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.

Showing 1-9 of 9 results.

A001586 Generalized Euler numbers, or Springer numbers.

Original entry on oeis.org

1, 1, 3, 11, 57, 361, 2763, 24611, 250737, 2873041, 36581523, 512343611, 7828053417, 129570724921, 2309644635483, 44110959165011, 898621108880097, 19450718635716001, 445777636063460643, 10784052561125704811, 274613643571568682777, 7342627959965776406281
Offset: 0

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From Peter Bala, Feb 02 2011: (Start)
The Springer numbers were originally considered by Glaisher (see references). They are a type B analog of the zigzag numbers A000111 for the group of signed permutations.
COMBINATORIAL INTERPRETATIONS
Several combinatorial interpretations of the Springer numbers are known:
1) a(n) gives the number of Weyl chambers in the principal Springer cone of the Coxeter group B_n of symmetries of an n dimensional cube. An example can be found in [Arnold - The Calculus of snakes...].
2) Arnold found an alternative combinatorial interpretation of the Springer numbers in terms of snakes. Snakes are a generalization of alternating permutations to the group of signed permutations. A signed permutation is a sequence (x_1,x_2,...,x_n) of integers such that {|x_1|,|x_2|,...,|x_n|} = {1,2,...,n}. They form a group, the hyperoctahedral group of order 2^n*n! = A000165(n), isomorphic to the group of symmetries of the n dimensional cube. A snake of type B_n is a signed permutation (x_1,x_2,...,x_n) such that 0 < x_1 > x_2 < ... x_n. For example, (3,-4,-2,-5,1,-6) is a snake of type B_6. a(n) gives the number of snakes of type B_n [Arnold]. The cases n=2 and n=3 are given in the Example section below.
3) The Springer numbers also arise in the study of the critical points of functions; they count the topological types of odd functions with 2*n critical values [Arnold, Theorem 35].
4) Let F_n be the set of plane rooted forests satisfying the following conditions:
... each root has exactly one child, and each of the other internal nodes has exactly two (ordered) children,
... there are n nodes labeled by integers from 1 to n, but some leaves can be non-labeled (these are called empty leaves), and labels are increasing from each root down to the leaves. Then a(n) equals the cardinality of F_n. An example and proof are given in [Verges, Theorem 4.5].
OTHER APPEARANCES OF THE SPRINGER NUMBERS
1) Hoffman has given a connection between Springer numbers, snakes and the successive derivatives of the secant and tangent functions.
2) For integer N the quarter Gauss sums Q(N) are defined by ... Q(N) := Sum_{r = 0..floor(N/4)} exp(2*Pi*I*r^2/N). In the cases N = 1 (mod 4) and N = 3 (mod 4) an asymptotic series for Q(N) as N -> inf that involves the Springer numbers has been given by Evans et al., see 1.32 and 1.33.
For a sequence of polynomials related to the Springer numbers see A185417. For a table to recursively compute the Springer numbers see A185418.
(End)
Similar to the way in which the signed Euler numbers A122045 are 2^n times the value of the Euler polynomials at 1/2, the generalized signed Euler numbers A188458 can be seen as 2^n times the value of generalized Euler polynomials at 1/2. These are the Swiss-Knife polynomials A153641. A recursive definition of these polynomials is given in A081658. - Peter Luschny, Jul 19 2012
a(n) is the number of reverse-complementary updown permutations of [2n]. For example, the updown permutation 241635 is reverse-complementary because its complement is 536142, which is the same as its reverse, and a(2)=3 counts 1324, 2413, 3412. - David Callan, Nov 29 2012
a(n) = |2^n G(n,1/2;-1)|, a specialization of the Appell sequence of polynomials umbrally formed by G(n,x;t) = (G(.,0;t) + x)^n from the Grassmann polynomials G(n,0;t) of A046802 enumerating the cells of the positive Grassmannians. - Tom Copeland, Oct 14 2015
Named "Springer numbers" after the Dutch mathematician Tonny Albert Springer (1926-2011). - Amiram Eldar, Jun 13 2021

Examples

			a(2) = 3: The three snakes of type B_2 are
  (1,-2), (2,1), (2,-1).
a(3) = 11: The 11 snakes of type B_3 are
  (1,-2,3), (1,-3,2), (1,-3,-2),
  (2,1,3), (2,-1,3), (2,-3,1), (2,-3,-1),
  (3,1,2), (3,-1,2), (3,-2,1), (3,-2,-1).
		

