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Question Number 4492 by RasheedSindhi last updated on 01/Feb/16
By murging three sequences a_1 ,a_2 ,...a_n , b_1 ,b_2 ,...,b_n & c_1 ,c_2 ,...,c_n a new sequence a_1 ,b_1 ,c_1 ,a_2 ,b_2 ,c_2 ,...,a_n ,b_n ,c_n is produced. Determine the single-formula general term of this new sequence. What if m sequences were murged in this manner...
$$ \\ $$$${By}\:{murging}\:{three}\:{sequences} \\ $$$$\:{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,…{a}_{{n}} \:,\:{b}_{\mathrm{1}} ,{b}_{\mathrm{2}} ,…,{b}_{{n}} \:\&\:{c}_{\mathrm{1}} ,{c}_{\mathrm{2}} ,…,{c}_{{n}} \\ $$$${a}\:{new}\:{sequence}\: \\ $$$${a}_{\mathrm{1}} ,{b}_{\mathrm{1}} ,{c}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,{b}_{\mathrm{2}} ,{c}_{\mathrm{2}} ,…,{a}_{{n}} ,{b}_{{n}} ,{c}_{{n}} \\ $$$${is}\:{produced}. \\ $$$${Determine}\:{the}\:\mathrm{single}-\mathrm{formula} \\ $$$$\mathrm{general}\:\mathrm{term}\:{of}\:{this}\:{new}\:{sequence}. \\ $$$${What}\:{if}\:{m}\:{sequences}\:{were}\:{murged} \\ $$$${in}\:{this}\:{manner}… \\ $$
Commented by Yozzii last updated on 01/Feb/16
u_1 =a_1 ×1+b_1 ×0+c_1 ×0=a_1 u_2 =a_1 ×0+b_1 ×1+c_1 ×0=b_1 u_3 =a_1 ×0+b_1 ×0+c_1 ×1=c_1 u_4 =a_2 ×1+b_2 ×0+c_2 ×0=a_2 u_5 =a_2 ×0+b_2 ×1+c_2 ×0=b_2 u_6 =a_2 ×0+b_2 ×0+c_2 ×1=c_2 f(n)=⌊((n+2)/3)⌋ f(1)=1,f(2)=1, f(3)=1, f(4)=2, f(5)=2,f(6)=2,... ∴ u_r =a_(⌊((r+2)/3)⌋) p(r)+b_(⌊((r+2)/3)⌋) q(r)+c_(⌊((r+2)/3)⌋) t(r)
$${u}_{\mathrm{1}} ={a}_{\mathrm{1}} ×\mathrm{1}+{b}_{\mathrm{1}} ×\mathrm{0}+{c}_{\mathrm{1}} ×\mathrm{0}={a}_{\mathrm{1}} \\ $$$${u}_{\mathrm{2}} ={a}_{\mathrm{1}} ×\mathrm{0}+{b}_{\mathrm{1}} ×\mathrm{1}+{c}_{\mathrm{1}} ×\mathrm{0}={b}_{\mathrm{1}} \\ $$$${u}_{\mathrm{3}} ={a}_{\mathrm{1}} ×\mathrm{0}+{b}_{\mathrm{1}} ×\mathrm{0}+{c}_{\mathrm{1}} ×\mathrm{1}={c}_{\mathrm{1}} \\ $$$${u}_{\mathrm{4}} ={a}_{\mathrm{2}} ×\mathrm{1}+{b}_{\mathrm{2}} ×\mathrm{0}+{c}_{\mathrm{2}} ×\mathrm{0}={a}_{\mathrm{2}} \\ $$$${u}_{\mathrm{5}} ={a}_{\mathrm{2}} ×\mathrm{0}+{b}_{\mathrm{2}} ×\mathrm{1}+{c}_{\mathrm{2}} ×\mathrm{0}={b}_{\mathrm{2}} \\ $$$${u}_{\mathrm{6}} ={a}_{\mathrm{2}} ×\mathrm{0}+{b}_{\mathrm{2}} ×\mathrm{0}+{c}_{\mathrm{2}} ×\mathrm{1}={c}_{\mathrm{2}} \\ $$$${f}\left({n}\right)=\lfloor\frac{{n}+\mathrm{2}}{\mathrm{3}}\rfloor \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1},{f}\left(\mathrm{2}\right)=\mathrm{1},\:{f}\left(\mathrm{3}\right)=\mathrm{1},\:{f}\left(\mathrm{4}\right)=\mathrm{2}, \\ $$$${f}\left(\mathrm{5}\right)=\mathrm{2},{f}\left(\mathrm{6}\right)=\mathrm{2},… \\ $$$$\therefore\:{u}_{{r}} ={a}_{\lfloor\frac{{r}+\mathrm{2}}{\mathrm{3}}\rfloor} {p}\left({r}\right)+{b}_{\lfloor\frac{{r}+\mathrm{2}}{\mathrm{3}}\rfloor} {q}\left({r}\right)+{c}_{\lfloor\frac{{r}+\mathrm{2}}{\mathrm{3}}\rfloor} {t}\left({r}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by 123456 last updated on 03/Feb/16
u_n =Σ_(i=1) ^m a_(i,⌊((n−1)/m)⌋+1) f(i,n) f(i,n)= { ((1 n≡k(mod m)∧i=k)),((0 n≡k(mod m)∧i≠k)),((1 n≡0(mod m)∧i=m)),((0 n≡0(mod m)∧i≠m)) :}∧k∈{1,2,...,m−1} a_(i,n) to denote sequence like a_(1,n) =a_1 ,a_2 ,a_3 ,... a_(2,n) =b_1 ,b_2 ,b_3 ,...
$${u}_{{n}} =\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}{a}_{{i},\lfloor\frac{{n}−\mathrm{1}}{{m}}\rfloor+\mathrm{1}} {f}\left({i},{n}\right) \\ $$$${f}\left({i},{n}\right)=\begin{cases}{\mathrm{1}\:\:\:\:\:\:\:\:\:{n}\equiv{k}\left(\mathrm{mod}\:{m}\right)\wedge{i}={k}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:{n}\equiv{k}\left(\mathrm{mod}\:{m}\right)\wedge{i}\neq{k}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:{n}\equiv\mathrm{0}\left(\mathrm{mod}\:{m}\right)\wedge{i}={m}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:{n}\equiv\mathrm{0}\left(\mathrm{mod}\:{m}\right)\wedge{i}\neq{m}}\end{cases}\wedge{k}\in\left\{\mathrm{1},\mathrm{2},…,{m}−\mathrm{1}\right\} \\ $$$${a}_{{i},{n}} \:\mathrm{to}\:\mathrm{denote}\:\mathrm{sequence}\:\mathrm{like} \\ $$$${a}_{\mathrm{1},{n}} ={a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,{a}_{\mathrm{3}} ,… \\ $$$${a}_{\mathrm{2},{n}} ={b}_{\mathrm{1}} ,{b}_{\mathrm{2}} ,{b}_{\mathrm{3}} ,… \\ $$

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