Pointless sequence question?

Question

1, 3, 16, 123, 1453, .......

list the next 5 terms.

easy, i know. sorry, i just need to burn the points.

bonus 1 : what is the most frequent last digit in this sequence?

bonus 2 : the THIRD cube number in this sequence is the ---th term. what is --- and the cuberoot?

Answer 1

It comes from the Fibonacci sequence.

1 = 1

1 + 2 = 3

3 + 5 + 8 = 16

13 + 21 + 34 + 55 = 123

89 + 144 + 233 + 377 + 610 = 1453

987 + 1597 + 2584 + 4181 + 6765 + 10946 = 27060

17711 + 28657 + 46368 + 75025 + 121393 + 196418 + 317811 = 803383

514229 + 832040 + 1346269 + 2178309 + 3524578 + 5702887 + 9227465 + 1493052 = 38256129

24157817 + 39088169 + 63245986 + 102334155 + 165580141 + 267914296 + 433494437 + 701408733 + 1134903170 = 2932126904



1836311903 + 2971215073 + 4807526976 + 7778742049 + 12586269025 + 20365011074 + 32951280099 + 53316291173 + 86267571272 + 139583862445 =362464081089

Answer 2

Not a clue.

Lame, I know. I need the 2 points.  

Answer 3

JonDihon answered the real question correctly.

Bonus 1: Just as there exists a 60-recurrence for a Fibonacci sequence, there exists a 120-recurrence for this row-sum-of-fibonacci-triangle sequence.

1/3 of the units digit is even, 2/3 of the units digit is odd.

The most frequently occuring units digit is a tie between 1 and 9, both occuring 18 times in the 120-recurrence.

Bonus 2: So far I've only been able to come up with the 1st cube number... it happens to be the 1st term, and 1 is the cuberoot =P

Just out of curiosity, how do you come up with these questions?