton, 1974, p. 282.
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 444.
  • H. Rademacher, Topics in Analytic Number Theory, Springer, 1973, Chap. 1, p. 20.
  • L. Seidel, Über eine einfache Entstehungsweise der Bernoullischen Zahlen und einiger verwandten Reihen, Sitzungsberichte der mathematisch-physikalischen Classe der königlich bayerischen Akademie der Wissenschaften zu München, volume 7 (1877), 157-187.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • E. van Fossen Conrad, Some continued fraction expansions of elliptic functions, PhD thesis, The Ohio State University, 2002, p. 28.
  • J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, pp. 267-268.

Crossrefs

A350972 is essentially the same sequence.
a(n)=2^(n-1)*A002105(n). Apart from signs, 2^(2n-2)*A001469(n) = n*a(n).
Cf. A001469, A002430, A036279, A000364 (secant numbers), A000111 (secant-tangent numbers), A024283, A009764. First diagonal of A059419 and of A064190.
Equals A002425(n) * 2^A101921(n).
Equals leftmost column of A162005. - Johannes W. Meijer, Jun 27 2009

Programs

  • Maple
    series(tan(x),x,40);
    with(numtheory): a := n-> abs(2^(2*n)*(2^(2*n)-1)*bernoulli(2*n)/(2*n));
    A000182_list := proc(n) local T,k,j; T[1] := 1;
    for k from 2 to n do T[k] := (k-1)*T[k-1] od;
       for k from 2 to n do
           for j from k to n do
               T[j] := (j-k)*T[j-1]+(j-k+2)*T[j] od od;
    seq(T[j], j=1..n)  end:
    A000182_list(15);  # Peter Luschny, Apr 02 2012
  • Mathematica
    Table[ Sum[2^(2*n + 1 - k)*(-1)^(n + k + 1)*k!*StirlingS2[2*n + 1, k], {k, 1, 2*n + 1}], {n, 0, 7}] (* Victor Adamchik, Oct 05 2005 *)
    v[1] = 2; v[n_] /; n >= 2 := v[n] = Sum[ Binomial[2 n - 3, 2 k - 2] v[k] v[n - k], {k, n - 1}]; Table[ v[n]/2, {n, 15}] (* Zerinvary Lajos, Jul 08 2009 *)
    Rest@ Union[ Range[0, 29]! CoefficientList[ Series[ Tan[x], {x, 0, 30}], x]] (* Harvey P. Dale, Oct 19 2011; modified by Robert G. Wilson v, Apr 02 2012 *)
    t[1, 1] = 1; t[1, 0] = 0; t[n_ /; n > 1, m_] := t[n, m] = m*(m+1)*Sum[t[n-1, k], {k, m-1, n-1}]; a[n_] := t[n, 1]; Table[a[n], {n, 1, 15}]  (* Jean-François Alcover, Jan 02 2013, after A064190 *)
    a[ n_] := If[ n < 1, 0, With[{m = 2 n - 1}, m! SeriesCoefficient[ Tan[x], {x, 0, m}]]]; (* Michael Somos, Mar 12 2015 *)
    a[ n_] := If[ n < 1, 0, ((-16)^n - (-4)^n) Zeta[1 - 2 n]]; (* Michael Somos, Mar 12 2015 *)
    Table[2 PolyGamma[2n - 1, 1/2]/Pi^(2n), {n, 1, 10}] (* Vladimir Reshetnikov, Oct 18 2015 *)
    a[ n_] := a[n] = If[ n < 2, Boole[n == 1], Sum[Binomial[2 n - 2, 2 k - 1] a[k] a[n - k], {k, n - 1}]]; (* Michael Somos, Aug 02 2018 *)
    a[n_] := (2^(2*n)*(2^(2*n) - 1)*Abs[BernoulliB[2*n]])/(2*n); a /@  Range[20]  (* Stan Wagon, Nov 21 2022 *)
  • Maxima
    a(n):=sum(sum(binomial(k,r)*sum(sum(binomial(l,j)/2^(j-1)*sum((-1)^(n)*binomial(j,i)*(j-2*i)^(2*n),i,0,floor((j-1)/2))*(-1)^(l-j),j,1,l)*(-1)^l*binomial(r+l-1,r-1),l,1,2*n)*(-1)^(1-r),r,1,k)/k,k,1,2*n); /* Vladimir Kruchinin, Aug 23 2010 */
    
  • Maxima
    a[n]:=if n=1 then 1 else 2*sum(sum(binomial(2*j,j+k)*(-4*k^2)^(n-1)*(-1)^k/(4^j),k,1,j),j,1,n-1);
    makelist(a[n],n,1,30); /* Tani Akinari, Sep 20 2023 */
    
