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 1-4 of 4 results.

A342195 a(n) = Sum_{k=0..floor(n/2)} A323833(n,k) if A323833 is read as a triangle.

Original entry on oeis.org

0, 1, 1, -5, -8, 61, 130, -1385, -3680, 50521, 160816, -2702765, -10026368, 199360981, 844583440, -19391512145, -92369507840, 2404879675441, 12722897618176, -370371188237525, -2154662195222528, 69348874393137901, 440001333689382400, -15514534163557086905, -106615331831035289600, 4087072509293123892361
Offset: 0

Views

Author

Petros Hadjicostas, Mar 04 2021

Keywords

Comments

Because A323833(n,n/2) = 0 for n even (if A323833 is read as a triangle), we also have a(n) = Sum_{k=0..ceiling((n-1)/2)} A323833(n,k).

Examples

			a(3) = -2 - 3 = -5.
a(4) = -5 - 3 = -8.
a(5) = 16 + 21 + 24 = 61.
a(6) = 61 + 45 + 24 = 130.
a(7) = -272 - 333 - 378 - 402 = -1385.
		

Crossrefs

Programs

  • PARI
    {b(n) = local(v=[1], t); if( n<0, 0, for(k=2, n+2, t=0; v = vector(k, i, if( i>1, t+= v[k+1-i]))); v[2])}; \\ Michael Somos's PARI program for A000111
    c(n) = if(n==0, 0, (-1)^(n+floor(n/2))*b(n))
    a(n) = sum(k=0, floor(n/2), sum(i=0, n-k, binomial(n-k,i)*(-1)^(k+i)*c(k+i)))

Formula

a(n) = Sum_{k=0..floor(n/2)} A323833(n-k,k) if A323833 is read as a square array (by upwards antidiagonals).
a(2*n+1) = -A028296(n+1).
a(n) = Sum_{k=0..floor(n/2)} Sum_{i=0..n-k} binomial(n-k,i) * (-1)^(k+i) * A163747(k+i).

A163747 Expansion of e.g.f. 2*exp(x)*(1-exp(x))/(1+exp(2*x)).

Original entry on oeis.org

0, -1, -1, 2, 5, -16, -61, 272, 1385, -7936, -50521, 353792, 2702765, -22368256, -199360981, 1903757312, 19391512145, -209865342976, -2404879675441, 29088885112832, 370371188237525, -4951498053124096, -69348874393137901
Offset: 0

Views

Author

Roger L. Bagula, Aug 03 2009

Keywords

Comments

The real part of the exponential expansion of 2*((1+i)/(1+i*exp(z))-1) = (-1-i)*z + (-1/2+i/2)*z^2 + (1/3+i/3)*z^3 + (5/24-5i/24)*z^4 + (-2/15-2i/15)*z^5 + ... where i is the imaginary unit.
From Paul Curtz, Mar 12 2013: (Start)
a(n) is an autosequence of the first kind; a(n) and successive differences are:
0, -1, -1, 2, 5, -16, -61;
-1, 0, 3, 3, -21, -45, 333;
1, 3, 0, -24, -24, 378, 780;
2, -3, -24, 0, 402, 402, -11214;
-5, -21, 24, 402, 0, -11616, -11616;
-16, 45, 378, -402, -11616, 0, 514608;
61, 333, -780, -11214, 11616, 514608, 0;
The main diagonal is A000004. The inverse binomial transform is the signed sequence.
The first two upper diagonals are A002832 (median Euler numbers) signed.
Sum of the antidiagonals: 0, -2, 0, 10, 0, ... = 2*A122045(n+1) (End)

Crossrefs

Variant: A163982.
Minus the zeroth column of A323833.

