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 1-3 of 3 results.

A000464 Expansion of e.g.f. sin(x)/cos(2*x).

Original entry on oeis.org

1, 11, 361, 24611, 2873041, 512343611, 129570724921, 44110959165011, 19450718635716001, 10784052561125704811, 7342627959965776406281, 6023130568334172003579011, 5858598896811701995459355761, 6667317162352419006959182803611, 8776621742176931117228228227924441
Offset: 0

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Keywords

Comments

From Peter Bala, Dec 22 2021: (Start)
Conjectures:
1) 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 [11, 4, 20, 10, 17, 1, 11, 4, 20, 10, 17, 1, 11, 4, 20, 10, 17, 1, ...] with an apparent period of length 6, which divides phi(21) = 12.
2) For i >= 0, define a_i(n) = a(n+i). Then for each 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 the expansion of exp(Sum_{n >= 1} a_i(n)*x^n/n) has integer coefficients.
3) a(m*n) == a(m)^n (mod 2^k) for k = 2*v_2(m) + 4, where v_p(i) denotes the p-adic valuation of i.
4)(i) a(2*m*n) == a(n)^(2*m) (mod 2^k) for k = v_2(m) + 4
(ii) a((2*m+1)*n) == a(n)^(2*m+1) (mod 2^k) for k = v_2(m) + 4. (End)

References

  • H. Cohen, Number Theory - Volume II: Analytic and Modern Tools, Graduate Texts in Mathematics. Springer-Verlag.
  • 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

Row 2 of A235606.
Cf. A064073. Bisection of A000822, A001586.

Programs

  • Maple
    a := n -> (-1)^n*2^(6*n+4)*(Zeta(0, -2*n-1, 5/8)-Zeta(0, -2*n-1, 7/8)):
    seq(a(n), n=0..12); # Peter Luschny, Oct 15 2015
  • Mathematica
    With[{nn=30},Take[CoefficientList[Series[Sin[x]/Cos[2x],{x,0,nn}],x] Range[0,nn-1]!,{2,-1,2}]] (* Harvey P. Dale, Mar 23 2012 *)
    nmax = 15; km0 = 10; d[n_, km_] := Round[(2^(4n-1/2) (2n-1)! Sum[ JacobiSymbol[2, 2k+1]/(2k+1)^(2n), {k, 0, km}])/Pi^(2n)]; dd[km_] := dd[km] = Table[d[n, km], {n, 1, nmax}]; dd[km0]; dd[km = 2*km0]; While[dd[km] != dd[km/2, km = 2*km]]; A000464 = dd[km] (* Jean-François Alcover, Feb 08 2016 *)
  • PARI
    a(n)=if(n<0, 0, n+=n+1; n!*polcoeff(sin(x+x*O(x^n))/cos(2*x+x*O(x^n)),n)) /* Michael Somos, Feb 09 2006 */

