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 1-5 of 5 results.

A060096 Numerator of coefficients of Euler polynomials (rising powers).

Original entry on oeis.org

1, -1, 1, 0, -1, 1, 1, 0, -3, 1, 0, 1, 0, -2, 1, -1, 0, 5, 0, -5, 1, 0, -3, 0, 5, 0, -3, 1, 17, 0, -21, 0, 35, 0, -7, 1, 0, 17, 0, -28, 0, 14, 0, -4, 1, -31, 0, 153, 0, -63, 0, 21, 0, -9, 1, 0, -155, 0, 255, 0, -126, 0, 30, 0, -5, 1, 691, 0, -1705, 0, 2805, 0, -231, 0, 165, 0, -11, 1, 0, 2073, 0, -3410, 0, 1683, 0, -396
Offset: 0

Views

Author

Wolfdieter Lang, Mar 29 2001

Keywords

Comments

From S. Roman, The Umbral Calculus (see the reference in A048854), p. 101, (4.2.10) (corrected): E(n,x)= sum(sum(binomial(n,m)*((-1/2)^j)*j!*S2(n-m,j),j=0..k)*x^m,m=0..n), with S2(n,m)=A008277(n,m) and S2(n,0)=1 if n=0 else 0 (Stirling2).
From Wolfdieter Lang, Oct 31 2011: (Start)
This is the Sheffer triangle (2/(exp(x)+1),x) (which would be called in the above mentioned S. Roman reference Appell for (exp(t)+1)/2) (see p. 27).
The e.g.f. for the row sums is 2/(1+exp(-x)). The row sums look like A198631(n)/A006519(n+1), n>=0.
The e.g.f. for the alternating row sums is 2/(exp(x)*(exp(x)+1)). These sums look like (-1)^n*A143074(n)/ A006519(n+1).
The e.g.f. for the a-sequence of this Sheffer array is 1. The z-sequence has e.g.f. (1-exp(x))/(2*x). This z-sequence is -1/(2*A000027(n))=-1/(2*(n+1)) (see the link under A006232 for the definition of a- and z-sequences). This leads to the recurrences given below.
The alternating power sums for the first n positive integers are given by sum((-1)^(n-j)*j^k,j=1..n) = (E(k, x=n+1)+(-1)^n*E(k, x=0))/2, k>=1, n>=1,with the row polynomials E(n, x)(see the Abramowitz-Stegun reference, p. 804, 23.1.4, and an addendum in the W. Lang link under A196837).
(End)

Examples

			n\m  0    1    2    3    4    5    6    7  8  ...
0:   1
1:  -1    1
2:   0   -1    1
3:   1    0   -3    1
4:   0    1    0   -2    1
5:  -1    0    5    0   -5    1
6:   0   -3    0    5    0   -3    1
7:  17    0  -21    0   35    0   -7    1
8:   0   17    0  -28    0   14    0   -4  1
...
The rational triangle a(n,m)/A060097(n,m) starts
n\m  0    1    2    3    4    5    6    7  8  ...
0:   1
1: -1/2   1
2:   0   -1    1
3:  1/4   0  -3/2   1
4:   0    1    0   -2    1
5: -1/2   0   5/2   0  -5/2   1
6:   0   -3    0    5    0   -3    1
7: 17/8   0 -21/2   0  35/4   0  -7/2   1
8:   0   17    0  -28    0   14    0   -4  1
...

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 809.
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 20, equations 20:4:1 - 20:4:8 at pages 177-178.

Links

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

Crossrefs

Cf. A060097.

Programs

  • Maple
    A060096 := proc(n,m) coeff(euler(n,x),x,m) ; numer(%) ;end proc:
seq(seq(A060096(n,m),m=0..n),n=0..12) ; # R. J. Mathar, Dec 21 2010
  • Mathematica
    Numerator[Flatten[Table[CoefficientList[EulerE[n, x], x], {n, 0, 12}]]] (* Jean-François Alcover, Apr 29 2011 *)

Formula

E(n, x)= sum((a(n, m)/b(n, m))*x^m, m=0..n), denominators b(n, m)= A060097(n, m).
From Wolfdieter Lang, Oct 31 2011: (Start)
E.g.f. for E(n, x) is 2*exp(x*z)/(exp(z)+1).
E.g.f. of column no. m, m>=0, is 2*x^{m+1}/(m!*(exp(x)+1)).
Recurrences for E(n,m):=a(n,m)/A060097(n,m) from the Sheffer a-and z-sequence:
E(n,m)=(n/m)*E(n-1,m-1), n>=1,m>=1.
E(n,0)=-n*sum(E(n-1,j)/(2*(j+1)),j=0..n-1), n>=1, E(0,0)=1.
(see the Sheffer comments above).
(End)
E(n,m) = binomial(n,m)*sum(((-1)^j)*j!*S2(n-m,j)/2^j ,j=0..n-m), 0<=m<=n, with S2 given by A008277. From S. Roman, The umbral calculus, reference under A048854, eq. (4.2.10), p. 101, with a=1, and a misprint corrected: replace 1/k! by binomial(n,k) (also in the two preceding formulas). - Wolfdieter Lang, Nov 03 2011
The first (m=0) column of the rational triangle is conjectured to be E(n,0) = ((-1)^n)*A198631(n) / A006519(n+1). See also the first column shown in A209308 (different signs). - Wolfdieter Lang, Jun 15 2015

Extensions

Table rewritten by Wolfdieter Lang, Oct 31 2011

A119881 Expansion of e.g.f. exp(3*x)*sech(x).