Programs

  • Maple
    A163747 := proc(n) exp(t)*(1-exp(t))/(1+exp(2*t)) ; coeftayl(%,t=0,n) ; 2*%*n! ; end proc: # R. J. Mathar, Sep 11 2011
    seq((euler(n) - 2^n*(2*euler(n,1)+euler(n,3/2)))/2 + 1, n=0..30); # Robert Israel, May 24 2016
    egf := (2 - 2*I)/(exp(-x) + I); ser := series(egf, x, 24):
    seq(n!*Re(coeff(ser, x, n)), n = 0..22); # Peter Luschny, Aug 09 2021
  • Mathematica
    f[t_] = (1 + I)/(1 + I*Exp[t]) - 1;
    Table[Re[2*n!*SeriesCoefficient[Series[f[t], {t, 0, 30}], n]], {n, 0, 30}]
    max = 20; Clear[g]; g[max + 2] = 1; g[k_] := g[k] = 1 - x + (4*k+3)*(k+1)*x^2 /( 1 + (4*k+5)*(k+1)*x^2 / g[k+1]); gf = -x/g[0]; CoefficientList[Series[gf, {x, 0, max}], x] (* Vaclav Kotesovec, Jan 22 2015, after Sergei N. Gladkovskii *)
    Table[(EulerE[n] - 2^n (2 EulerE[n, 1] + EulerE[n, 3/2]))/2 + 1, {n, 0, 20}] (* Benedict W. J. Irwin, May 24 2016 *)

Formula

G.f.: -x/W(0), where W(k) = 1 - x + (4*k+3)*(k+1)*x^2 / (1 + (4*k+5)*(k+1)*x^2 / W(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Jan 22 2015
a(n) ~ n! * (cos(Pi*n/2) - sin(Pi*n/2)) * 2^(n+2) / Pi^(n+1). - Vaclav Kotesovec, Apr 23 2015
a(n) = (A122045(n) - 2^n(2*Euler(n,1) + Euler(n,3/2)))/2 + 1, where Euler(n,x) is the n-th Euler polynomial. - Benedict W. J. Irwin, May 24 2016
a(n) = 2*4^n*(HurwitzZeta(-n, 1/4) - HurwitzZeta(-n, 3/4)) + HurwitzZeta(-n, 1)*(4^(n+1) - 2^(n+1)). - Peter Luschny, Jul 21 2020
a(n) = 2^n*(Euler(n, 1/2) - Euler(n, 1)). - Peter Luschny, Mar 19 2021
a(n) = ((-2)^(n + 1)*(1 - 2^(n + 1))*Bernoulli(n + 1))/(n + 1) + Euler(n). - Peter Luschny, May 06 2021
a(n) = n!*Re([x^n]((2 - 2*i)/(i + exp(-x)))). - Peter Luschny, Aug 09 2021

A002832 Median Euler numbers.

Original entry on oeis.org

1, 3, 24, 402, 11616, 514608, 32394624, 2748340752, 302234850816, 41811782731008, 7106160248346624, 1455425220196234752, 353536812021243273216, 100492698847094242603008, 33045185784774350171111424
Offset: 1

Views

Author

N. J. A. Sloane, Dec 11 1996

Keywords

Comments

There are two kinds of Euler median numbers, the 'right' median numbers (this sequence), and the 'left' median numbers (A000657).
Apparently all terms (except the initial 1) have 3-valuation 1. - F. Chapoton, Aug 02 2021

Crossrefs

Cf. A000657.
See related polynomials in A098277.
A diagonal of A323833.

Programs

  • Maple
    rr := array(1..40,1..40):rr[1,1] := 0:for i from 1 to 39 do rr[i+1,1] := (subs(x=0,diff((exp(x)-1)/cosh(x),x$i))):od: for i from 2 to 40 do for j from 2 to i do rr[i,j] := rr[i,j-1]-rr[i-1,j-1]:od:od: seq(rr[2*i-1,i-1],i=2..20); # Barbara Haas Margolius (margolius(AT)math.csuohio.edu) Feb 16 2001, corrected by R. J. Mathar, Dec 22 2010
    # alternative
    A002832 := proc(n)
        abs(A323833(n-1,n)) ;
    end proc:
    seq(A002832(n),n=1..40) ; # R. J. Mathar, Jun 11 2025
  • Mathematica
    max = 20; rr[1, 1] = 0; For[i = 1, i <= 2*max - 1, i++, rr[i + 1, 1] = D[(Exp[x] - 1)/Cosh[x], {x, i}] /. x -> 0]; For[i = 2, i <= 2*max, i++, For[j = 2, j <= i, j++, rr[i, j] = rr[i, j - 1] - rr[i - 1, j - 1]]]; Table[(-1)^i*rr[2*i - 1, i - 1], {i, 2, max}] (* Jean-François Alcover, Jul 10 2012, after Maple *)

Formula

G.f.: Sum_{n>=0} a(n)*x^n = 1/(1-1*3x/(1-1*5x/(1-2*7x/(1-2*9x/(1-3*11x/...))))).
G.f.: -1/G(0) where G(k)= x*(8*k^2+8*k+3) - 1 - (4*k+5)*(4*k+3)*(k+1)^2*x^2/G(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Aug 08 2012
a(n) ~ 2^(4*n+3/2) * n^(2*n-1/2) / (exp(2*n) * Pi^(2*n-1/2)). - Vaclav Kotesovec, Apr 23 2015

Extensions

More terms from Barbara Haas Margolius (margolius(AT)math.csuohio.edu), Feb 16 2001
Terms corrected by R. J. Mathar, Dec 22 2010

A323834 A Seidel matrix A(n,k) read by antidiagonals downwards.

