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Unsigned row sums of triangle A114700.
2

%I #10 Feb 22 2017 03:42:46

%S 1,2,2,4,2,4,4,8,10,20,32,64,112,224,408,816,1514,3028,5680,11360,

%T 21472,42944,81644,163288,311896,623792,1196132,2392264,4602236,

%U 9204472,17757184,35514368,68680170,137360340,266200112,532400224,1033703056

%N Unsigned row sums of triangle A114700.

%C Both triangles A112555 and A114700 have the property that the m-th matrix power of the triangles satisfy T^m = I + m*(T - I). So it is curious that the row squared sums of A112555 is a bisection of the unsigned row sums of A114700.

%F G.f.: (1+2*x)*( 2*(1+x^2)/(1-x^2) + x^2/(1-4*x^2)^(1/2) )/(2+x^2). Also, a(2*n+1) = 2*a(2*n), a(2*n) = A112556(n), where A112556 equals the row squared sums of triangle A112555.

%t CoefficientList[Series[(1 + 2*x)*(2*(1 + x^2)/(1 - x^2) + x^2/(1 - 4*x^2)^(1/2))/(2 + x^2), {x, 0, 50}], x] (* _Wesley Ivan Hurt_, Feb 21 2017 *)

%o (PARI) a(n)=local(x=X+X*O(X^n)); polcoeff((1+2*x)*(2*(1+x^2)/(1-x^2)+x^2/(1-4*x^2)^(1/2))/(2+x^2),n,X)

%o (PARI) /* a(n) as the unsigned row sums of A114700 */ a(n)=sum(k=0,n,abs(polcoeff(polcoeff(1/(1-x*y)+ x*(1+x-2*x^2*y)/(1-x)/(1+x+x*y+x*O(x^n)+y*O(y^k))/(1-x*y),n,x),k,y)))

%Y Cf. A112556, A112555, A114700.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Feb 19 2006