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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A173253 Partial sums of A000111.

Original entry on oeis.org

1, 2, 3, 5, 10, 26, 87, 359, 1744, 9680, 60201, 413993, 3116758, 25485014, 224845995, 2128603307, 21520115452, 231385458428, 2636265133869, 31725150246701, 402096338484226, 5353594391608322, 74702468784746223, 1090126355291598575, 16604660518848685480
Offset: 0

Views

Author

Jonathan Vos Post, Feb 14 2010

Keywords

Comments

Partial sums of Euler or up/down numbers. Partial sums of expansion of sec x + tan x. Partial sums of number of alternating permutations on n letters.

Examples

			a(22) = 1 + 1 + 1 + 2 + 5 + 16 + 61 + 272 + 1385 + 7936 + 50521 + 353792 + 2702765 + 22368256 + 199360981 + 1903757312 + 19391512145 + 209865342976 + 2404879675441 + 29088885112832 + 370371188237525 + 4951498053124096 + 69348874393137901.
		

Crossrefs

Programs

  • Maple
    b:= proc(u, o) option remember;
          `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
        end:
    a:= proc(n) option remember;
          `if`(n<0, 0, a(n-1))+ b(n, 0)
        end:
    seq(a(n), n=0..25);  # Alois P. Heinz, Oct 27 2017
  • Mathematica
    With[{nn=30},Accumulate[CoefficientList[Series[Sec[x]+Tan[x],{x,0,nn}],x] Range[0,nn]!]] (* Harvey P. Dale, Feb 26 2012 *)
  • Python
    from itertools import accumulate
    def A173253(n):
        if n<=1:
            return n+1
        c, blist = 2, (0,1)
        for _ in range(n-1):
            c += (blist := tuple(accumulate(reversed(blist),initial=0)))[-1]
        return c # Chai Wah Wu, Apr 16 2023

Formula

a(n) = SUM[i=0..n] A000111(i) = SUM[i=0..n] (2^i|E(i,1/2)+E(i,1)| where E(n,x) are the Euler polynomials).
G.f.: (1 + x/Q(0))/(1-x),m=+4,u=x/2, where Q(k) = 1 - 2*u*(2*k+1) - m*u^2*(k+1)*(2*k+1)/( 1 - 2*u*(2*k+2) - m*u^2*(k+1)*(2*k+3)/Q(k+1) ) ; (continued fraction). - Sergei N. Gladkovskii, Sep 24 2013
G.f.: 1/(1-x) + T(0)*x/(1-x)^2, where T(k) = 1 - x^2*(k+1)*(k+2)/(x^2*(k+1)*(k+2) - 2*(1-x*(k+1))*(1-x*(k+2))/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 20 2013
a(n) ~ 2^(n+2)*n!/Pi^(n+1). - Vaclav Kotesovec, Oct 27 2016

A236935 The infinite Seidel matrix H read by antidiagonals upwards; H = (H(n,k): n,k >= 0).

Original entry on oeis.org

1, 0, -1, -1, -1, 0, 0, 1, 2, 2, 5, 5, 4, 2, 0, 0, -5, -10, -14, -16, -16, -61, -61, -56, -46, -32, -16, 0, 0, 61, 122, 178, 224, 256, 272, 272, 1385, 1385, 1324, 1202, 1024, 800, 544, 272, 0, 0, -1385, -2770, -4094, -5296, -6320, -7120, -7664, -7936, -7936, -50521, -50521, -49136, -46366, -42272, -36976, -30656, -23536, -15872, -7936, 0
Offset: 0

Views

Author

N. J. A. Sloane, Feb 17 2014

Keywords

Comments

This is, in essence, a signed version of the triangle in A008280.

Examples

			Array begins:
     1   -1    0    2    0 -16   0 272 0 ...
     0   -1    2    2  -16 -16 272 272 ...
    -1    1    4  -14  -32 256 544 ...
     0    5  -10  -46  224 800 ...
     5   -5  -56  178 1024 ...
     0  -61  122 1202 ...
   -61   61 1324 ...
     0 1385 ...
  1385 ...
  ...
		