Original entry on oeis.org

1, 3, 8, 18, 32, 48, 128, 528, 512, -6912, 2048, 357888, 8192, -22351872, 32768, 1903822848, 131072, -209865080832, 524288, 29088886161408, 2097152, -4951498048929792, 8388608, 1015423886523629568, 33554432, -246921480190140874752, 134217728, 70251601603944228323328
Offset: 0

Views

Author

Paul Barry, May 26 2006

Keywords

Comments

Transform of 3^n under the matrix A119879.
Also the Swiss-Knife polynomials A153641 evaluated at x=3. - Peter Luschny, Nov 23 2012

Links

Crossrefs

Cf. A027641, A027642, A119879, A119880, A153641.

Programs

  • Magma
    EulerPoly:= func< n,x | (&+[ (&+[ (-1)^j*Binomial(k,j)*(x+j)^n : j in [0..k]])/2^k: k in [0..n]]) >;
A119881:= func< n| (-2)^n*EulerPoly(n,-1) >;
[A119881(n): n in [0..40]]; // G. C. Greubel, Jun 07 2023
  • Maple
    a := proc(n) add(binomial(n,k)*bernoulli(k,1)*2^(n+k)/(n-k+1),k=0..n) end: # Peter Luschny, Dec 14 2008
a := n -> 2^n*abs(euler(n,-1)):  # Peter Luschny, Jan 25 2009
P := proc(n,x) option remember; if n = 0 then 1 else
   (n*x+2*(1-x))*P(n-1,x)+x*(1-x)*diff(P(n-1,x),x);
   expand(%) fi end:
A119881 := n -> subs(x=-1,P(n,x)):
seq(A119881(n), n=0..27);  # Peter Luschny, Mar 07 2014
  • Mathematica
    Table[2^(n+1) (Zeta[-n] (2^(n+1)-1)+1), {n,0,27}] (* Peter Luschny, Jul 16 2013 *)
Range[0, 30]! CoefficientList[Series[Exp[3 x] Sech[x], {x, 0, 30}], x] (* Vincenzo Librandi, Mar 08 2014 *)
  • PARI
    my(x='x+O('x^66)); Vec(serlaplace(exp(3*x)/cosh(x))) \\ Joerg Arndt, Apr 20 2013
  • Sage
    def skp(n, x):
    A = lambda k: 0 if (k+1)%4 == 0 else (-1)^((k+1)//4)*2^(-(k//2))
    return add(A(k)*add((-1)^v*binomial(k,v)*(v+x+1)^n for v in (0..k)) for k in (0..n))
A119881 = lambda n: skp(n,3)
[A119881(n) for n in (0..27)]  # Peter Luschny, Nov 23 2012

Formula

a(n) = Sum_{k=0..n} A119879(n,k)*3^k.
a(n) = Sum_{k=0..n} binomial(n,k)*B(k,1)*2^(n+k)/(n-k+1). Here B(k,1) are the Bernoulli number A027641(k)/A027642(k) with the exception B(1,1)=1/2. - Peter Luschny, Dec 14 2008
a(n) = 2^n |E(n,-1)| where E(n,x) are the Euler polynomials. - Peter Luschny, Jan 25 2009
The odd part of a(n) = numerator(Euler(n,2)/2) = 1, 3, 1, 9, 1, 3, 1, 33, 1, -27, 1, 699, ... (compare A143074). - Peter Luschny, Nov 23 2012
G.f.: 1/Q(0), where Q(k) = 1 - 2*x - x*(k+1)/(1+x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 19 2013
G.f.: 1/Q(0), where Q(k) = 1 - 4*x + x*(k+1)/(1-x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 20 2013
a(n) = 2^(n+1)*(zeta[-n]*(2^(n+1)-1)+1). - Peter Luschny, Jul 16 2013
E.g.f.: 2/Q(0), where Q(k) = 1 + 2^k/( 1 - 2*x/( 2*x - 2^k*(k+1)/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Dec 16 2013
a(n) = 2^(n+1)*(1+(-1)^n*(2^(n+1)-1)*Bernoulli(n+1)/(n+1)). - Vladimir Reshetnikov, Oct 21 2015

A244213 Inverse binomial transform of -2 followed by A000032(n+1).