References

  • V. I. Arnold, Springer numbers and Morsification spaces. J. Algebraic Geom., Vol. 1, No. 2 (1992), pp. 197-214.
  • J. W. L. Glaisher, "On the coefficients in the expansions of cos x/cos 2x and sin x/cos 2x", Quart. J. Pure and Applied Math., Vol. 45 (1914), pp. 187-222.
  • J. W. L. Glaisher, On the Bernoullian function, Q. J. Pure Appl. Math., Vol. 29 (1898), pp. 1-168.
  • Ulrike Sattler, Decidable classes of formal power series with nice closure properties, Diplomarbeit im Fach Informatik, Univ. Erlangen - Nürnberg, Jul 27 1994.
  • 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).
  • Tonny Albert Springer, Remarks on a combinatorial problem, Nieuw Arch. Wisk., Vol. 19, No. 3 (1971), pp. 30-36.

Crossrefs

Row 2 of A349271.
Bisections are A000281 and A000464. Overview in A349264.
Related polynomials are given in A098432, A081658 and A153641.
Cf. A046802.

Programs

  • Maple
    a := proc(n) local k; (-1)^iquo(n,2)*add(2^k*binomial(n,k)*euler(k),k=0..n) end; # Peter Luschny, Jul 08 2009
    a := n -> (-1)^(n+iquo(n,2))*2^(3*n+1)*(Zeta(0,-n,1/8) - Zeta(0,-n,5/8)):
    seq(a(n),n=0..21); # Peter Luschny, Mar 11 2015
  • Mathematica
    n=21; CoefficientList[Series[1/(Cos[x]-Sin[x]), {x, 0, n}], x] * Table[k!, {k, 0, n}] (* Jean-François Alcover, May 18 2011 *)
    Table[Abs[Numerator[EulerE[n,1/4]]],{n,0,35}] (* Harvey P. Dale, May 18 2011 *)
  • PARI
    {a(n) = if(n<0, 0, n! * polcoeff( 1 / (cos(x + x * O(x^n)) - sin(x + x * O(x^n))), n))}; /* Michael Somos, Feb 03 2004 */
    
  • PARI
    {a(n) = my(an); if(n<2, n>=0, an = vector(n+1, m, 1); for(m=2, n, an[m+1] = 2*an[m] + an[m-1] + sum(k=0, m-3, binomial(m-2, k) * (an[k+1] * an[m-1-k] + 2*an[k+2] * an[m-k] - an[k+3] * an[m-1-k]))); an[n+1])}; /* Michael Somos, Feb 03 2004 */
    
  • PARI
    /* Explicit formula by Peter Bala: */
    {a(n)=((1+I)/2)^n*sum(k=0,n,((1-I)/(1+I))^k*sum(j=0,k,(-1)^(k-j)*binomial(n+1,k-j)*(2*j+1)^n))}
    
  • Sage
    @CachedFunction
    def p(n,x) :
        if n == 0 : return 1
        w = -1 if n%2 == 0 else 0
        v =  1 if n%2 == 0 else -1
        return v*add(p(k,0)*binomial(n,k)*(x^(n-k)+w) for k in range(n)[::2])
    def A001586(n) : return abs(2^n*p(n, 1/2))
    [A001586(n) for n in (0..21)] # Peter Luschny, Jul 19 2012

