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The number sequence is 

73, 70, 49, 52, 35, ?, ?, ?, ?, ?, 72,.....

Can you fill in the gaps and predict what happens after 72?

I have updated the hints below.

Hint 1

What are the factors of each number? Calculate factors for each number!

Hint 2

This is a mathematical problem, but does not require complicated maths

Hint 3

If a sequence using exactly the same mathematical rules is started with a different number a different pattern develops. For example; [ 1, 1, 1, 1, ........ ] and [ 48, 48, 48, 48, ........]

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The 5 missing values are:

24, 12, 3, 9, 81

After 72:

Each subsequent value will be 9 less until it gets to 9, then start over at 81, repeating forever.

The method:

To calculate the next value, add the digits of the current value and multiply by the first digit of the current value.

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  • $\begingroup$ Great Job - Well done :-) $\endgroup$
    – tom
    Mar 2 '18 at 23:47
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The gaps:

34, 21, 16, 7, 0

Reasoning:

Every other number is in octal. When converted to decimal, the values would be:
73, 56, 49, 42, 35, 28, 21, 14, 7, 0
Which (after 73) is of course multiples of 7 descending

What comes after 72 (less sure about this part):

I think the 73, 72, is another countdown. The first number in each sub-sequence is the largest multiple of 7 less than octal 73 or octal 72, etc.

So the sequence will continue...

73, 70, 49, 52, 35, 34, 21, 16, 7, 0
72, 70, 49, 52, 35, 34, 21, 16, 7, 0
71, 70, 49, 52, 35, 34, 21, 16, 7, 0
70, 61, 42, 43, 28, 25, 14, 7, 0
(Sub-sequence has shifted left and octal/decimal values are flipped)

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  • $\begingroup$ Nice reasoning and ideas, but very sorry this is not correct.... $\endgroup$
    – tom
    Mar 1 '18 at 23:45
  • $\begingroup$ ...you are correct to be working along mathematical lines, but I am afraid that this particular approach is not correct $\endgroup$
    – tom
    Mar 1 '18 at 23:47
  • $\begingroup$ The idea of every other number in octal giving the sequence is brilliant, but sadly not correct.... --- so +1 for a great idea, but sorry not the correct approach $\endgroup$
    – tom
    Mar 1 '18 at 23:50
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    $\begingroup$ @tom haha, thank you :) no need to apologize $\endgroup$
    – ferret
    Mar 1 '18 at 23:55

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