A028296 Expansion of e.g.f. Gudermannian(x) = 2*arctan(exp(x)) - Pi/2.
1, -1, 5, -61, 1385, -50521, 2702765, -199360981, 19391512145, -2404879675441, 370371188237525, -69348874393137901, 15514534163557086905, -4087072509293123892361, 1252259641403629865468285, -441543893249023104553682821, 177519391579539289436664789665
Offset: 0
Examples
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
References
- 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.
Links
- 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.
Programs
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Maple
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
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Mathematica
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 *)
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Maxima
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 */
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PARI
a(n) = 2*imag(polylog(-2*n, I)); \\ Michel Marcus, May 30 2018
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PARI
a(n)=eulerfrac(2*n) \\ Charles R Greathouse IV, Mar 23 2022
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Python
from sympy import euler def A028296(n): return euler(n<<1) # Chai Wah Wu, Apr 16 2023
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Sage
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))
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Sage
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
Formula
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
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