Formula

E.g.f.: 1/(cos(x) - sin(x)).
Values at 1 of polynomials Q_n() defined in A104035. - N. J. A. Sloane, Nov 06 2009
a(n) = numerator of abs(Euler(n,1/4)). - N. J. A. Sloane, Nov 07 2009
Let B_n(x) = Sum_{k=0.. n*(n-1)/2} b(n,k)*x^k, where b(n,k) is number of n-node acyclic digraphs with k arcs, cf. A081064; then a(n) = |B_n(-2)|. - Vladeta Jovovic, Jan 25 2005
G.f. A(x) = y satisfies y'^2 = 2y^4 - y^2, y''y = y^2 + 2y'^2. - Michael Somos, Feb 03 2004
a(n) = (-1)^floor(n/2) Sum_{k=0..n} 2^k C(n,k) Euler(k). - Peter Luschny, Jul 08 2009
From Peter Bala, Feb 02 2011: (Start)
(1)... a(n) = ((1 + i)/2)^n*B(n,(1 - i)/(1 + i)), where i = sqrt(-1) and {B(n,x)}n>=0 = [1, 1 + x, 1 + 6*x + x^2, 1 + 23*x + 23*x^2 + x^3, ...] is the sequence of type B Eulerian polynomials - see A060187.
This yields the explicit formula
(2)... a(n) = ((1 + i)/2)^n*Sum_{k = 0..n} ((1 - i)/(1 + i))^k * Sum_{j = 0..k} (-1)^(k-j)*binomial(n+1,k-j)*(2*j + 1)^n.
The result (2) can be used to find congruences satisfied by the Springer numbers. For example, for odd prime p
(3)
... a(p) = 1 (mod p) when p = 4*n + 1
... a(p) = -1 (mod p) when p = 4*n + 3.
(End)
E.g.f.: 1/Q(0) where Q(k) = 1 - x/((2k+1)-x*(2k+1)/(x+(2k+2)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 19 2011
E.g.f.: 2/U(0) where U(k) = 1 + 1/(1 + x/(2*k + 1 -x - (2*k+1)/(2 - x/(x+ (2*k+2)/U(k+1))))); (continued fraction, 5-step). - Sergei N. Gladkovskii, Sep 24 2012
E.g.f.: 1/G(0) where G(k) = 1 - x/(4*k+1 - x*(4*k+1)/(4*k+2 + x + x*(4*k+2)/(4*k+3 - x - x*(4*k+3)/(x + (4*k+4)/G(k+1) )))); (continued fraction, 3rd kind, 5-step). - Sergei N. Gladkovskii, Oct 02 2012
G.f.: 1/G(0) where G(k) = 1 - x*(2*k+1) - 2*x^2*(k+1)*(k+1)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 11 2013
a(n) = | 2*4^n*lerchphi(-1, -n, 1/4) |. - Peter Luschny, Apr 27 2013
a(n) ~ 4 * n^(n+1/2) * (4/Pi)^n / (sqrt(Pi)*exp(n)). - Vaclav Kotesovec, Oct 07 2013
G.f.: T(0)/(1-x), where T(k) = 1 - 2*x^2*(k+1)^2/( 2*x^2*(k+1)^2 - (1-x-2*x*k)*(1-3*x-2*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 15 2013
a(n) = (-1)^C(n+1,2)*2^(3*n+1)*(Zeta(-n,1/8)-Zeta(-n,5/8)), where Zeta(a,z) is the generalized Riemann zeta function. - Peter Luschny, Mar 11 2015
E.g.f. A(x) satisfies: A(x) = exp( Integral A(x)/A(-x) dx ). - Paul D. Hanna, Feb 04 2017
E.g.f. A(x) satisfies: A'(x) = A(x)^2/A(-x). - Paul D. Hanna, Feb 04 2017

Extensions

More terms from Vladeta Jovovic, Jan 25 2005

A000281 Expansion of cos(x)/cos(2x).

Original entry on oeis.org

1, 3, 57, 2763, 250737, 36581523, 7828053417, 2309644635483, 898621108880097, 445777636063460643, 274613643571568682777, 205676334188681975553003, 184053312545818735778213457, 193944394596325636374396208563
Offset: 0

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Comments

a(n) is (2n)! times the coefficient of x^(2n) in the Taylor series for cos(x)/cos(2x).

Examples

			cos x / cos 2*x = 1 + 3*x^2/2 + 19*x^4/8 + 307*x^6/80 + ...
		

References

  • J. W. L. Glaisher, "On the coefficients in the expansions of cos x / cos 2x and sin x / cos 2x", Quart. J. Pure and Applied Math., 45 (1914), 187-222.
  • I. J. Schwatt, Intro. to Operations with Series, Chelsea, p. 278.
  • 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

Programs

  • Maple
    a := n -> (-1)^n*2^(6*n+1)*(Zeta(0,-2*n,1/8)-Zeta(0,-2*n,5/8)):
    seq(a(n), n=0..13); # Peter Luschny, Mar 11 2015
  • Mathematica
    With[{nn=30},Take[CoefficientList[Series[Cos[x]/Cos[2x],{x,0,nn}],x] Range[0,nn]!,{1,-1,2}]] (* Harvey P. Dale, Oct 06 2011 *)
  • PARI
    {a(n) = if( n<0, 0, n*=2; n! * polcoeff( cos(x + x * O(x^n)) / cos(2*x + x * O(x^n)), n))}; /* Michael Somos, Feb 09 2006 */