  • PARI
    {a(n) = if( n<1, 0, ((-4)^n - (-16)^n) * bernfrac(2*n) / (2*n))};
    
  • PARI
    {a(n) = my(an); if( n<2, n==1, an = vector(n, m, 1); for( m=2, n, an[m] = sum( k=1, m-1, binomial(2*m - 2, 2*k - 1) * an[k] * an[m-k])); an[n])}; /* Michael Somos */
    
  • PARI
    {a(n) = if( n<1, 0, (2*n - 1)! * polcoeff( tan(x + O(x^(2*n + 2))), 2*n - 1))}; /* Michael Somos */
    
  • PARI
    {a(n) = my(X=x+x*O(x^n),Egf); Egf = x*sum(m=0,n, prod(k=1,m, tanh(2*k*X))); (n-1)!*polcoeff(Egf,n)} /* Paul D. Hanna, May 11 2010 */
    
  • PARI
    /* Continued Fraction for the e.g.f. tan(x), from Paul D. Hanna: */
    {a(n)=local(CF=1+O(x)); for(i=1, n, CF=1/(2*(n-i+1)-1-x^2*CF)); (2*n-1)!*polcoeff(x*CF, 2*n-1)}
    
  • PARI
    /* O.g.f. Sum_{n>=1} a(n)*x^n, from Paul D. Hanna Feb 05 2013: */
    {a(n)=polcoeff( x+2*x*sum(m=1, n, x^m*prod(k=1, m, (2*k-1)^2/(1+(2*k-1)^2*x +x*O(x^n))) ), n)}
    
  • Python
    # The objective of this implementation is efficiency.
    # n -> [0, a(1), a(2), ..., a(n)] for n > 0.
    def A000182_list(n):
        T = [0 for i in range(1, n+2)]
        T[1] = 1
        for k in range(2, n+1):
            T[k] = (k-1)*T[k-1]
        for k in range(2, n+1):
            for j in range(k, n+1):
                T[j] = (j-k)*T[j-1]+(j-k+2)*T[j]
        return T
    print(A000182_list(100)) # Peter Luschny, Aug 07 2011
    
  • Python
    from sympy import bernoulli
    def A000182(n): return abs(((2-(2<<(m:=n<<1)))*bernoulli(m)<Chai Wah Wu, Apr 14 2023
    
  • Sage
    # Algorithm of L. Seidel (1877)
    # n -> [a(1), ..., a(n)] for n >= 1.
    def A000182_list(len) :
        R = []; A = {-1:0, 0:1}; k = 0; e = 1
        for i in (0..2*len-1) :
            Am = 0; A[k + e] = 0; e = -e
            for j in (0..i) : Am += A[k]; A[k] = Am; k += e
            if e > 0 : R.append(A[i//2])
        return R
    A000182_list(15) # Peter Luschny, Mar 31 2012

