A181926 Diagonal sums of Fibonomial triangle A010048.

1, 1, 2, 2, 4, 6, 13, 27, 70, 191, 609, 2130, 8526, 38156, 194000, 1109673, 7176149, 52238676, 429004471, 3970438003, 41454181730, 488046132076, 6482590679282, 97134793638750, 1641654359781521, 31285014253070731, 672372121341768918, 16299021330860540657

Offset: 0

Author

Emanuele Munarini, Apr 02 2012

Keywords

• nonn
• easy

Comments

Cf. A000045 (Fibonacci) as diagonal sums of A007318 (Pascal's Triangle). For Fibonacci numbers, the ratio A000045(i+1)/A000045(i) approaches the golden ratio (1+sqrt(5))/2 as i increases. For this sequence, it appears that (a(i+5)/a(i+4))/(a(i+1)/a(i)) approaches the golden ratio. - Dale Gerdemann, Apr 23 2015

This sequence can be interpreted as counting colored, square-domino tilings of a 1xn board, where the number of colors available for a domino with k squares to the left is Fib(k+1) and the number of colors available for a square with k dominoes to the left is Fib(k-1). "Fib(n)" here is A000045 (Fibonacci), extended so that Fib(-1) = 1, Fib(0) = 0,... . As an example, let d be a domino, s be a square an consider the uncolored tilings of length 5: sssss, sssd, ssds, sdss, dsss, sdd, dsd, dds. Then, after each 's' or 'd', write the number of colors available: s1s1s1s1s1, s1s1s1d3, s1s1d2s0, s1d1s0s0, d1s0s0s0, s1d1d1, d1s0d1, d1d1s1. So the number of colorings for these tilings is: 1,3,0,0,0,1,0,1 and the total number of colored tilings is 6 (= a(5)). - Dale Gerdemann, Apr 30 2015

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..195
• Vaclav Kotesovec, Graph - The asymptotic ratio

Crossrefs

Cf. A003266, A010048, A056569, A062073.

Programs

Mathematica

Table[Sum[Product[Fibonacci[k-j+1]/Fibonacci[j],{j,1,n-k}],{k,Ceiling[n/2],n}],{n,0,30}] (* Vaclav Kotesovec, Apr 29 2015 *)
(* Or, since version 10 *) Table[Sum[Fibonorial[k]/Fibonorial[2k-n]/Fibonorial[n-k],{k,Ceiling[n/2],n}],{n,0,30}] (* Vaclav Kotesovec, Apr 30 2015 *)
(* List of the multiplicative constants from an asymptotic formula: *) {N[EllipticTheta[3, 0, GoldenRatio^(-2)]/QPochhammer[-(GoldenRatio^2)^(-1)], 80], N[Sum[GoldenRatio^(-2*(j + 1/4)^2), {j, -Infinity, Infinity}]/QPochhammer[-(GoldenRatio^2)^(-1)], 80], N[EllipticTheta[2, 0, GoldenRatio^(-2)]/QPochhammer[-(GoldenRatio^2)^(-1)], 80]} (* Vaclav Kotesovec, Apr 30 2015 *)

Maxima

ffib(n):=prod(fib(k),k,1,n);
fibonomial(n,k):=ffib(n)/(ffib(k)*ffib(n-k));
makelist(sum(fibonomial(k,n-k),k,ceiling(n/2),n),n,0,30);

Formula

a(n) = sum(fibonomial(k,n-k),k=ceiling(n/2)..n).

From Vaclav Kotesovec, Apr 29 2015: (Start)

a(n) ~ c * ((1+sqrt(5))/2)^(n^2/8), where

c = 1.472885929099569314607134281503815932269629515265... if mod(n,4)=0,

c = 1.472782295338429619549807628338486893461428897618... if mod(n,4)=1 or 3,

c = 1.472678661577289942545896597162143392952724631588... if mod(n,4)=2.

Or c = Sum_{j} ((1+sqrt(5))/2)^(-2*(j+(1-cos(Pi*n/2))/4)^2) / A062073, where A062073 = 1.2267420107203532444176302... is the Fibonacci factorial constant.

(End)

a(n) = Sum_{k=ceiling(n/2)..n} A003266(k) / (A003266(2*k-n) * A003266(n-k)). - Vaclav Kotesovec, Apr 30 2015

Extensions

a(14) corrected by Vaclav Kotesovec, Apr 29 2015