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.

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A226158 a(n) = 2*n*(2^n - 1)*zeta(1-n) where in the case n=0 the limit is understood, zeta(s) the Riemann zeta function.

Original entry on oeis.org

0, -1, -1, 0, 1, 0, -3, 0, 17, 0, -155, 0, 2073, 0, -38227, 0, 929569, 0, -28820619, 0, 1109652905, 0, -51943281731, 0, 2905151042481, 0, -191329672483963, 0, 14655626154768697, 0, -1291885088448017715, 0, 129848163681107301953
Offset: 0

Views

Author

Peter Luschny, Jun 28 2013

Keywords

Comments

Also known as the Genocchi numbers, apart from a(0) and a(1) same as A036968.
Consider the difference table of a(n), which is a variant of Seidel's Genocchi table A014781:
0 -1 -1 0 1 0 -3 0 17
-1 0 1 1 -1 -3 3 17 -17
1 1 0 -2 -2 6 14 -34 -138
0 -1 -2 0 8 8 -48 -104 448
-1 -1 2 8 0 -56 -56 552 1160
0 3 6 -8 -56 0 608 608 -8832
3 3 -14 -48 56 608 0 -9440 -9440
0 -17 -34 104 552 -608 -9440 0 198272
-17 -17 138 448 -1160 -8832 9440 198272 0
a(n) is an autosequence: its inverse binomial transform is the sequence signed (see A181722). The first column (inverse binomial transform) is 0, followed by -A036968. - Paul Curtz, Jul 22 2013
a(n+1) = p(0) where p(x) is the unique degree-n polynomial such that p(k) = a(k) for k = 1, ..., n+1. - Michael Somos, Apr 23 2014

Examples

			G.f. = - x - x^2 + x^4 - 3*x^6 + 17*x^8 - 155*x^10 + 2073*x^12 - 38227*x^14 + ...

Links

Crossrefs

Cf. A000367, A005439, A014781, A036968, A090648, A181722, A227577.

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 50); [0] cat Coefficients(R!(Laplace( -2*x/(1+Exp(-x)) ))); // G. C. Greubel, Apr 22 2023
  • Maple
    seq(n!*coeff(series(-2*x/(1+exp(-x)), x, 34), x, n), n=0..32);
# Second program:
A226158 := proc(n) local f; f := z -> Zeta(1-z)*2*z*(2^z-1);
if n=0 then limit(f(z), z=0) else f(n) fi end: seq(A226158(n), n=0..32);
  • Mathematica
    a[0]=0; a[1]= -1; a[n_]:= n*EulerE[n-1, 0]; Table[a[n], {n,0,32}] (* Jean-François Alcover, Sep 12 2013 *)
(* Programs from Michael Somos, Apr 23 2014 *)
a[n_]:= If[n<1, 0, -n*EulerE[n-1, 1]];
a[n_]:= If[n<0, 0, 2*(1-2^n)*BernoulliB[n,1]]; (* End *)
Table[2*n*PolyLog[1-n, -1], {n,0,32}] (* Peter Luschny, Aug 17 2021 *)
  • PARI
    my(x='x+O('x^40)); concat([0], Vec(serlaplace(-2*x/(1+exp(-x))))) \\ G. C. Greubel, Jan 19 2018
  • Sage
    def A226158(n): return -2*n*zeta(1-n)*(1-2^n) if n != 0 else 0
[A226158(n) for n in (0..32)]
# Alternatively:
def A226158_list(len):
    e, f, R, C = 4, 1, [0], [1]+[0]*(len-1)
    for n in (2..len-1):
        for k in range(n, 0, -1):
            C[k] = -C[k-1] / (k+1)
        C[0] = -sum(C[k] for k in (1..n))
        R.append((2-e)*f*C[0])
        f *= n; e *= 2
    return R
print(A226158_list(34)) # Peter Luschny, Feb 22 2016