Formula

E.g.f.: log(sec x) = Sum_{n > 0} a(n)*x^(2*n)/(2*n)!.
E.g.f.: tan x = Sum_{n >= 0} a(n+1)*x^(2*n+1)/(2*n+1)!.
E.g.f.: (sec x)^2 = Sum_{n >= 0} a(n+1)*x^(2*n)/(2*n)!.
2/(exp(2x)+1) = 1 + Sum_{n>=1} (-1)^(n+1) a(n) x^(2n-1)/(2n-1)! = 1 - x + x^3/3 - 2*x^5/15 + 17*x^7/315 - 62*x^9/2835 + ...
a(n) = 2^(2*n) (2^(2*n) - 1) |B_(2*n)| / (2*n) where B_n are the Bernoulli numbers (A000367/A002445 or A027641/A027642).
Asymptotics: a(n) ~ 2^(2*n+1)*(2*n-1)!/Pi^(2*n).
Sum[2^(2*n + 1 - k)*(-1)^(n + k + 1)*k!*StirlingS2[2*n + 1, k], {k, 1, 2*n + 1}]. - Victor Adamchik, Oct 05 2005
a(n) = abs[c(2*n-1)] where c(n)= 2^(n+1) * (1-2^(n+1)) * Ber(n+1)/(n+1) = 2^(n+1) * (1-2^(n+1)) * (-1)^n * Zeta(-n) = [ -(1+EN(.))]^n = 2^n * GN(n+1)/(n+1) = 2^n * EP(n,0) = (-1)^n * E(n,-1) = (-2)^n * n! * Lag[n,-P(.,-1)/2] umbrally = (-2)^n * n! * C{T[.,P(.,-1)/2] + n, n} umbrally for the signed Euler numbers EN(n), the Bernoulli numbers Ber(n), the Genocchi numbers GN(n), the Euler polynomials EP(n,t), the Eulerian polynomials E(n,t), the Touchard / Bell polynomials T(n,t), the binomial function C(x,y) = x!/[(x-y)!*y! ] and the polynomials P(j,t) of A131758. - Tom Copeland, Oct 05 2007
a(1) = A094665(0,0)*A156919(0,0) and a(n) = Sum_{k=1..n-1} 2^(n-k-1)*A094665(n-1, k)*A156919(k,0) for n = 2, 3, .., see A162005. - Johannes W. Meijer, Jun 27 2009
G.f.: 1/(1-1*2*x/(1-2*3*x/(1-3*4*x/(1-4*5*x/(1-5*6*x/(1-... (continued fraction). - Paul Barry, Feb 24 2010
From Paul Barry, Mar 29 2010: (Start)
G.f.: 1/(1-2x-12x^2/(1-18x-240x^2/(1-50x-1260x^2/(1-98x-4032x^2/(1-162x-9900x^2/(1-... (continued fraction);
coefficient sequences given by 4*(n+1)^2*(2n+1)*(2n+3) and 2(2n+1)^2 (see Van Fossen Conrad reference). (End)
E.g.f.: x*Sum_{n>=0} Product_{k=1..n} tanh(2*k*x) = Sum_{n>=1} a(n)*x^n/(n-1)!. - Paul D. Hanna, May 11 2010 [corrected by Paul D. Hanna, Sep 28 2023]
a(n) = (-1)^(n+1)*Sum_{j=1..2*n+1} j!*Stirling2(2*n+1,j)*2^(2*n+1-j)*(-1)^j for n >= 0. Vladimir Kruchinin, Aug 23 2010: (Start)
If n is odd such that 2*n-1 is prime, then a(n) == 1 (mod (2*n-1)); if n is even such that 2*n-1 is prime, then a(n) == -1 (mod (2*n-1)). - Vladimir Shevelev, Sep 01 2010
Recursion: a(n) = (-1)^(n-1) + Sum_{i=1..n-1} (-1)^(n-i+1)*C(2*n-1,2*i-1)* a(i). - Vladimir Shevelev, Aug 08 2011
E.g.f.: tan(x) = Sum_{n>=1} a(n)*x^(2*n-1)/(2*n-1)! = x/(1 - x^2/(3 - x^2/(5 - x^2/(7 - x^2/(9 - x^2/(11 - x^2/(13 -...))))))) (continued fraction from J. H. Lambert - 1761). - Paul D. Hanna, Sep 21 2011
From Sergei N. Gladkovskii, Oct 31 2011 to Oct 09 2013: (Start)
Continued fractions:
E.g.f.: (sec(x))^2 = 1+x^2/(x^2+U(0)) where U(k) = (k+1)*(2k+1) - 2x^2 + 2x^2*(k+1)*(2k+1)/U(k+1).
E.g.f.: tan(x) = x*T(0) where T(k) = 1-x^2/(x^2-(2k+1)*(2k+3)/T(k+1)).
E.g.f.: tan(x) = x/(G(0)+x) where G(k) = 2*k+1 - 2*x + x/(1 + x/G(k+1)).
E.g.f.: tanh(x) = x/(G(0)-x) where G(k) = k+1 + 2*x - 2*x*(k+1)/G(k+1).
E.g.f.: tan(x) = 2*x - x/W(0) where W(k) = 1 + x^2*(4*k+5)/((4*k+1)*(4*k+3)*(4*k+5) - 4*x^2*(4*k+3) + x^2*(4*k+1)/W(k+1)).
E.g.f.: tan(x) = x/T(0) where T(k) = 1 - 4*k^2 + x^2*(1 - 4*k^2)/T(k+1).
