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Max Alekseyev
Max Alekseyev
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First notice that $$\frac{{m-1\choose{k}} {n-1\choose{k}}}{ {m+n-1\choose{2k+1}} {2k\choose{k}}} = \frac{(m-1)!(n-1)!(m+n-2-2k)!}{(2k+1)(n-1-k)!(m-1-k)!(m+n-1)!} = \frac{(2k+1)\binom{m+n-2-2k}{n-1-k}}{(m+n-1)\binom{m+n-2}{n-1}}.$$ Then $\binom{m+n-2-2k}{n-1-k}$ can be expressed as $$\binom{m+n-2-2k}{n-1-k}=[x^{n-1-k}]\ \frac{1}{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^{m-n},$$ where $[x^d]$ is the operator taking the coefficient of $x^d$.

ForSimilarly we can express $(2k+1)F_{k+1}$ we haveas $$(2k+1)F_{k+1}=[x^{2k}]\ \frac{d}{dx}\frac{x}{1-x^2-x^4} = [x^k]\ \frac{1+x+3x^2}{(1-x-x^2)^2}.$$$$(2k+1)F_{k+1}=[x^{2k}]\ \frac{\partial}{\partial x}\frac{x}{1-x^2-x^4} = [x^{2k}]\ \frac{1+x^2+3x^4}{(1-x^2-x^4)^2} = [x^k]\ \frac{1+x+3x^2}{(1-x-x^2)^2}.$$ Hence, $$S(m,n) = \frac{1}{(m+n-1)\binom{m+n-2}{n-1}}\cdot [x^{n-1}]\ \frac{1+x+3x^2}{(1-x-x^2)^2\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^{m-n}$$ $$=\frac{1}{(m+n-1)\binom{m+n-2}{n-1}}\cdot [y^{n-1}]\ \frac{1+y+2y^2-6y^3+3y^4}{(1-y+2y^3-y^4)^2(1-y)^m},$$ where the latter expression is obtained with Lagrange inversion. The asymptotic of the coefficients of this generating function can now be obtained with the standard methods.

First notice that $$\frac{{m-1\choose{k}} {n-1\choose{k}}}{ {m+n-1\choose{2k+1}} {2k\choose{k}}} = \frac{(m-1)!(n-1)!(m+n-2-2k)!}{(2k+1)(n-1-k)!(m-1-k)!(m+n-1)!} = \frac{(2k+1)\binom{m+n-2-2k}{n-1-k}}{(m+n-1)\binom{m+n-2}{n-1}}.$$ Then $\binom{m+n-2-2k}{n-1-k}$ can be expressed as $$\binom{m+n-2-2k}{n-1-k}=[x^{n-1-k}]\ \frac{1}{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^{m-n},$$ where $[x^d]$ is the operator taking the coefficient of $x^d$.

For $(2k+1)F_{k+1}$ we have $$(2k+1)F_{k+1}=[x^{2k}]\ \frac{d}{dx}\frac{x}{1-x^2-x^4} = [x^k]\ \frac{1+x+3x^2}{(1-x-x^2)^2}.$$ Hence, $$S(m,n) = \frac{1}{(m+n-1)\binom{m+n-2}{n-1}}\cdot [x^{n-1}]\ \frac{1+x+3x^2}{(1-x-x^2)^2\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^{m-n}$$ $$=\frac{1}{(m+n-1)\binom{m+n-2}{n-1}}\cdot [y^{n-1}]\ \frac{1+y+2y^2-6y^3+3y^4}{(1-y+2y^3-y^4)^2(1-y)^m},$$ where the latter expression is obtained with Lagrange inversion. The asymptotic of the coefficients of this generating function can now be obtained with the standard methods.

First notice that $$\frac{{m-1\choose{k}} {n-1\choose{k}}}{ {m+n-1\choose{2k+1}} {2k\choose{k}}} = \frac{(m-1)!(n-1)!(m+n-2-2k)!}{(2k+1)(n-1-k)!(m-1-k)!(m+n-1)!} = \frac{(2k+1)\binom{m+n-2-2k}{n-1-k}}{(m+n-1)\binom{m+n-2}{n-1}}.$$ Then $\binom{m+n-2-2k}{n-1-k}$ can be expressed as $$\binom{m+n-2-2k}{n-1-k}=[x^{n-1-k}]\ \frac{1}{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^{m-n},$$ where $[x^d]$ is the operator taking the coefficient of $x^d$.