Formula

E.g.f.: -2*x/(1+exp(-x)).
a(2n) = -A000367(n)*A090648(n). - Paul Curtz, Jul 22 2013
E.g.f.: -2*x/(1+exp(-x))= -2 - 2*T(0), where T(k) = 4*k-1 + x/( 2 - x/( 4*k+1 + x/( 2 - x/T(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Nov 23 2013
G.f.: conjecture: -x/Q(0),where Q(k) = 1 - x*(k+1)/(1 + x*(k+1)/(1 - x*(k+1)/(1 + x*(k+1)/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Nov 23 2013
a(n) = 2*(1 - 2^n)*Bernoulli(n, 1). - Peter Luschny, Apr 16 2014
a(n) = -n*Euler(n - 1, 1). - Michael Somos, Apr 23 2014
a(n) = 2^n*(Bernoulli(n, 1/2) - Bernoulli(n, 1)). - Peter Luschny, Jul 10 2020
a(n) = 2*n*PolyLog[1 - n, -1] - Peter Luschny, Aug 17 2021

A189731 a(n) = numerator of B(0,n) where B(n,n) = 0, B(n-1,n) = 1/n, and B(m,n) = B(m-1,n+1) - B(m-1,n).

Original entry on oeis.org

0, 1, 1, 3, 2, 17, 4, 23, 25, 61, 18, 107, 40, 421, 1363, 1103, 210, 5777, 492, 7563, 24475, 19801, 2786, 103681, 33552, 135721, 146401, 355323, 39650, 1860497, 97108, 2435423, 2627065, 6376021, 20633238, 11128427, 1459960, 43701901
Offset: 0

Views

Author

Paul Curtz, Apr 26 2011

Keywords

Comments

Square array B(m,n) begins:
0, 1/1, 1/1, 3/2, 2/1, 17/6, ...
1/1, 0, 1/2, 1/2, 5/6, 7/6, ...
-1/1, 1/2, 0, 1/3, 1/3, 7/12, ...
3/2, -1/2, 1/3, 0, 1/4, 1/4, ...
-2/1, 5/6, -1/3, 1/4, 0, 1/5, ...
17/6, -7/6, 7/12, -1/4, 1/5, 0, ...
The inverse binomial transform of B(0,n) gives B(n,0) and thus it is an eigensequence in the sense that it remains the same (up to a sign) under inverse binomial transform.
The bisection of B(0,n) (odd part) gives A175385/A175386, and thus a(2*n+1) = A175385(n+1).

Links

Crossrefs

Cf. A000204, A242926 (denominators).
Cf. A174341, A177690, A181722.

Programs

  • Maple
    B:= proc(m, n) option remember;
      if m=n then 0
    elif n=m+1 then 1/n
    elif n>m then B(m, n-1) +B(m+1, n-1)
    else B(m-1, n+1) -B(m-1, n)
      fi
    end:
a:= n-> numer(B(0, n)):
seq(a(n), n=0..50);  # Alois P. Heinz, Apr 29 2011
  • Mathematica
    Rest[Numerator[Abs[CoefficientList[Normal[Series[Log[1 - x^2/(1 + x)], {x, 0, 40}]], x]]]] (* Vaclav Kotesovec, Jul 07 2020 *)
Table[Numerator[(LucasL[n]-1)/n],{n,1,38}] (* Artur Jasinski, Oct 21 2022 *)

Formula

Numerator of (A000204(n) - 1)/n. - Artur Jasinski, Oct 21 2022

A227577 Square array read by antidiagonals, A(n,k) the numerators of the elements of the difference table of the Euler polynomials evaluated at x=1, for n>=0, k>=0.

Original entry on oeis.org

1, -1, 1, 0, -1, 0, 1, 1, -1, -1, 0, 1, 1, 1, 0, -1, -1, -1, 1, 1, 1, 0, -1, -1, -5, -1, -1, 0, 17, 17, 13, 5, -5, -13, -17, -17, 0, 17, 17, 47, 13, 47, 17, 17, 0, -31, -31, -107, -73, -13, 13, 73, 107, 31, 31, 0, -31, -31, -355
Offset: 0