Formula

a(n) = Sum_{k=0..n} (-1)^k*binomial(2n, 2k)*A000364(n-k)*4^(n-k). - Philippe Deléham, Jan 26 2004
E.g.f.: Sum_{k>=0} a(k)x^(2k)/(2k)! = cos(x)/cos(2x).
a(n-1) is approximately 2^(4*n-3)*(2*n-1)!*sqrt(2)/((Pi^(2*n-1))*(2*n-1)). The approximation is quite good a(250) is of the order of 10^1181 and this formula is accurate to 238 digits. - Simon Plouffe, Jan 31 2007
G.f.: 1 / (1 - 1*3*x / (1 - 4*4*x / (1 - 5*7*x / (1 - 8*8*x / (1 - 9*11*x / ... ))))). - Michael Somos, May 12 2012
G.f.: 1/E(0) where E(k) = 1 - 3*x - 16*x*k*(2*k+1) - 16*x^2*(k+1)^2*(4*k+1)*(4*k+3)/E(k+1) (continued fraction, 1-step). - Sergei N. Gladkovskii, Sep 17 2012
G.f.: T(0)/(1-3*x), where T(k) = 1 - 16*x^2*(4*k+1)*(4*k+3)*(k+1)^2/( 16*x^2*(4*k+1)*(4*k+3)*(k+1)^2 - (32*x*k^2+16*x*k+3*x-1 )*(32*x*k^2+80*x*k+51*x -1)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 11 2013
From Peter Bala, Mar 09 2015: (Start)
a(n) = (-1)^n*4^(2*n)*E(2*n,1/4), where E(n,x) denotes the n-th Euler polynomial.
O.g.f.: Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 + x*(4*k + 1)^2) = 1 + 3*x + 57*x^2 + 2763*x^3 + ....
We appear to have 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 + .... See A093954.
Bisection of A001586. See also A188458 and A212435. Second row of A235605 (read as a square array).
The expansion of exp( Sum_{n >= 1} a(n)*x^n/n ) appears to have integer coefficients. See A255883. (End)
From Peter Luschny, Mar 11 2015: (Start)
a(n) = ((-64)^n/((n+1/2)))*(B(2*n+1,7/8)-B(2*n+1,3/8)), B(n,x) Bernoulli polynomials.
a(n) = 2*(-16)^n*LerchPhi(-1, -2*n, 1/4).
a(n) = (-1)^n*Sum_{0..2*n} 2^k*C(2*n,k)*E(k), E(n) the Euler secant numbers A122045.
a(n) = (-4)^n*SKP(2*n,1/2) where SKP are the Swiss-Knife polynomials A153641.
a(n) = (-1)^n*2^(6*n+1)*(Zeta(-2*n,1/8) - Zeta(-2*n,5/8)), where Zeta(a,z) is the generalized Riemann zeta function. (End)
From Peter Bala, May 13 2017: (Start)
G.f.: 1/(1 + x - 4*x/(1 - 12*x/(1 + x - 40*x/(1 - 56*x/(1 + x - ... - 4*n(4*n - 3)*x/(1 - 4*n(4*n - 1)*x/(1 + x - ...
G.f.: 1/(1 + 9*x - 12*x/(1 - 4*x/(1 + 9*x - 56*x/(1 - 40*x/(1 + 9*x - ... - 4*n(4*n - 1)*x/(1 - 4*n(4*n - 3)*x/(1 + 9*x - .... (End)
From Peter Bala, Nov 08 2019: (Start)
a(n) = sqrt(2)*4^n*Integral_{x = 0..inf} x^(2*n)*cosh(Pi*x/2)/cosh(Pi*x) dx. Cf. A002437.
The L-series 1 + 1/3^(2*n+1) - 1/5^(2*n+1) - 1/7^(2*n+1) + + - - ... = sqrt(2)*(Pi/4)^(2*n+1)*a(n)/(2*n)! (see Shanks), which gives a(n) ~ (1/sqrt(2))*(2*n)!*(4/Pi)^(2*n+1). (End)

A181049 Decimal expansion of (Pi/2 - log(1+sqrt(2)))/(2*sqrt(2)) = Sum_{k>=0} (-1)^k/(4k+3).

Original entry on oeis.org

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

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Author

Jonathan D. B. Hodgson, Oct 01 2010, Oct 05 2010

Keywords

Comments

Let N be a positive integer divisible by 4. We have the asymptotic expansion 2*((Pi/2 - log(1 + sqrt(2)))/(2*sqrt(2)) - Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 3)) ~ 1/N - 1/N^2 - 3/N^3 + 11/N^4 + 57/N^5 - ..., where the sequence of coefficients [1, -1, -3, 11, 57, ...] is A212435. This follows from Borwein et al., Lemma 2 with f(x) = 1/x and then set x = N/4 and h = 3/4. An example is given below. Cf. A181048. - Peter Bala, Sep 23 2016

Examples

			0.2437477471996805241799750836323027110...
From _Peter Bala_, Sep 23 2016: (Start)
At N = 100000 the truncated series 2*Sum_{k = 0..N/4 - 1} (-1)^k/(4*k + 3) = 0.4874(8)5494(4)9936(4)048(24)99(444)67(625)6... to 32 digits. The bracketed numbers show where this decimal expansion differs from that of 2*A181049. The numbers 1, -1, -3, 11, 57, -361 must be added to the bracketed numbers to give the correct decimal expansion to 32 digits: 2*( Pi/2 - log(1+sqrt(2)))/(2*sqrt(2) ) = 0.4874(9)5494(3)9936(1)048(35)99(501)67(264)6.... (End)
		