Similarly we can express $(2k+1)F_{k+1}$ as $$(2k+1)F_{k+1}=[x^{2k}]\ \frac{\partial}{\partial x}\frac{x}{1-x^2-x^4} = [x^{2k}]\ \frac{1+x^2+3x^4}{(1-x^2-x^4)^2} = [x^k]\ \frac{1+x+3x^2}{(1-x-x^2)^2}.$$ Hence, $$S(m,n) = \frac{1}{(m+n-1)\binom{m+n-2}{n-1}}\cdot [x^{n-1}]\ \frac{1+x+3x^2}{(1-x-x^2)^2\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^{m-n}$$ $$=\frac{1}{(m+n-1)\binom{m+n-2}{n-1}}\cdot [y^{n-1}]\ \frac{1+y+2y^2-6y^3+3y^4}{(1-y+2y^3-y^4)^2(1-y)^m},$$ where the latter expression is obtained with Lagrange inversion. The asymptotic of the coefficients of this generating function can now be obtained with the standard methods.

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Max Alekseyev
Max Alekseyev
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First notice that $$\frac{{m-1\choose{k}} {n-1\choose{k}}}{ {m+n-1\choose{2k+1}} {2k\choose{k}}} = \frac{(m-1)!(n-1)!(m+n-2-2k)!}{(2k+1)(n-1-k)!(m-1-k)!(m+n-1)!} = \frac{(2k+1)\binom{m+n-2-2k}{n-1-k}}{(m+n-1)\binom{m+n-2}{n-1}}.$$ Then $\binom{m+n-2-2k}{n-1-k}$ is the coefficient of $x^{n-1-k}$ incan be expressed as $$\frac{1}{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^{m-n}.$$$$\binom{m+n-2-2k}{n-1-k}=[x^{n-1-k}]\ \frac{1}{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^{m-n},$$ The Fibonacci numberwhere $(2k+1)F_{k+1}$$[x^d]$ is the operator taking the coefficient of $x^{2k}$ in$x^d$.

For $$\frac{d}{dx}\frac{x}{1-x^2-x^4} = \frac{1+x^2+3x^4}{(1-x^2-x^4)^2}.$$$(2k+1)F_{k+1}$ we have $$(2k+1)F_{k+1}=[x^{2k}]\ \frac{d}{dx}\frac{x}{1-x^2-x^4} = [x^k]\ \frac{1+x+3x^2}{(1-x-x^2)^2}.$$ Hence, $$S(m,n) = \frac{1}{(m+n-1)\binom{m+n-2}{n-1}}\cdot [x^{n-1}]\ \frac{1+x+3x^2}{(1-x-x^2)^2\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^{m-n}$$ $$=\frac{1}{(m+n-1)\binom{m+n-2}{n-1}}\cdot [y^{n-1}]\ \frac{1+y+2y^2-6y^3+3y^4}{(1-y+2y^3-y^4)^2(1-y)^m},$$ where the latter expression is obtained with Lagrange inversion. The asymptotic of the coefficients of this generating function can now be obtained with the standard methodsstandard methods.

