0,4

If b(0)=1 and b(n+1) = -sum(u(k)*binomial(n,k)*2^(n-k-1),k=0..n-1) then a(n) = abs(b(n)) (in fact b(n) = 1,1,0,-2,0,16,0,-272,...). - Robert FERREOL (ferreol(AT)mathcurve.com), Dec 30 2006

Sum_{k, 0<=k<=n}A075263(n,k)*2^k = 1,-1,0,2,0,-16,0,272,0,-7936,0,...for n=0, 1, 2, 3, 4, ...respectively . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Aug 20 2007

R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 1, 1997; See Exercise 1.41(d).

Kwang-Wu Chen, An Interesting Lemma for Regular C-fractions, J. Integer Seqs., Vol. 6, 2003.

Let b(n) be a(n) shifted one place to the left with b(2+4k)=-a(3+4k), k=0, 1, .. Then b(n) is the expansion of sech(x)^2. - Mario Catalani (mario.catalani(AT)unito.it), Feb 08 2003

g(x)=x+x^2-2*x^4+16*x^6-272*x^8+... satisfies g(x/(1+2x))=-g(-x).

E.g.f.: 1+tan(x).

E.g.f. exp(x)sech(x) is 1,1,0,-2,0,16,0,-272,... - Paul Barry (pbarry(AT)wit.ie), Mar 15 2006

a(n)= 2^n*abs(Euler(n,0)) where Euler(n,x) is the n-th Eulerian polynomial. - Robert FERREOL (ferreol(AT)mathcurve.com), Dec 30 2006

u:=proc(n) if n=0 then 1 else -add(u(k)*binomial(n, k)/2*2^(n-k), k=0..n-1) fi end; seq(u(n), n=0..15); - Robert FERREOL (ferreol(AT)mathcurve.com), Dec 30 2006

1+Tan[ x ]

a[m_] := Abs[Sum[(-2)^(m-k) k! StirlingS2[m, k], {k, 0, m}]]; Table[a[i], {i, 0, 20}] [From Peter Luschny (peter(AT)luschny.de), Apr 29 2009]

(PARI) a(n)=if(n<1, n==0, n!*polcoeff(tan(x+x*O(x^n)), n))

A000182(n)=a(2n-1).

Adjacent sequences: A009003 A009004 A009005 this_sequence A009007 A009008 A009009

Sequence in context: A111978 A146558 A025600 this_sequence A155585 A057375 A009045

nonn

R. H. Hardin (rhhardin(AT)att.net)

Reformatted Mar 15 1997