Views

Author

Paul Curtz, Jul 16 2013

Keywords

Comments

The difference table of the Euler polynomials evaluated at x=1:
1, 1/2, 0, -1/4, 0, 1/2, 0, -17/8, ...
-1/2, -1/2, -1/4, 1/4, 1/2, -1/2, -17/8, 17/8, ...
0, 1/4, 1/2, 1/4; -1, -13/8, 17/4, 107/8, ...
1/4, 1/4, -1/4, -5/4, -5/8, 47/8, 73/8, -355/8, ...
0, -1/2, -1, 5/8 13/2, 13/4, -107/2, -655/8, ...
-1/2, -1/2, 13/8, 47/8, -13/4, -227/4, -227/8, 5687/8, ...
0, 17/8, 17/4, -73/8, -107/2, 227/8, 2957/4, 2957/8, ...
17/8, 17/8, -107/8, -355/8, 655/8, 5687/8, -2957/8, -107125/8, ...
To compute the difference table, take
1, 1/2;
-1/2;
The next term is always half of the sum of the antidiagonals. Hence (-1/2 + 1/2 = 0)
1, 1/2, 0;
-1/2, -1/2;
0;
The first column (inverse binomial transform) lists the numbers (1, -1/2, 0, 1/4, ..., not in the OEIS; corresponds to A027641/A027642). See A209308 and A060096.
A198631(n)/A006519(n+1) is an autosequence. See A181722.
Note the main diagonal: 1, -1/2, 1/2, -5/4, 13/2, -227/4, 2957/4, -107125/8, .... (See A212196/A181131.)
This twice the first upper diagonal. The autosequence is of the second kind.
From 0, -1, the algorithm gives A226158(n), full Genocchi numbers, autosequence of the first kind.
The difference table of the Bernoulli polynomials evaluated at x=1 is (apart from signs) A085737/A085738 and its analysis by Ludwig Seidel was discussed in the Luschny link. - Peter Luschny, Jul 18 2013

Examples

			Read by antidiagonals:
    1;
  -1/2,  1/2;
    0,  -1/2,   0;
   1/4,  1/4, -1/4, -1/4;
    0,   1/4,  1/2,  1/4,   0;
  -1/2, -1/2, -1/4,  1/4,  1/2,  1/2;
    0,  -1/2, - 1,  -5/4,  -1,  -1/2,   0;
  ...
Row sums: 1, 0, -1/2, 0, 1, 0, -17/4, 0, ... = 2*A198631(n+1)/A006519(n+2).
Denominators: 1, 1, 2, 1, 1, 1, 4, 1, ... = A160467(n+2)?

Links

Crossrefs

Cf. A164555/A027642 in A190339.

Programs

  • Maple
    DifferenceTableEulerPolynomials := proc(n) local A,m,k,x;
A := array(0..n,0..n); x := 1;
for m from 0 to n do for k from 0 to n do A[m,k]:= 0 od od;
for m from 0 to n do A[m,0] := euler(m,x);
   for k from m-1 by -1 to 0 do
      A[k,m-k] := A[k+1,m-k-1] - A[k,m-k-1] od od;
LinearAlgebra[Transpose](convert(A, Matrix)) end:
DifferenceTableEulerPolynomials(7);  # Peter Luschny, Jul 18 2013
  • Mathematica
    t[0, 0] = 1; t[0, k_] := EulerE[k, 1]; t[n_, 0] := -t[0, n]; t[n_, k_] := t[n, k] = t[n-1, k+1] - t[n-1, k]; Table[t[n-k, k] // Numerator, {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 18 2013 *)
  • Sage
    def DifferenceTableEulerPolynomialsEvaluatedAt1(n) :
    @CachedFunction
    def ep1(n):          # Euler polynomial at x=1
        if n < 2: return 1 - n/2
        s = add(binomial(n,k)*ep1(k) for k in (0..n-1))
        return 1 - s/2
    T = matrix(QQ, n)
    for m in range(n) :  # Compute difference table
        T[m,0] = ep1(m)
        for k in range(m-1,-1,-1) :
            T[k,m-k] = T[k+1,m-k-1] - T[k,m-k-1]
    return T
def A227577_list(m):
    D = DifferenceTableEulerPolynomialsEvaluatedAt1(m)
    return [D[k,n-k].numerator() for n in range(m) for k in (0..n)]
A227577_list(12)  # Peter Luschny, Jul 18 2013

Extensions

Corrected by Jean-François Alcover, Jul 17 2013
Showing 1-3 of 3 results.

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