Original entry on oeis.org

0, 1, 1, 1, 2, 3, -2, -1, 1, 4, -5, -7, -8, -7, -3, 16, 11, 4, -4, -11, -14, 61, 77, 88, 92, 88, 77, 63, -272, -211, -134, -46, 46, 134, 211, 274, -1385, -1657, -1868, -2002, -2048, -2002, -1868, -1657, -1383, 7936, 6551, 4894, 3026, 1024, -1024, -3026, -4894, -6551, -7934, 50521, 58457, 65008, 69902, 72928, 73952, 72928, 69902, 65008, 58457, 50523
Offset: 0

Views

Author

N. J. A. Sloane, Feb 03 2019

Keywords

Comments

The first row is a signed version of the Euler numbers A000111.
Other rows are defined by A(n+1,k) = A(n,k) + A(n,k+1).

Examples

			Read as triangle T(n,k) = A(k, n-k) (n >= 0, k = 0..n), the first few antidiagonals of the square array A are:
     0,
     1,    1,
     1,    2,    3,
    -2,   -1,    1,   4,
    -5,   -7,   -8,  -7,  -3,
    16,   11,    4,  -4, -11, -14,
    61,   77,   88,  92,  88,  77,  63,
  -272, -211, -134, -46,  46, 134, 211, 274,
  ...
From _Petros Hadjicostas_, Mar 02 2021: (Start)
Square array A(n,k) (with rows n >= 0 and columns k >= 0) begins:
     0,   1,   1,    -2,    -5,    16,     61,     -272,    -1385, ...
     1,   2,  -1,    -7,    11,    77,   -211,    -1657,     6551, ...
     3,   1,  -8,     4,    88,  -134,  -1868,     4894,    65008, ...
     4,  -7,  -4,    92,   -46, -2002,   3026,    69902,  -179806, ...
    -3, -11,  88,    46, -2048,  1024,  72928,  -109904, -3784448, ...
   -14,  77, 134, -2002, -1024, 73952, -36976, -3894352,  5860016, ...
   ... (End)
		

Crossrefs

Cf. A000111, A000657 (next-to-main diagonal), A323833.

Programs

  • PARI
    {b(n) = local(v=[1], t); if( n<0, 0, for(k=2, n+2, t=0; v = vector(k, i, if( i>1, t+= v[k+1-i]))); v[2])}; \\ Michael Somos's PARI program for A000111.
    c(n) = if(n==0, 0, (-1)^floor((n-1)/2)*b(n))
    A(n, k) = sum(i=0, n, binomial(n, i)*c(k+i)) \\ Petros Hadjicostas, Mar 02 2021

Formula

From Petros Hadjicostas, Mar 02 2021: (Start)
Formulas for the square array A(n,k) (n, k >= 0):
A(0,k) = (-1)^floor((k-1)/2)*A000111(k) for k > 0 with A(0,0) = 0.
A(n,k) = Sum_{i=0..n} binomial(n, i)*A(0,k+i) for n, k >= 0.
A(n,n)/2 = A(n+1,n) = +/- A000657(n) for n > 0.
Bivariate e.g.f.: Sum_{n,k >= 0} A(n,k)*(x^n/n!)*(y^k/k!) = (-sech(x + y) + tanh(x + y) + 1)*exp(x).
Formulas for the triangular array T(n,k) = A(k,n-k) (n >= 0, 0 <= k <= n):
T(n,k) = T(n-1,k-1) + T(n,k-1) for 1 <= k <= n with T(n,0) = (-1)^floor((n-1)/2) * A000111(n) for n > 0 and T(0,0) = 0.
T(n,k) = Sum_{i=0..k} binomial(k,i)*T(n-k+i,0) for 0 <= k <= n. (End)

Extensions

Typo corrected by and more terms from Petros Hadjicostas, Mar 02 2021
Showing 1-4 of 4 results.