Crossrefs

Programs

  • Magma
    C := ComplexField(); [(Pi(C)/2 - Log(1+Sqrt(2)))/(2*Sqrt(2))]; // G. C. Greubel, Nov 28 2017
  • Mathematica
    First@ RealDigits[N[(Pi/2 - Log[1 + Sqrt@ 2])/(2 Sqrt@ 2), 105]] (* Michael De Vlieger, Oct 07 2015 *)
  • PARI
    default(realprecision, 106);
    eval(vecextract(Vec(Str(sumalt(n=0, (-1)^(n)/(4*n+3)))), "3..-2")) \\ Gheorghe Coserea, Oct 06 2015
    
  • PARI
    (Pi/2 - log(1+sqrt(2)))/(2*sqrt(2)) \\ G. C. Greubel, Nov 28 2017
    

Formula

Equals Integral_{x=0..1} (x^2 dx)/(1+x^4).
Equals (1/2) * Integral_{x = 0..Pi/4} sqrt(tan(x)) dx. Cf. A247719. - Peter Bala, Sep 23 2016
Equals Sum_{n >= 0} 2^(n-1)*n!/(Product_{k = 0..n} 4*k + 3) = Sum_{n >= 0} 2^(n-1)*n!/A008545(n+1) (apply Euler's series transformation to Sum_{k >= 0} (-1)^k/(4*k+3)). - Peter Bala, Dec 01 2021
From Peter Bala, Mar 03 2024: (Start)
Continued fraction: 1/(3 + 3^2/(4 + 7^2/(4 + 11^2/(4 + 15^2/(4 + ... ))))) due to Euler.
Equals (1/3)*hypergeom([3/4, 1], [7/4], -1).
Gauss's continued fraction: 1/(3 + 3^2/(7 + 4^2/(11 + 7^2/(15 + 8^2/(19 + 11^2/(23 + 12^2/(27 + 15^2/(31 + 16^2/(35 + 19^2/(39 + ... )))))))))). (End)
Equals Integral_{x=1..oo, y=1..oo} 1/(x^4 + y^4) dx. - Vaclav Kotesovec, Jun 13 2024

A188514 Expansion of exp( Sum_{n >= 1} A188458(n)*x^n/n ).

Original entry on oeis.org

1, 1, -1, -5, 11, 91, -391, -4115, 27971, 357331, -3353731, -50789375, 607914581, 10692083221, -155442170521, -3120028100285, 53341649623091, 1204301220497011, -23663734574555011, -593828627529030095, 13182525824990398001
Offset: 0

Author

Paul D. Hanna, Apr 02 2011

Keywords

Comments

The e.g.f. of A188458 is exp(x)/cosh(2*x).
The e.g.f. of this sequence is the product of the e.g.f. of A188458 and an even function (see formula section).
From Peter Bala, Mar 10 2015: (Start)
Note exp( Sum_{n >= 1} A212435(n)*x^n/n ) = exp( -x - 3*x^2/2 + 11*x^3/3 + 57*x^4/4 - ... ) = 1 - x - x^2 + 5*x^3 + 11*x^4 - 91*x^5 - 391*x^6 + + - - ... appears to give this sequence but with a different pattern of signs.
More generallly, it appears that when h is an integer and k is a nonzero integer, the expansion of exp( Sum_{n >= 1} (4*k)^n*E(n,h/(4*k))*x^n/n ) has integer coefficients, where E(n,x) denotes the n-th Euler polynomial. (End)

Examples

			O.g.f.: A(x) = 1 + x - x^2 - 5*x^3 + 11*x^4 + 91*x^5 - 391*x^6 +...
Illustration of the properties of the exponential generating function.
E.g.f.: E(x) = 1 + x - x^2/2! - 5*x^3/3! + 11*x^4/4! + 91*x^5/5! - 391*x^6/6! +...
Note that E(x)*cosh(2*x)/exp(x) is an even function:
E(x)*cosh(2*x)/exp(x) = 1 + 2*x^2/2! - 10*x^4/4! + 212*x^6/6! - 10330*x^8/8! + 926972*x^10/10! +...+ A092635(2*n)*x^(2*n)/(2*n)! +...
which equals (G(x)+G(-x))/2 with G(x) being the e.g.f of A092635:
G(x) = 1 - 2*x + 2*x^2/2! + 4*x^3/3! - 10*x^4/4! - 92*x^5/5! + 212*x^6/6! +...
		

