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%I #23 Apr 16 2023 20:27:55
%S 1,2,3,5,10,26,87,359,1744,9680,60201,413993,3116758,25485014,
%T 224845995,2128603307,21520115452,231385458428,2636265133869,
%U 31725150246701,402096338484226,5353594391608322,74702468784746223,1090126355291598575,16604660518848685480
%N Partial sums of A000111.
%C 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.
%H Alois P. Heinz, <a href="/A173253/b173253.txt">Table of n, a(n) for n = 0..485</a>
%F 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).
%F 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
%F 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
%F a(n) ~ 2^(n+2)*n!/Pi^(n+1). - _Vaclav Kotesovec_, Oct 27 2016
%e 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.
%p b:= proc(u, o) option remember;
%p `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
%p end:
%p a:= proc(n) option remember;
%p `if`(n<0, 0, a(n-1))+ b(n, 0)
%p end:
%p seq(a(n), n=0..25); # _Alois P. Heinz_, Oct 27 2017
%t With[{nn=30},Accumulate[CoefficientList[Series[Sec[x]+Tan[x],{x,0,nn}],x] Range[0,nn]!]] (* _Harvey P. Dale_, Feb 26 2012 *)
%o (Python)
%o from itertools import accumulate
%o def A173253(n):
%o if n<=1:
%o return n+1
%o c, blist = 2, (0,1)
%o for _ in range(n-1):
%o c += (blist := tuple(accumulate(reversed(blist),initial=0)))[-1]
%o return c # _Chai Wah Wu_, Apr 16 2023
%Y Cf. A000111, A000364, A000182, A008280, A008281, A008282, A010094, A059720, A008970, A109449, A162170.
%K nonn
%O 0,2
%A _Jonathan Vos Post_, Feb 14 2010
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