Original entry on oeis.org

-2, 3, -1, 0, 3, -7, 14, -25, 43, -72, 119, -195, 318, -517, 839, -1360, 2203, -3567, 5774, -9345, 15123, -24472, 39599, -64075, 103678, -167757, 271439, -439200, 710643, -1149847, 1860494, -3010345, 4870843, -7881192
Offset: 0

Views

Author

Paul Curtz, Jun 23 2014

Keywords

Comments

A simple transform of a(n) is a(n) with -a(0) instead of nonzero a(0) (or -a(0) followed by a(n+1)). Example: -1 followed by A198631(n+1)/A006519(n+2). Its inverse binomial transform is -1, 3/2, -2, 9/4, -2, 3/2, -2,... = -(-1)^n*A143074(n).
Difference table of -2 followed by A000032(n+1):
-2, 1, 3, 4, 7, 11, 18,...
3, 2, 1, 3, 4, 7, 11,...
-1, -1, 2, 1, 3, 4, 7,...
0, 3, -1, 2, 1, 3, 4,...
3, -4, 3, -1, 2, 1, 3,...
-7, 7, -4, 3, -1, 2, 1,...
14, -11, 7, -4, 3, -1, 2,...
etc.
a(n) is the first column.

Links

Crossrefs

Cf. A000045, A000032, A000204.

Programs

  • PARI
    Vec(-(5*x^2-x-2)/((x+1)*(x^2-x-1)) + O(x^100)) \\ Colin Barker, Jun 23 2014

Formula

a(n) = -2, 3, -1, followed by -(-1)^n*A206417(n).
a(n) = (-1)^n* (A000032(n) - 4).
a(n+3) = -a(n) -(-1)^n*A022112(n).
a(n) = -2*a(n-1) + a(n-3). - Colin Barker, Jun 23 2014
G.f.: -(5*x^2-x-2) / ((x+1)*(x^2-x-1)). - Colin Barker, Jun 23 2014

A292706 a(n) = 1/2*((-1)^n*E(2*n-1,n) - E(2*n-1,0)), where E(n,x) is the Euler polynomial.

Original entry on oeis.org

0, 1, -31, 2060, -242972, 44808921, -11905513623, 4306834677808, -2035350070549744, 1217544864812657225, -899267301542329562375, 803729476432302540694956, -854933675015747706872042556, 1067328531318200947345698975505, -1545426104859564195269842899644047
Offset: 1

Views

Author

Vladimir Shevelev, Sep 21 2017

Keywords

References

  • M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions, 1972, Ch. 23.

Crossrefs

Cf. A143074, A157805.

Programs

  • Mathematica
    Table[((-1)^n EulerE[2n-1,n]-EulerE[2n-1,0])/2,{n,10}]
Map[Total[(Map[(-1)^# (#-1)&,Range[#]])^(2#-1)]&,Range[10]]
(* Peter J. C. Moses, Sep 21 2017 *)
  • PARI
    a(n) = sum(k=1, n-1, (-1)^(k+1)*k^(2*n-1)); \\ Michel Marcus, Sep 22 2017

Formula

a(n) = 1^(2*n-1) - 2^(2*n-1) + ... + (-1)^n*(n-1)^(2*n-1).
|a(n)| ~ 1/(1+e^(-2))*(n-1)^(2*n-1) = 0.88079707...*(n-1)^(2*n-1) as n goes to infinity.

Extensions

More terms from Peter J. C. Moses, Sep 21 2017

A244237 Numerators of the inverse binomial transform of (-1 followed by A164555(n+1)/A027642(n+1)).

Original entry on oeis.org

-1, 3, -11, 2, -61, 2, -83, 2, -61, 2, -127, 2, -6151, 2, -5, 2, -4637, 2, 42271, 2, -175241, 2, 854237, 2, -236369551, 2, 8553091, 2, -23749462769, 2, 8615841247361, 2, -7709321042237, 2, 2577687858355, 2, -26315271553057315753, 2
Offset: 0

Views

Author

Paul Curtz, Jun 23 2014

Keywords

Comments

See A244213. (The binomial transform of A198631(n)/A006519(n+1) is A143074(n)/A006519(n+1)).
Difference table of -1 followed by A164555(n+1)/A027642(n+1), see A190339:
-1, 1/2, 1/6, 0, -1/30, 0, 1/42, 0,...
3/2, -1/3, -1/6, -1/30, 1/30, 1/42, -1/42,...
-11/6, 1/6, 2/15, 1/15, -1/105, -1/21,...
2, -1/30, -1/15, -8/105, -4/105,...
-61/30, -1/30, -1/105, 4/105,...
2, 1/42, 1/21,...
-83/42, 1/42,...
2,...
etc.
The corresponding denominators to a(n) are A027642(n). See A085738.
From the second Bernoulli numbers.

Crossrefs

Cf. A190339, A235774, A244213, A085738, A164555, A027642, A143074, A006519.

Formula

(A164555(n+2) - a(n+2))/A027642(n+2) = (-1)^n*2.

Extensions

a(12)-a(37) from Jean-François Alcover
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