Formula

E.g.f.: Sum_{k>=0} a(k)x^(2k+1)/(2k+1)! = sin(x)/cos(2x).
a(n) = (-1)^n*L(X,-2n+1) where L(X,z) is the Dirichlet L-function L(X,z) = Sum_{k>=0} X(k)/k^z and where X(k) is the Dirichlet character Legendre(k,2) which begins 1,0,-1,0,-1,0,1,0,1,0,-1,0,-1,0,1,0,1,0,-1,0.... - Benoit Cloitre, Mar 22 2009 [This Dirichlet character is A091337. - Jianing Song, Oct 22 2023]
From Peter Bala, Mar 24 2009: (Start)
Basic hypergeometric generating function:
2*exp(-t)*Sum_{n = 0..inf} (Product_{k = 1..n} (1-exp(-16*k*t))/Product_{k = 1..n+1} (1+exp(-(16*k-8)*t))) = 1 + 11*t + 361*t^2/2! + 24611*t^3/3! + .... For other sequences with generating functions of a similar type see A000364, A002105, A002439, A079144 and A158690.
a(n) = (-1)^(n+1)*L(-2*n-1), where L(s) is the Dirichlet L-function L(s) = 1 - 1/3^s - 1/5^s + 1/7^s + - - + ... [Andrews et al., Theorem 5]. (End)
From Peter Bala, Jun 18 2009: (Start)
a(n) = (-1)^n*B_(2*n+2)(X)/(2*n+2), where B_n(X) denotes the X-Bernoulli number with X a Dirichlet character modulus 8 given by X(8*n+1) = X(8*n+7) = 1, X(8*n+3) = X(8*n+5) = -1 and X(2*n) = 0. See A161722 for the values of B_n(X).
For the theory and properties of the generalized Bernoulli numbers B_n(X) and the associated generalized Bernoulli polynomials B_n(X,x) see [Cohen, Section 9.4].
The present sequence also occurs in the evaluation of the finite sum of powers Sum_{i = 0..m-1} {(8*i+1)^n - (8*i+3)^n - (8*i+5)^n + (8*i+7)^n}, n = 1,2,... - see A151751 for details. (End)
G.f. 1/G(0) where G(k) = 1 + x - x*(4*k+3)*(4*k+4)/(1 - (4*k+4)*(4*k+5)*x/G(k+1)); (continued fraction, 2-step). - Sergei N. Gladkovskii, Aug 11 2012
G.f.: 1/E(0) where E(k) = 1 - 11*x - 32*x*k*(k+1) - 16*x^2*(k+1)^2*(4*k+3)*(4*k+5)/E(k+1) (continued fraction, 1-step). - Sergei N. Gladkovskii, Sep 17 2012
a(n) ~ (2*n+1)! * 2^(4*n+7/2) / Pi^(2*n+2). - Vaclav Kotesovec, May 03 2014
a(n) = (-1)^n*2^(6*n+4)*(Zeta(-2*n-1,5/8)-Zeta(-2*n-1,7/8)). - Peter Luschny, Oct 15 2015
From Peter Bala, May 11 2017: (Start)
G.f. A(x) = 1 + 11*x + 361*x^2 + ... = 1/(1 + x - 12*x/(1 - 20*x/(1 + x - 56*x/(1 - 72*x/(1 + x - ... - 4*n*(4*n - 1)*x/(1 - 4*n*(4*n + 1)*x/((1 + x) - ...))))))).
A(x) = 1/(1 + 9*x - 20*x/(1 - 12*x/(1 + 9*x - 72*x/(1 - 56*x/(1 + 9*x - ... - 4*n*(4*n + 1)*x/(1 - 4*n*(4*n - 1)*x/(1 + 9*x - ...))))))).
It follows that the first binomial transform of A(x) and the ninth binomial transform of A(x) have continued fractions of Stieltjes-type (S-fractions). (End)
a(n) = (-1)^(n+1)*4^(2*n+1)*E(2*n+1,1/4), where E(n,x) is the n-th Euler polynomial. Cf. A002439. - Peter Bala, Aug 13 2017
From Peter Bala, Dec 04 2021: (Start)
F(x) = exp(x)*(exp(2*x) - 1)/(exp(4*x) + 1) = x - 11*x^3/3! + 361x^5/5! - 24611*x^7/7! + ... is the e.g.f. for the sequence [1, 0, -11, 0, 361, 0, -24611, 0, ...], a signed and aerated version of this sequence.
The binomial transform exp(x)*F(x) = x + 2*x^2/2! - 8*x^3/3! - 40*x^4/4! + + - - is an e.g.f. for a signed version of A000828 (omitting the initial term). (End)
From Peter Bala, Dec 22 2021: (Start)
a(1) = 1, a(n) = (-1)^(n-1) - Sum_{k = 1..n} (-4)^k*C(2*n-1,2*k)*a(n-k).
a(n) == 1 (mod 10); a(5*n+1) == 0 mod(11);
a(n) == - 23^(n+1) (mod 108); a(n) == (7^2)*59^n (mod 144);
a(n) == 11^n (mod 240); a(n) == (11^2)*131^n (mod 360). (End)

Extensions

Better description, new reference, Aug 15 1995

A117442 Number triangle read by rows, related to exp(x)/(cos(x) + sin(x)).

Original entry on oeis.org

1, -1, 1, 3, -2, 1, -11, 9, -3, 1, 57, -44, 18, -4, 1, -361, 285, -110, 30, -5, 1, 2763, -2166, 855, -220, 45, -6, 1, -24611, 19341, -7581, 1995, -385, 63, -7, 1, 250737, -196888, 77364, -20216, 3990, -616, 84, -8, 1, -2873041, 2256633, -885996, 232092, -45486, 7182, -924, 108, -9, 1
Offset: 0

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Author

Paul Barry, Mar 16 2006

Keywords

Examples

			Triangle begins
       1;
      -1,     1;
       3,    -2,     1;
     -11,     9,    -3,    1;
      57,   -44,    18,   -4,    1;
    -361,   285,  -110,   30,   -5,  1;
    2763, -2166,   855, -220,   45, -6,  1;
  -24611, 19341, -7581, 1995, -385, 63, -7, 1;
		

Crossrefs

Inverse of A117440.
Second column contains A161722 as subsequence.

