## Calculus 8th Edition

The relationship between these two sequences is $B_{n+9}=A_{n}$. The sequence always ends with $2, 1, 4$, regardless of the value of $a_1$.
Find first 40 terms with $a_{1}=11:$ $a_{1}$ =11 $a_{11}$ =16 $a_{21}$ =1 $a_{31}$ =4 $a_{2}$ =34 $a_{12}$ =8 $a_{22}$ =4 $a_{32}$ =2 $a_{3}$ =17 $a_{13}$ =4 $a_{23}$ =2 $a_{33}$ =1 $a_{4}$ =52 $a_{14}$ =2 $a_{24}$ = 1 $a_{34}$ =4 $a_{5}$ =26 $a_{15}$ = 1 $a_{25}$ =4 $a_{35}$ = 2 $a_{6}$ =13 $a_{16}$ =4 $a_{26}$ =2 $a_{36}$ =1 $a_{7}$ =40 $a_{17}$ =2 $a_{27}$ = 1 $a_{37}$ =4 $a_{8}$ =20 $a_{18}$ =1 $a_{28}$ =4 $a_{38}$ = 2 $a_{9}$ =10 $a_{19}$ =4 $a_{29}$ =2 $a_{39}$ =1 $a_{10}$ =5 $a_{20}$ =2 $a_{30}$ =1 $a_{40}$ =4 Similarly, find the first 40 terms with $a_{1}=25:$ $a_{1}$ =25 $a_{11}$ =34 $a_{21}$ =8 $a_{31}$ =4 $a_{2}$ =76 $a_{12}$ =17 $a_{22}$ =4 $a_{32}$ =2 $a_{3}$ =38 $a_{13}$ =52 $a_{23}$ = 2 $a_{33}$ = 1 $a_{4}$ =19 $a_{14}$ =26 $a_{24}$ =1 $a_{34}$ =4 $a_{5}$ =58 $a_{15}$ = 13 $a_{25}$ =4 $a_{35}$ =2 $a_{6}$ =29 $a_{16}$ =40 $a_{26}$ =2 $a_{36}$ =1 $a_{7}$ =88 $a_{17}$ =20 $a_{27}$ = 1 $a_{37}$ =4 $a_{8}$ =44 $a_{18}$ =10 $a_{28}$ =4 $a_{38}$ =2 $a_{9}$ =22 $a_{19}$ =5 $a_{29}$ =2 $a_{39}$ =1 $a_{10}$ =11 $a_{20}$ =16 $a_{30}$ =1 $a_{40}$ =4 We can see that table 1 is contained in table 2 because it starts at $a_{10}$. In other words for $a_{1}=11$, $a_{1}=11$ and for $a_{1}=25$, $a_{10}=11$ Let the first sequence be $A_{n}$ and the second sequence is $B_{n}$. Then the relationship between these two sequences is $B_{n+9}=A_{n}$ Put another way, the sequence always ends with $2, 1, 4$, regardless of the value of $a_1$.