E.g.f.: tan(x) = -3*x/(T(0)+3*x^2) where T(k)= 64*k^3 + 48*k^2 - 4*k*(2*x^2 + 1) - 2*x^2 - 3 - x^4*(4*k -1)*(4*k+7)/T(k+1).
G.f.: 1/G(0) where G(k) = 1 - 2*x*(2*k+1)^2 - x^2*(2*k+1)*(2*k+2)^2*(2*k+3)/G(k+1).
G.f.: 2*Q(0) - 1 where Q(k) = 1 + x^2*(4*k + 1)^2/(x + x^2*(4*k + 1)^2 - x^2*(4*k + 3)^2*(x + x^2*(4*k + 1)^2)/(x^2*(4*k + 3)^2 + (x + x^2*(4*k + 3)^2)/Q(k+1) )).
G.f.: (1 - 1/G(0))*sqrt(-x), where G(k) = 1 + sqrt(-x) - x*(k+1)^2/G(k+1).
G.f.: Q(0), where Q(k) = 1 - x*(k+1)*(k+2)/( x*(k+1)*(k+2) - 1/Q(k+1)). (End)
O.g.f.: x + 2*x*Sum_{n>=1} x^n * Product_{k=1..n} (2*k-1)^2 / (1 + (2*k-1)^2*x). - Paul D. Hanna, Feb 05 2013
a(n) = (-4)^n*Li_{1-2*n}(-1). - Peter Luschny, Jun 28 2012
a(n) = (-4)^n*(4^n-1)*Zeta(1-2*n). - Jean-François Alcover, Dec 05 2013
Asymptotic expansion: 4*((2*(2*n-1))/(Pi*e))^(2*n-1/2)*exp(1/2+1/(12*(2*n-1))-1/(360*(2*n-1)^3)+1/(1260*(2*n-1)^5)-...). (See Luschny link.) - Peter Luschny, Jul 14 2015
From Peter Bala, Sep 11 2015: (Start)
The e.g.f. A(x) = tan(x) satisfies the differential equation A''(x) = 2*A(x)*A'(x) with A(0) = 0 and A'(0) = 1, leading to the recurrence a(0) = 0, a(1) = 1, else a(n) = 2*Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i) for the aerated sequence [0, 1, 0, 2, 0, 16, 0, 272, ...].
Note, the same recurrence, but with the initial conditions a(0) = 1 and a(1) = 1, produces the sequence n! and with a(0) = 1/2 and a(1) = 1 produces A080635. Cf. A002105, A234797. (End)
a(n) = 2*polygamma(2*n-1, 1/2)/Pi^(2*n). - Vladimir Reshetnikov, Oct 18 2015
a(n) = 2^(2n-2)*|p(2n-1,-1/2)|, where p_n(x) are the shifted row polynomials of A019538. E.g., a(2) = 2 = 2^2 * |1 + 6(-1/2) + 6(-1/2)^2|. - Tom Copeland, Oct 19 2016
From Peter Bala, May 05 2017: (Start)
With offset 0, the o.g.f. A(x) = 1 + 2*x + 16*x^2 + 272*x^3 + ... has the property that its 4th binomial transform 1/(1 - 4*x) A(x/(1 - 4*x)) has the S-fraction representation 1/(1 - 6*x/(1 - 2*x/(1 - 20*x/(1 - 12*x/(1 - 42*x/(1 - 30*x/(1 - ...))))))), where the coefficients [6, 2, 20, 12, 42, 30, ...] in the partial numerators of the continued fraction are obtained from the sequence [2, 6, 12, 20, ..., n*(n + 1), ...] by swapping adjacent terms. Compare with the S-fraction associated with A(x) given above by Paul Barry.
A(x) = 1/(1 + x - 3*x/(1 - 4*x/(1 + x - 15*x/(1 - 16*x/(1 + x - 35*x/(1 - 36*x/(1 + x - ...))))))), where the unsigned coefficients in the partial numerators [3, 4, 15, 16, 35, 36,...] come in pairs of the form 4*n^2 - 1, 4*n^2 for n = 1,2,.... (End)
a(n) = Sum_{k = 1..n-1} binomial(2*n-2, 2*k-1) * a(k) * a(n-k), with a(1) = 1. - Michael Somos, Aug 02 2018
a(n) = 2^(2*n-1) * |Euler(2*n-1, 0)|, where Euler(n,x) are the Euler polynomials. - Daniel Suteu, Nov 21 2018 (restatement of one of Copeland's 2007 formulas.)
x - Sum_{n >= 1} (-1)^n*a(n)*x^(2*n)/(2*n)! = x - log(cosh(x)). The series reversion of x - log(cosh(x)) is (1/2)*x - (1/2)*log(2 - exp(x)) = Sum_{n >= 0} A000670(n)*x^(n+1)/(n+1)!. - Peter Bala, Jul 11 2022
For n > 1, a(n) = 2*Sum_{j=1..n-1} Sum_{k=1..j} binomial(2*j,j+k)*(-4*k^2)^(n-1)*(-1)^k/(4^j). - Tani Akinari, Sep 20 2023
a(n) = A110501(n) * 4^(n-1) / n (Han and Liu, 2018). - Amiram Eldar, May 17 2024