Crossrefs

Programs

  • PARI
    a(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 */
    H(n,k) = sum(i=0, k, (-1)^i*binomial(k,i)*a(n+k-i)) /* Petros Hadjicostas, Feb 21 2021 */
    /* Second PARI program (same a(n) for A122045 as above) */
    H(n,k) = (-1)^(n+k)*sum(i=0, k, binomial(k,i)*a(n+i)) /* Petros Hadjicostas, Feb 21 2021 */

Formula

From Petros Hadjicostas, Feb 20 2021: (Start)
H(n,0) = A122045(n).
H(0,k) = (-1)^n*A155585(n).
H(n,k) = Sum_{i=0..n} binomial(n,i)*H(0,k+i).
H(n,k) = Sum_{i=0..k} (-1)^i*binomial(k,i)*H(n+k-i,0).
H(n,n) = A099023(n).
Bivariate e.g.f.: Sum_{n,k>=0} H(n,k)*(x^n/n!)*(y^k/k!) = 2*exp(x)/(1 + exp(2*(x+y))).
H(n,k) = (-1)^(n+k)*A239005(n+k,k), where the latter is a triangle.
H(n,k) = -A008280(n+k,k) if ((n+k) mod 4) == 1 or 2, and H(n,k) = A008280(n+k,k) if ((n+k) mod 4) == 3 or 0, provided A008280 is read as a triangle. (End)

Extensions

More terms from Petros Hadjicostas, Feb 21 2021

A256679 A signed triangle of V. I. Arnold for the Springer numbers (A001586).

Original entry on oeis.org

1, 1, 0, -2, -3, -3, -8, -6, -3, 0, 40, 48, 54, 57, 57, 256, 216, 168, 114, 57, 0, -1952, -2208, -2424, -2592, -2706, -2763, -2763, -17408, -15456, -13248, -10824, -8232, -5526, -2763, 0, 177280, 194688, 210144, 223392, 234216, 242448, 247974, 250737, 250737
Offset: 0

Views

Author

Vladimir Kruchinin, Apr 07 2015

Keywords

Comments

This triangle is denoted R(b) on page 6 of the Arnold reference.
Unsigned version of triangle is triangle of A202818.
First column (m=0) is A000828. - Robert Israel, Apr 08 2015

Examples

			Triangle begins:
    1;
    1,   0;
   -2,  -3,  -3;
   -8,  -6,  -3,   0;
   40,  48,  54,  57, 57;
  256, 216, 168, 114, 57, 0;
		

Crossrefs

Programs

  • Maple
    T:= (n,m) -> add(add(4^i*euler(2*i)*binomial(2*k,2*i)*binomial(n-m,2*k-m),i=0..k),k=floor(m/2)..floor(n/2)):
    seq(seq(T(n,m),m=0..n),n=0..10); # Robert Israel, Apr 08 2015
    # Second program, about 100 times faster than the first for the first 100 rows.
    Triangle := proc(len) local s, A, n, k;
    A := Array(0..len-1,[1]); lprint(A[0]);
    for n from 1 to len-1 do
       if n mod 2 = 1 then s := 0 else
       s := 2^(3*n+1)*(Zeta(0,-n,1/8)-Zeta(0,-n,5/8)) fi;
       A[n] := s;
       for k from n-1 by -1 to 0 do
           s := s + A[k]; A[k] := s od;
       lprint(seq(A[k], k=0..n));
    od end:
    Triangle(100); # Peter Luschny, Apr 08 2015
  • Mathematica
    T[n_, m_] := Sum[4^i EulerE[2i] Binomial[2k, 2i] Binomial[n-m, 2k-m], {k, Floor[m/2], n/2}, {i, 0, k}];
    Table[T[n, m], {n, 0, 8}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jul 12 2019 *)
  • Maxima
    T(n,m):=(sum((sum(4^i*euler(2*i)*binomial(2*k,2*i),i,0,k))*binomial(n-m,2*k-m),k,floor(m/2),n/2));
    
  • Sage
    def triangle(len):
        L = [1]; print(L)
        for n in range(1,len):
            if is_even(n):
                s = 2^(3*n+1)*(hurwitz_zeta(-n,1/8)-hurwitz_zeta(-n,5/8))
            else: s = 0
            L.append(s)
            for k in range(n-1,-1,-1):
                s = s + L[k]; L[k] = s
            print(L)
    triangle(7) # Peter Luschny, Apr 08 2015

Formula

E.g.f.: cosh(x+y)/cosh(2*(x+y))*exp(x).
T(n,m) = Sum_{k=floor(m/2)..floor(n/2)} Sum_{i=0..k} 4^i*E(2*i)*C(2*k,2*i)*C(n-m,2*k-m), where E(n) are the Euler secant numbers A122045.
Previous Showing 11-13 of 13 results.