Question 92400
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What are the next five terms in this sequence?:
3,10,5,16,8,4,2,1,4.... ???

The recursive rule seems to be:

a<sub>1</sub> = 3
If a<sub>n</sub> is odd, use formula a<sub>n+1</sub> = 3a<sub>n</sub>+1
If a<sub>n</sub> is even, use the formula a<sub>n+1</sub> = a<sub>n</sub>/2

So,

a<sub>1</sub> = 3, which is odd, so a<sub>2</sub> = 3a<sub>1</sub>+1 = 3(3)+1 = 9+1 = 10
a<sub>2</sub> = 10, which is even, so a<sub>3</sub> = a<sub>2</sub>/2 = 10/2 = 5
a<sub>3</sub> = 5, which is odd, so a<sub>4</sub> = 3a<sub>3</sub>+1 = 3(5)+1 = 15+1 = 16
a<sub>4</sub> = 16, which is even, so a<sub>5</sub> = a<sub>4</sub>/2 = 16/2 = 8
a<sub>5</sub> = 8, which is even, so a<sub>6</sub> = a<sub>5</sub>/2 = 8/2 = 4
a<sub>6</sub> = 4, which is even, so a<sub>7</sub> = a<sub>6</sub>/2 = 4/2 = 2
a<sub>7</sub> = 2, which is even, so a<sub>8</sub> = a<sub>7</sub>/2 = 2/2 = 1
a<sub>8</sub> = 1, which is odd, so a<sub>9</sub> = 3a<sub>8</sub>+1 = 3(1)+1 = 3+1 = 4
a<sub>9</sub> = 4, which is even, so a<sub>10</sub> = a<sub>9</sub>/2 = 4/2 = 2
a<sub>10</sub> = 2, which is even, so a<sub>11</sub> = a<sub>10</sub>/2 = 2/2 = 1
a<sub>11</sub> = 1, which is odd, so a<sub>12</sub> = 3a<sub>11</sub>+1 = 3(1)+1 = 3+1 = 4
a<sub>12</sub> = 4, which is even, so a<sub>13</sub> = a<sub>12</sub>/2 = 4/2 = 2
a<sub>13</sub> = 2, which is even, so a<sub>14</sub> = a<sub>13</sub>/2 = 2/2 = 1
a<sub>14</sub> = 1, which is odd, so a<sub>15</sub> = 3a<sub>14</sub>+1 = 3(1)+1 = 3+1 = 4
a<sub>15</sub> = 4, which is even, so a<sub>16</sub> = a<sub>15</sub>/2 = 4/2 = 2

After a<sub>5</sub>, which = 8, the terms go 4,2,1, over and over forever.

Edwin</pre>