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
Keywords
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
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
Comments