Crossrefs

Programs

  • Maple
    exp(add(4^n*euler(n, 3/4)*x^n/n, n = 1 .. 20)): seq(coeftayl(%, x = 0, n), n = 0 .. 20); # Peter Bala, Mar 09 2015
  • Mathematica
    A188458:= With[{nn = 160}, CoefficientList[Series[E^x/Cosh[2*x], {x, 0, nn}], x]*Range[0, nn]!]; a:= With[{nmax = 80}, CoefficientList[ Series[Exp[Sum[A188458[[k + 1]]*x^(k)/(k), {k, 1, 75}]], {x, 0, nmax}], x]]; Table[a[[n]], {n, 1, 51}] (* G. C. Greubel, Aug 26 2018 *)
  • PARI
    {A188458(n)=local(X=x+x*O(x^n));n!*polcoeff(exp(X)/cosh(2*X),n)}
    {a(n)=polcoeff(exp(sum(m=1,n,A188458(m)*x^m/m)+x*O(x^n)),n)}
    
  • PARI
    {A092635(n)=if(n<0, 0, polcoeff(exp(intformal(serlaplace(-1/cosh(x*2+x*O(x^n))^2*2))), n))} /* Michael Somos */
    {a(n)=n!*polcoeff(exp(-x+x*O(x^n))*sum(m=0,n,A092635(m)*(-x)^m/m!),n)}

Formula

G.f.: A(x) = 1/(1-x/(1+2*x/(1 -3*x/(1+3*x/(1+x -5*x/(1+5*x/(1+x -7*x/(1+7*x/(1+x -9*x/(1+9*x/(1+x -11*x/(1+11*x/(1+x -... ))))))))))))) (continued fraction).
Let E(x) be the e.g.f. of this sequence, and let G(x) be the e.g.f of A092635 such that G(x) = G(-x)*exp(-4*x), then E(x) and G(x) are related by:
(1) E(x) = exp(-x) * G(-x),
(2) E(x) = exp(x)/cosh(2*x) * (G(x)+G(-x))/2.

A225147 a(n) = Im((1-I)^(1-n)*A_{n, 3}(I)) where A_{n, k}(x) are the generalized Eulerian polynomials.

Original entry on oeis.org

-1, 2, 5, -46, -205, 3362, 22265, -515086, -4544185, 135274562, 1491632525, -54276473326, -718181418565, 30884386347362, 476768795646785, -23657073914466766, -417370516232719345, 23471059057478981762, 465849831125196593045, -29279357851856595135406
Offset: 0

Author

Peter Luschny, Apr 30 2013

Keywords

Crossrefs

Cf. A000810 (real part (up to sign)), A212435 (k=2), A122045 (k=1), A002439.
Bisections are A002438 and A000191.

Programs

  • Maple
    B := proc(n, u, k) option remember;
    if n = 1 then if (u < 0) or (u >= 1) then 0 else 1 fi
    else k*u*B(n-1, u, k) + k*(n-u)*B(n-1, u-1, k) fi end:
    EulerianPolynomial := proc(n, k, x) local m; if x = 0 then RETURN(1) fi;
    add(B(n+1, m+1/k, k)*u^m, m = 0..n); subs(u=x, %) end:
    seq(Im((1-I)^(1-n)*EulerianPolynomial(n, 3, I)), n=0..19);
  • Mathematica
    CoefficientList[Series[-E^(-2*x)*Sech[3*x],{x,0,20}],x] * Range[0,20]! (* Vaclav Kotesovec, Sep 29 2014 after Sergei N. Gladkovskii *)
    Table[-6^n EulerE[n,1/6], {n,0,19}] (* Peter Luschny, Nov 16 2016 after Peter Bala *)
  • Sage
    from mpmath import mp, polylog, im
    mp.dps = 32; mp.pretty = True
    def A225147(n): return im(-2*I*(1+add(binomial(n,j)*polylog(-j,I)*3^j for j in (0..n))))
    [int(A225147(n)) for n in (0..19)]

Formula

a(n) = Im(-2*i*(1+Sum_{j=0..n}(binomial(n,j)*Li{-j}(i)*3^j))).
For a recurrence see the Maple program.
G.f.: conjecture -T(0)/(1+2*x), where T(k) = 1 - 9*x^2*(k+1)^2/(9*x^2*(k+1)^2 + (1+2*x)^2/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 12 2013
a(n) = -(-3)^n*skp(n, 2/3), where skp(n,x) are the Swiss-Knife polynomials A153641. - Peter Luschny, Apr 19 2014
G.f.: A225147 = -1/T(0), where T(k) = 1 + 2*x + (k+1)^2*(3*x)^2/ T(k+1); (continued fraction). - Sergei N. Gladkovskii, Sep 29 2014
E.g.f.: -exp(-2*x)*sech(3*x). - Sergei N. Gladkovskii, Sep 29 2014
a(n) ~ n! * (sqrt(3)*sin(Pi*n/2) - cos(Pi*n/2)) * 2^(n+1) * 3^n / Pi^(n+1). - Vaclav Kotesovec, Sep 29 2014
From Peter Bala, Nov 13 2016: (Start)
a(n) = - 6^n*E(n,1/6), where E(n,x) denotes the Euler polynomial of order n.
a(2*n) = (-1)^(n+1)*A002438(n); a(2*n+1) = (1/2)*(-1)^n*A002439(n). (End)

