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Question Number 4492 by RasheedSindhi last updated on 01/Feb/16

B y m u r g i n g t h r e e s e q u e n c e s

a_{1}, a_{2}, \dots a_{n}, b_{1}, b_{2}, \dots, b_{n} & c_{1}, c_{2}, \dots, c_{n}

a n e w s e q u e n c e

a_{1}, b_{1}, c_{1}, a_{2}, b_{2}, c_{2}, \dots, a_{n}, b_{n}, c_{n}

i s p r o d u c e d .

D e t e r m i n e t h e single - formula

general term o f t h i s n e w s e q u e n c e .

W h a t i f m s e q u e n c e s w e r e m u r g e d

i n t h i s m a n n e r \dots

Commented by Yozzii last updated on 01/Feb/16

u_{1} = a_{1} \times 1 + b_{1} \times 0 + c_{1} \times 0 = a_{1}

u_{2} = a_{1} \times 0 + b_{1} \times 1 + c_{1} \times 0 = b_{1}

u_{3} = a_{1} \times 0 + b_{1} \times 0 + c_{1} \times 1 = c_{1}

u_{4} = a_{2} \times 1 + b_{2} \times 0 + c_{2} \times 0 = a_{2}

u_{5} = a_{2} \times 0 + b_{2} \times 1 + c_{2} \times 0 = b_{2}

u_{6} = a_{2} \times 0 + b_{2} \times 0 + c_{2} \times 1 = c_{2}

f (n) = ⌊ \frac{n + 2}{3} ⌋

f (1) = 1, f (2) = 1, f (3) = 1, f (4) = 2,

f (5) = 2, f (6) = 2, \dots

∴ u_{r} = a_{⌊ \frac{r + 2}{3} ⌋} p (r) + b_{⌊ \frac{r + 2}{3} ⌋} q (r) + c_{⌊ \frac{r + 2}{3} ⌋} t (r)

Answered by 123456 last updated on 03/Feb/16

u_{n} = \underset{i = 1}{\sum^{m}} a_{i, ⌊ \frac{n - 1}{m} ⌋ + 1} f (i, n)

f (i, n) = {\begin{cases} 1 n \equiv k (\mod m) \land i = k \\ 0 n \equiv k (\mod m) \land i \neq k \\ 1 n \equiv 0 (\mod m) \land i = m \\ 0 n \equiv 0 (\mod m) \land i \neq m \end{cases} \land k \in {1, 2, \dots, m - 1}

a_{i, n} to denote sequence like

a_{1, n} = a_{1}, a_{2}, a_{3}, \dots

a_{2, n} = b_{1}, b_{2}, b_{3}, \dots

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