%I #45 Apr 30 2015 20:58:04

%S 1,1,2,2,4,6,13,27,70,191,609,2130,8526,38156,194000,1109673,7176149,

%T 52238676,429004471,3970438003,41454181730,488046132076,6482590679282,

%U 97134793638750,1641654359781521,31285014253070731,672372121341768918,16299021330860540657

%N Diagonal sums of Fibonomial triangle A010048.

%C 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

%C 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

%H Vaclav Kotesovec, <a href="/A181926/b181926.txt">Table of n, a(n) for n = 0..195</a>

%H Vaclav Kotesovec, <a href="/A181926/a181926.jpg">Graph - The asymptotic ratio</a>

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

%F From _Vaclav Kotesovec_, Apr 29 2015: (Start)

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

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

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

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

%F 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.

%F (End)

%F a(n) = Sum_{k=ceiling(n/2)..n} A003266(k) / (A003266(2*k-n) * A003266(n-k)). - _Vaclav Kotesovec_, Apr 30 2015

%t 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 *)

%t (* 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 *)

%t (* 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 *)

%o (Maxima) ffib(n):=prod(fib(k),k,1,n);

%o fibonomial(n,k):=ffib(n)/(ffib(k)*ffib(n-k));

%o makelist(sum(fibonomial(k,n-k),k,ceiling(n/2),n),n,0,30);

%Y Cf. A003266, A010048, A056569, A062073.

%K nonn,easy

%O 0,3

%A _Emanuele Munarini_, Apr 02 2012

%E a(14) corrected by _Vaclav Kotesovec_, Apr 29 2015