cp's OEIS Frontend

a(n) = ((n+1)*F(2*n+3)+(2*n+3)*F(2*(n+1)))/5 with F(n)=A000045(n) (Fibonacci numbers). One half of odd-indexed A001629(n), n >= 2 (Fibonacci convolution).

Convolution of F(2n+1) (A001519) and F(2n+2) (A001906(n+1)). - Graeme McRae, Jun 07 2006

Number of reentrant corners along the lower contours of all directed column-convex polyominoes of area n+3 (a reentrant corner along the lower contour is a vertical step that is followed by a horizontal step). a(n) = Sum_{k=0..ceiling((n+1)/2)} k*A121466(n+3,k). - Emeric Deutsch, Aug 02 2006

From Wolfdieter Lang, Jan 02 2012: (Start)

a(n) = A024458(2*n), n >= 1 (bisection, even arguments).

a(n) is also the odd part of the bisection of the half-convolution of the sequence A000045(n+1), n >= 0, with itself. See a comment on A201204 for the definition of the half-convolution of a sequence with itself. There one also finds the rule for the o.g.f. which in this case is Chato(x)/2 with the o.g.f. Chato(x) = 2*(1-x)/(1-3*x+x^2)^2 of A001629(2*n+3), n >= 0.

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Or, multiplication table of the positive Fibonacci numbers read by antidiagonals.

Or, triangle of products of nonzero Fibonacci numbers.

Or, a two-dimensional square Fibonacci array read by antidiagonals, with offset 1: T(1,1) = T(1,2) = T(2,1) = T(2,2) = 1; thereafter T(m,n) = max {T(m,n-2) + T(m,n-1), T(m-2,n) + T(m-1,n), T(m-2,n-2) + T(m-1,n-1)}. If "max" is changed to "min" we get A283845. - N. J. A. Sloane, Mar 31 2017

Row sums are A001629 (Fibonacci numbers convolved with themselves.). The main diagonal and first subdiagonal are Fibonacci numbers, for other entries T(n,k) = T(n-1,k) + T(n-2,k). The central numbers form A006498. - Gerald McGarvey, Jun 02 2005

Alternating row sums = (1, 0, 3, 0, 8, ...), given by Fibonacci(2n) if n even, else zero.

Row n = edge-counting vector for the Fibonacci cube F(n+1) embedded in the natural way in the hypercube Q(n+1). - Emanuele Munarini, Apr 01 2008

The augmentation of A058071 is the triangle A193595. To fit the definition of augmented triangle at A103091, it is helpful to represent A058071 using p(n,k)=F(k+1)*F(n+1-k) for 0<=k<=n. - Clark Kimberling, Jul 31 2011

T(n,k) = number of appearances of a(k) in p(n) in the n-th convergent p(n)/q(n) of the formal infinite continued fraction [a(0), a(1), ...]; e.g., p(3) = a(0)*a(1)*a(2)*a(3) + a(0)*a(1) + a(0)*a(3) + a(2)*a(3) + 1. Also, T(n,k) = number of appearances of a(k+1) in q(n+1); e.g., q(3) = a(1)*a(2)*a(3) + a(1) + a(3). - Clark Kimberling, Dec 21 2015

Each row is a palindrome, and the central term of row 2n is the square of the F(n+1), where F = A000045 (Fibonacci numbers). - Clark Kimberling, Dec 21 2015

Also called Hosoya's triangle, after the Japanese chemist Haruo Hosoya (b. 1936). - Amiram Eldar, Jun 10 2021

From Wolfdieter Lang, Jan 02 2012: (Start)

a(n) = A024458(2*n-1), n>=1 (bisection, odd arguments).

chate(n):=a(n+1), n>=0, is the even part of the bisection of the half-convolution of the sequence A000045(n+1), n>=0, with itself. See a comment on A201204 for the definition of half-convolution. There one finds also the rule for the o.g.f.s of the bisection. Here the o.g.f. of the sequence chate(n), n>=0, is Chate(x):= (Ce(x)+U2(x))/2 with Ce(x)=(1-x+x^2)/(1-3*x+x^2)^2, the o.g.f. of A054444(n), and

U2(x)=(1-x)/((1+x)*(1-3*x+x^2)), the o.g.f. of A007598(n+1), n>=0. This results (after multiplying with x) in the o.g.f. given below in the formula section. It is equivalent to the explicit formula given there, as can be seen after a partial fraction decomposition of the o.g.f.

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