1-2-2-4-3-8-6-16-11-32-20-64-37-128-

Question Number 136489 by liberty last updated on 22/Mar/21

$$\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{8}}+\frac{\mathrm{6}}{\mathrm{16}}+\frac{\mathrm{11}}{\mathrm{32}}+\frac{\mathrm{20}}{\mathrm{64}}+\frac{\mathrm{37}}{\mathrm{128}}+…\:=? \\ $$

Answered by Olaf last updated on 22/Mar/21

Tribonacci numbers : T_0 = 0, T_1 = 1, T_2 = 0 T_n = T_(n−1) +T_(n−2) +T_(n−3) ⇒ T_n = {0, 1, 0, 1, 2, 3, 6, 11, 20, 37...} S = Σ_(n=3) ^∞ (T_n /2^(n−2) ) = 4Σ_(n=3) ^∞ (T_n /2^n ) = 4Σ_(n=0) ^∞ (T_n /2^n )−2 The generator function of {T_n } is given by : F(X) = ((X−X^2 )/(1−X−X^2 −X^3 )) = Σ_(n=0) ^∞ T_n X^n ⇒ S = 4F((1/2))−2 S = 4((1/4)/(1−(1/2)−(1/4)−(1/8)))−2 = 6

$$\mathrm{Tribonacci}\:\mathrm{numbers}\:: \\ $$$$\mathrm{T}_{\mathrm{0}} \:=\:\mathrm{0},\:\mathrm{T}_{\mathrm{1}} \:=\:\mathrm{1},\:\mathrm{T}_{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{T}_{{n}} \:=\:\mathrm{T}_{{n}−\mathrm{1}} +\mathrm{T}_{{n}−\mathrm{2}} +\mathrm{T}_{{n}−\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{T}_{{n}} \:=\:\left\{\mathrm{0},\:\mathrm{1},\:\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{6},\:\mathrm{11},\:\mathrm{20},\:\mathrm{37}…\right\} \\ $$$$\mathrm{S}\:=\:\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\frac{\mathrm{T}_{{n}} }{\mathrm{2}^{{n}−\mathrm{2}} }\:=\:\mathrm{4}\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\frac{\mathrm{T}_{{n}} }{\mathrm{2}^{{n}} }\:=\:\mathrm{4}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{T}_{{n}} }{\mathrm{2}^{{n}} }−\mathrm{2} \\ $$$$\mathrm{The}\:\mathrm{generator}\:\mathrm{function}\:\mathrm{of}\:\left\{\mathrm{T}_{{n}} \right\}\:\mathrm{is} \\ $$$$\mathrm{given}\:\mathrm{by}\:: \\ $$$$\mathrm{F}\left(\mathrm{X}\right)\:=\:\frac{\mathrm{X}−\mathrm{X}^{\mathrm{2}} }{\mathrm{1}−\mathrm{X}−\mathrm{X}^{\mathrm{2}} −\mathrm{X}^{\mathrm{3}} }\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{T}_{{n}} \mathrm{X}^{{n}} \\ $$$$\Rightarrow\:\mathrm{S}\:=\:\mathrm{4F}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{2} \\ $$$$\mathrm{S}\:=\:\mathrm{4}\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}}−\mathrm{2}\:=\:\mathrm{6} \\ $$

Commented by Olaf last updated on 22/Mar/21