A173253 Partial sums of A000111.

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

Author

Jonathan Vos Post, Feb 14 2010

Keywords

• nonn

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.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..485

Crossrefs

Cf. A000111, A000364, A000182, A008280, A008281, A008282, A010094, A059720, A008970, A109449, A162170.

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