A377666 Array read by ascending antidiagonals: A(n, k) = Sum_{j = 0..k} binomial(k, j) * Euler(j, 0) *(2*n)^j.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, -1, -1, 1, 1, -2, -3, 0, 1, 1, -3, -5, 11, 5, 1, 1, -4, -7, 46, 57, 0, 1, 1, -5, -9, 117, 205, -361, -61, 1, 1, -6, -11, 236, 497, -3362, -2763, 0, 1, 1, -7, -13, 415, 981, -15123, -22265, 24611, 1385, 1
Offset: 0

Author

Peter Luschny, Nov 05 2024

Keywords

Examples

			Array A(n, k) starts:
  [0]  1,  1,   1,   1,    1,       1,        1, ...  A000012
  [1]  1,  0,  -1,   0,    5,       0,      -61, ...  A122045
  [2]  1, -1,  -3,  11,   57,    -361,    -2763, ...  A212435
  [3]  1, -2,  -5,  46,  205,   -3362,   -22265, ...  A225147
  [4]  1, -3,  -7, 117,  497,  -15123,   -95767, ...  A156201
  [5]  1, -4,  -9, 236,  981,  -47524,  -295029, ...  A377665
  [6]  1, -5, -11, 415, 1705, -120125,  -737891, ...
  [7]  1, -6, -13, 666, 2717, -262086, -1599793, ...
		

Crossrefs

Cf. A377663 (column 3), A377664 (main diagonal), A363393 (column polynomials).

Programs

  • Maple
    GHZeta := (k, n, m) -> m^(k+1)*Zeta(0, -k, 1/(m*n)):
    A := (n, k) -> ifelse(n = 0, 1, n^k*(GHZeta(k, n, 4) - GHZeta(k, n, 2))):
    for n from 0 to 7 do lprint(seq(A(n, k), k = 0..7)) od;
    # Alternative:
    P := proc(n, k) local j; 2*I*(1 + add(binomial(k, j)*polylog(-j, I)*n^j, j = 0..k)) end:
    A := n -> Im(P(n, k)): seq(lprint(seq(A(n, k), k = 0..7)), n = 0..7);
    # Computing the transpose using polynomials P from A363393.
    P := n -> add(binomial(n + 1, j)*bernoulli(j, 1)*(4^j - 2^j)*x^(j-1), j = 0..n+1)/(n + 1):
    Column := (k, n) -> subs(x = -n, P(k)):
    for k from 0 to 6 do seq(Column(k, n), n = 0..9) od;
    # According to the definition:
    A := (n, k) -> local j; add(binomial(k, j)*euler(j, 0)*(2*n)^j, j = 0..k):
    seq(lprint(seq(A(n, k), k = 0..6)), n = 0..7);
  • Mathematica
    A[n_, k_] := n^k (4^(k+1) HurwitzZeta[-k, 1/(4n)] - 2^(k + 1) HurwitzZeta[-k, 1/(2n)]);
  • SageMath
    from mpmath import *
    mp.dps = 32; mp.pretty = True
    def T(n, k):
        p = 2*I*(1+sum(binomial(k, j)*polylog(-j, I)*n^j for j in range(k+1)))
        return int(imag(p))
    for n in range(8): print([T(n, k) for k in range(7)])

Formula

A(n, k) = n^k*(GHZeta(k, n, 4) - GHZeta(k, n, 2)) where GHZeta(k, n, m) = m^(k+1) * HurwitzZeta(-k, 1/(m*n)) for n > 0, and T(0, k) = 1.
A(n, k) = Im(P(n, k)) where P(n, k) = 2*i*(1 + Sum_{j=0..k} binomial(k, j)*polylog(-j, i)*n^j).
A(n, k) = substitute(x = -n, P(k, x)) where P(n, x) = (1/(n + 1)) * Sum_{j=0..n+1} binomial(n + 1, j) * Bernoulli(j, 1) * (4^j - 2^j)*x^(j-1).

A339058 a(n) = 4^n*Euler(n, 1/4)*2^(valuation_{2}(n + 1)).