Programs

  • Maple
    A117442_row := proc(n) 2^n*add(binomial(n,k)*euler(k)*((x+1)/2)^(n-k), k=0..n);
    seq((-1)^(n-j)*abs(coeff(%,x,j)),j=0..n) end:
    seq(print(A117442_row(n)),n=0..5);  # Peter Luschny, Jun 08 2013
  • Mathematica
    row[n_] := row[n] = 2^n Sum[Binomial[n, k] EulerE[k] ((x+1)/2)^(n-k), {k, 0, n}];
    T[n_, k_] := (-1)^(n-k) Abs[Coefficient[row[n], x, k]];
    Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* Jean-François Alcover, Jun 13 2019, from Maple *)
    Table[(-1)^(n-k)*Binomial[n, k]*Abs[Numerator[EulerE[n-k, 1/4]]], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Jun 02 2021 *)
  • PARI
    E(n) = 2^n*2^(n+1)*(subst(bernpol(n+1, x), x, 3/4) - subst(bernpol(n+1, x), x, 1/4))/(n+1); \\ A122045
    p(n) = 2^n*sum(k=0, n, binomial(n,k)*E(k)*((x+1)/2)^(n-k));
    row(n) = my(rp=p(n)); vector(n+1, k, k--; (-1)^(n-k)*abs(polcoeff(rp, k))); \\ Michel Marcus, Nov 16 2020
    
  • Sage
    def f(n): return (1/4)^n*sum( binomial(n, j)*2^j*euler_number(j) for j in (0..n)) # f(n) = Euler(n, 1/4)
    def A117442(n,k): return (-1)^(n+k)*binomial(n,k)*abs(numerator(f(n-k)))
    flatten([[A117442(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 02 2021

Formula

T(n, 0) = (-1)^n*A001586(n).
Sum_{k=0..n} T(n, k) = A117443(n).
Column k has e.g.f. (x^k/k!)/(cos(x) + sin(x)).
Apart from signs the T(n,k) are the coefficients of the polynomials p(n, x) = 2^n*Sum_{k=0..n} binomial(n,k)*euler(k)*((x+1)/2)^(n-k). - Peter Luschny, Jun 08 2013
From G. C. Greubel, Jun 02 2021: (Start)
T(n, k) = (-1)^(n+k) * binomial(n, k) * abs(numerator( Euler(n-k, 1/4) )), where Euler(n, x) is the Euler number polynomial.
T(n, n) = 1.
T(n, n-1) = -A000027(n) = -binomial(n+1, 1).
T(n, n-2) = A045943(n+1) = 3*binomial(n+2, 2).
T(n, n-3) = -A111080(n) = -11*binomial(n+3, 3).
T(j, k) = (-1)^k * binomial(j+k, k) * abs(numerator( Euler(k, 1/4) )) (columns).
T(n, n-j) = (-1)^n * binomial(n+j, j) * abs(numerator( Euler(n, 1/4) )) (downward diagonals). (End)
The pair of triangles P*((I + P^4)/2)^(-1) and P^3*((I + P^4)/2)^(-1), where P denotes Pascal's triangle A007318, give the present triangle but with a different pattern of signs. - Peter Bala, Mar 07 2024

A151751 Triangle of coefficients of generalized Bernoulli polynomials associated with a Dirichlet character modulus 8.

