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Predict the next three members of the sequence below and explain what the relationship is.

2, 10, 44, 1012, 248,

This number sequence does not appear in the Online Encyclopedia of Integer Sequences

Hint #1

The series above is closely related to the following series of numbers

2, 3, 5, 22, 17, 250, 134, 262

So if you can work out the mathematical rule for calculating the members of this series then you will be very close to the mathematical rule for the sequence above.

Hint #2

I thought this was an almost perfect number sequence...

I don't think another hint will be required, but I will add another if necessary

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    $\begingroup$ Does the relationship need to explain all the elements in the set that you provide? Or, for instance, is it enough to present a relationship that uses 2, 10, 44 and 1012 to get 248, and use that relationship to predict the next 3? $\endgroup$ – cinico May 24 at 11:11
  • $\begingroup$ @cinico There is a mathematical rule that can be used to predict each member of the series. So this rule gives 2, 10, 44.... and all the other numbers in the series. $\endgroup$ – tom May 25 at 22:24
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The nth number in the sequence is

the nth positive integer such that its deficiency is 2^(n-1).

That might have been a bit confusing, so here's examples:

2 is the 1st number where the sum of the proper divisors is 1 less than the number: the sum is 1 and 1=2-1.

10 is the second where the sum of proper divisors is 2 less: 3 is the first (3-1=2) and 10 is the second (10-(1+2+5)=2).

44 is the 3rd where the sum of proper divisors is 4 less: 5 is the first, 14 is the second (14-(1+2+7)=4) and 44 is the 3rd (44-(1+2+4+11+22)=4.)

This makes the next 3 terms

the 6th number with deficiency 32, the 7th with deficiency 64, and the 8th with deficiency 128, which are 396916, 4064, and 27064 according to Mathematica.

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  • $\begingroup$ Great., well done.. but can you predict the next three terms.... $\endgroup$ – tom May 27 at 13:32
  • $\begingroup$ @tom added them. $\endgroup$ – ev3commander May 27 at 13:40
  • $\begingroup$ Ah... ever so slight error with your formula and calculation - consider the case n=1 for the first member of the series..... minor mod required to your formula and then this will have an effect on the next 3 numbers $\endgroup$ – tom May 27 at 13:53
  • $\begingroup$ oh, i see, fixed it $\endgroup$ – ev3commander May 27 at 13:55
  • $\begingroup$ Good job, too early to award bounty - should be able to do it this time tomorrow - you are welcome to remind me - good job :-) $\endgroup$ – tom May 27 at 14:01

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