Original entry on oeis.org

1, -2, -3, 44, 57, -722, -2763, 196888, 250737, -5746082, -36581523, 2049374444, 7828053417, -259141449842, -2309644635483, 705775346640176, 898621108880097, -38901437271432002, -445777636063460643, 43136210244502819244, 274613643571568682777, -14685255919931552812562
Offset: 0

Author

Peter Luschny, Nov 27 2020

Keywords

Examples

			The array of the general case starts:
[k]
[1] 1,  1,  0, -1,   0,     1,     0,     -17,       0, ... [A198631]
[2] 1,  0, -1,  0,   5,     0,   -61,       0,    1385, ... [A122045]
[3] 1, -1, -2, 13,  22,  -121,  -602,   18581,   30742, ... [A156179]
[4] 1, -2, -3, 44,  57,  -722, -2763,  196888,  250737, ... [this sequence]
[5] 1, -3, -4, 99, 116, -2523, -8764, 1074243, 1242356, ... [A156182]
...
		

Crossrefs

Note the difference from A001586, A188458, and A212435.

Programs

  • Maple
    a := n -> 4^n*euler(n, 1/4)*2^padic[ordp](n+1, 2): seq(a(n), n=0..9);
  • Mathematica
    Array[4^#*EulerE[#, 1/4]*2^IntegerExponent[# + 1, 2] &, 22, 0] (* Michael De Vlieger, Mar 15 2022 *)
  • SageMath
    def euler_sum(n):
        return (-1)^n*sum(2^k*binomial(n, k)*euler_number(k) for k in (0..n))
    def a(n): return euler_sum(n) << valuation(n + 1, 2)
    print([a(n) for n in range(22)])

A214554 Triangle read by rows, coefficients of polynomials related to the Springer numbers A001586.

Original entry on oeis.org

1, -1, 2, -3, -4, 4, 11, -18, -12, 8, 57, 88, -72, -32, 16, -361, 570, 440, -240, -80, 32, -2763, -4332, 3420, 1760, -720, -192, 64, 24611, -38682, -30324, 15960, 6160, -2016, -448, 128, 250737, 393776, -309456, -161728, 63840, 19712, -5376, -1024, 256
Offset: 0

Author

Peter Luschny, Jul 30 2012

Keywords

Comments

The polynomials might be called Springer polynomials because both p{n}(0) and p{n}(1) are signed versions of the Springer numbers. p{n}(0) is the first column of the triangle (A212435 with e.g.f. exp(-x)/cosh(2x)) and p{n}(1) are the row sums (A188458 with e.g.f. exp(x)/cosh(2x)).

Examples

			[0]     1,
[1]    -1,      2,
[2]    -3,     -4,      4,
[3]    11,    -18,    -12,     8,
[4]    57,     88,    -72,   -32,   16,
[5]  -361,    570,    440,  -240,  -80,    32,
[6] -2763,  -4332,   3420,  1760, -720,  -192,   64,
[7] 24611, -38682, -30324, 15960, 6160, -2016, -448, 128.
		

Programs

  • Sage
    @CachedFunction
    def SpringerPoly(n,x) :
        if n == 0 : return 1
        return add(2^(n-k)*SpringerPoly(k,1/2)*binomial(n,k)*((x-1/2)^(n-k)+n%2-1) for k in range(n)[::2])
    R = PolynomialRing(ZZ, 'x')
    def A214554_row(n) : return R(SpringerPoly(n,x)).coeffs()
    for n in (0..7) : A214554_row(n)

Formula

p{0}(x) = 1 and for n>0
p{n}(x) = Sum_{0<=k

A326325 a(n) = 2^n*n!*([z^n] exp(x*z)*tanh(z))(1/2).

Original entry on oeis.org

0, 2, 4, -10, -56, 362, 2764, -24610, -250736, 2873042, 36581524, -512343610, -7828053416, 129570724922, 2309644635484, -44110959165010, -898621108880096, 19450718635716002, 445777636063460644, -10784052561125704810, -274613643571568682776, 7342627959965776406282
Offset: 0

Author

Peter Luschny, Jun 28 2019

Keywords

Crossrefs

Programs

  • Maple
    seq(1 - 4^n*euler(n, 1/4), n=0..21);
  • Mathematica
    p := CoefficientList[Series[Exp[x z] Tanh[z], {z, 0, 21}], z];
    norm := Table[2^n n!, {n, 0, 21}]; norm  (p /. x -> 1/2)

Formula

a(n) = 1 - 4^n*Euler(n, 1/4).
Let p(n, x) = -x^n + Sum_{k=0..n} binomial(n,k)*Euler(k)*(x+1)^(n-k) (the polynomials defined in A162660), then a(n) = 2^n*p(n, 1/2).
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