First notice that $$\frac{{m-1\choose{k}} {n-1\choose{k}}}{ {m+n-1\choose{2k+1}} {2k\choose{k}}} = \frac{(m-1)!(n-1)!(m+n-2-2k)!}{(2k+1)(n-1-k)!(m-1-k)!(m+n-1)!} = \frac{(2k+1)\binom{m+n-2-2k}{n-1-k}}{(m+n-1)\binom{m+n-2}{n-1}}.$$ Then $\binom{m+n-2-2k}{n-1-k}$ is the coefficient of $x^{n-1-k}$ in $$\frac{1}{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^{m-n}.$$ The Fibonacci number $(2k+1)F_{k+1}$ is the coefficient of $x^{2k}$ in $$\frac{d}{dx}\frac{x}{1-x^2-x^4} = \frac{1+x^2+3x^4}{(1-x^2-x^4)^2}.$$ Hence, $$S(m,n) = \frac{1}{(m+n-1)\binom{m+n-2}{n-1}}\cdot [x^{n-1}]\ \frac{1+x+3x^2}{(1-x-x^2)^2\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^{m-n}$$ $$=\frac{1}{(m+n-1)\binom{m+n-2}{n-1}}\cdot [y^{n-1}]\ \frac{1+y+2y^2-6y^3+3y^4}{(1-y+2y^3-y^4)^2(1-y)^m},$$ where the latter expression is obtained with Lagrange inversion. The asymptotic of the coefficients of this generating function can now be obtained with the standard methods.

First notice that $$\frac{{m-1\choose{k}} {n-1\choose{k}}}{ {m+n-1\choose{2k+1}} {2k\choose{k}}} = \frac{(m-1)!(n-1)!(m+n-2-2k)!}{(2k+1)(n-1-k)!(m-1-k)!(m+n-1)!} = \frac{(2k+1)\binom{m+n-2-2k}{n-1-k}}{(m+n-1)\binom{m+n-2}{n-1}}.$$ Then $\binom{m+n-2-2k}{n-1-k}$ can be expressed as $$\binom{m+n-2-2k}{n-1-k}=[x^{n-1-k}]\ \frac{1}{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^{m-n},$$ where $[x^d]$ is the operator taking the coefficient of $x^d$.

For $(2k+1)F_{k+1}$ we have $$(2k+1)F_{k+1}=[x^{2k}]\ \frac{d}{dx}\frac{x}{1-x^2-x^4} = [x^k]\ \frac{1+x+3x^2}{(1-x-x^2)^2}.$$ Hence, $$S(m,n) = \frac{1}{(m+n-1)\binom{m+n-2}{n-1}}\cdot [x^{n-1}]\ \frac{1+x+3x^2}{(1-x-x^2)^2\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^{m-n}$$ $$=\frac{1}{(m+n-1)\binom{m+n-2}{n-1}}\cdot [y^{n-1}]\ \frac{1+y+2y^2-6y^3+3y^4}{(1-y+2y^3-y^4)^2(1-y)^m},$$ where the latter expression is obtained with Lagrange inversion. The asymptotic of the coefficients of this generating function can now be obtained with the standard methods.

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Max Alekseyev
Max Alekseyev
  • 34.3k
  • 5
  • 74
  • 152

First notice that $$\frac{{m-1\choose{k}} {n-1\choose{k}}}{ {m+n-1\choose{2k+1}} {2k\choose{k}}} = \frac{(m-1)!(n-1)!(m+n-2-2k)!}{(2k+1)(n-1-k)!(m-1-k)!(m+n-1)!} = \frac{(2k+1)\binom{m+n-2-2k}{n-1-k}}{(m+n-1)\binom{m+n-2}{n-1}}.$$ Then $\binom{m+n-2-2k}{n-1-k}$ is the coefficient of $x^{n-1-k}$ in $$\frac{1}{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^{m-n}.$$ The Fibonacci number $(2k+1)F_{k+1}$ is the coefficient of $x^{2k}$ in $$\frac{d}{dx}\frac{x}{1-x^2-x^4} = \frac{1+x^2+3x^4}{(1-x^2-x^4)^2}.$$ Hence, $$S(m,n) = \frac{1}{(m+n-1)\binom{m+n-2}{n-1}}\cdot [x^{n-1}]\ \frac{1+x+3x^2}{(1-x-x^2)^2\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^{m-n}$$ $$=\frac{1}{(m+n-1)\binom{m+n-2}{n-1}}\cdot [y^{n-1}]\ \frac{1+y+2y^2-6y^3+3y^4}{(1-y+2y^3-y^4)^2(1-y)^m},$$ where the latter expression is obtained with Lagrange inversion. The asymptotic of the coefficients of this generating function can now be obtained with the standard methods.