Original entry on oeis.org

2, 0, 6, -44, 0, 12, 0, -220, 0, 20, 2166, 0, -660, 0, 30, 0, 15162, 0, -1540, 0, 42, -196888, 0, 60648, 0, -3080, 0, 56, 0, -1771992, 0, 181944, 0, -5544, 0, 72, 28730410, 0, -8859960, 0, 454860, 0, -9240, 0, 90
Offset: 2

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Author

Peter Bala, Jun 17 2009

Keywords

Comments

Let X be a periodic arithmetical function with period m. The generalized Bernoulli polynomials B_n(X,x) attached to X are defined by means of the generating function
(1)... t*exp(t*x)/(exp(m*t)-1) * sum {r = 0..m-1} X(r)*exp(r*t)
= sum {n = 0..inf} B_n(X,x)*t^n/n!.
For the theory and properties of these polynomials see [Cohen, Section 9.4]. In the present case, X is chosen to be the Dirichlet character modulus 8 given by
(2)... X(8*n+1) = X(8*n+7) = 1; X(8*n+3) = X(8*n+5) = -1; X(2*n) = 0.
Cf. A153641.

Examples

			The triangle begins
n\k|........0.......1........2.......3......4.......5.......6
=============================================================
.2.|........2
.3.|........0.......6
.4.|......-44.......0.......12
.5.|........0....-220........0......20
.6.|.....2166.......0.....-660.......0......30
.7.|........0...15162........0...-1540.......0.....42
.8.|..-196888.......0....60648.......0...-3080......0......56
...
		

References

  • H. Cohen, Number Theory - Volume II: Analytic and Modern Tools, Graduate Texts in Mathematics. Springer-Verlag.

Crossrefs

Programs

  • Maple
    with(gfun):
    for n from 2 to 10 do
    Genbernoulli(n,x) := 8^(n-1)*(bernoulli(n,(x+1)/8)-bernoulli(n,(x+3)/8)-bernoulli(n,(x+5)/8)+bernoulli(n,(x+7)/8));
    seriestolist(series(Genbernoulli(n,x),x,10))
    end do;

Formula

TABLE ENTRIES
(1)... T(2*n,2*k+1) = 0, T(2*n+1,2*k) = 0;
(2)... T(2*n,2*k) = (-1)^(n-k-1)*C(2*n,2*k)*2*(n-k)*A000464(n-k-1);
(3)... T(2*n+1,2*k+1) = (-1)^(n-k-1)*C(2*n+1,2*k+1)*2*(n-k)*A000464(n-k-1);
where C(n,k) = binomial(n,k).
GENERATING FUNCTION
The e.g.f. for these generalized Bernoulli polynomials is
(4)... t*exp(x*t)*(exp(t)-exp(3*t)-exp(5*t)+exp(7*t))/(exp(8*t)-1)
= sum {n = 2..inf} B_n(X,x)*t^n/n! = 2*t^2/2! + 6*x*t^3/3! + (12*x^2 - 44)*t^4/4! + ....
In terms of the ordinary Bernoulli polynomials B_n(x)
(5)... B_n(X,x) = 8^(n-1)*{B_n((x+1)/8) - B_n((x+3)/8) - B_n((x+5)/8) + B_n((x+7)/8)}.
The B_n(X,x) are Appell polynomials of the form
(6)... B_n(X,x) = sum {j = 0..n} binomial(n,j)*B_j(X,0)*x*(n-j).
The sequence of generalized Bernoulli numbers
(7)... [B_n(X,0)]n>=2 = [2,0,-44,0,2166,0,...]
has the e.g.f.
(8)... t*(exp(t)-exp(3*t)-exp(5*t)+exp(7*t))/(exp(8*t)-1),
which simplifies to
(9)... t*sinh(t)/cosh(2*t).
Hence
(10)... B_(2*n)(X,0) = (-1)^(n+1)*2*n*A000464(n-1); B_(2*n+1)(X,0) = 0.
The sequence {B_(2*n)(X,0)}n>=2 is A161722.
RELATION WITH TWISTED SUMS OF POWERS
The generalized Bernoulli polynomials may be used to evaluate sums of k-th powers twisted by the function X(n). For the present case the result is
(11)... sum{n = 0..8*N-1} X(n)*n^k = 1^k-3^k-5^k+7^k- ... +(8*N-1)^k
= [B_(k+1)(X,8*N) - B_(k+1)(X,0)]/(k+1)
For the proof, apply [Cohen, Corollary 9.4.17 with m = 8 and x = 0].
MISCELLANEOUS
(12)... Row sums [2, 6, -32, ...] = (-1)^(1+binomial(n,2))*A109572(n)
= (-1)^(1+binomial(n,2))*n*A000828(n-1) = (-1)^(1+binomial(n,2))*n* 2^(n-2)*A000111(n-1).
Showing 1-3 of 3 results.