0,3

More generally coefficients of (1+m*x-sqrt(m^2*x^2-(2*m+2)*x+1))/(2*m*x) are given by : a(n)=sum(k=0,n,(m+1)^k*N(n,k))

a(n) is the series reversion of x(1-5x)/(1-4x); a(n+1) is the series reversion of x/(1+6x+5x^2); a(n+1) counts (6,5)-Motzkin paths of length n, where there are 6 colors available for the H(1,0) steps and 5 for the U(1,1) steps. - Paul Barry (pbarry(AT)wit.ie), May 19 2005

The Hankel transform of this sequence is 5^C(n+1,2) . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Oct 29 2007

Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.

G.f. (1+4*x-sqrt(16*x^2-12*x+1))/(10*x)

a(n) = Sum_{k=0..n} A088617(n, k)*5^k*(-4)^(n-k) . - DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Jan 21 2004

With offset 1 : a(1)=1, a(n)=-4*a(n-1)+5*sum(i=1, n-1, a(i)*a(n-i)) - Benoit Cloitre (benoit7848c(AT)orange.fr), Mar 16 2004

a(n+1)=sum{k=0..floor(n/2), C(n, 2k)C(k)6^(n-2k)*5^k}; - Paul Barry (pbarry(AT)wit.ie), May 19 2005

a(n) = [6(2n-1)a(n-1) - 16(n-2)a(n-2)] / (n+1) for n>=2, a(0) = a(1) = 1 . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Aug 19 2005

(PARI) a(n)=sum(k=0, n, 5^k/n*binomial(n, k)*binomial(n, k+1))

Cf. A001003, A007564, A059231.

Sequence in context: A122371 A083067 A000402 this_sequence A127848 A083161 A077147

Adjacent sequences: A078006 A078007 A078008 this_sequence A078010 A078011 A078012

nonn

Benoit Cloitre (benoit7848c(AT)orange.fr), May 10 2003