Problem 44337. Sums of Distinct Powers

Created by James

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You will be given three numbers: base, nstart, and nend. Write a MATLAB script that will compute the sum of a sequence of both the distinct powers of base as well as sums of distinct powers of base. Your sequence should start with the 'nstart'th term and end with the 'nend'th term. For example:<ul><li>base=4</li><li>nstart=2</li><li>nend=6</li></ul>The first several sums of the distinct powers of 4 are:<ul><li>1 (4^0)</li><li>4 (4^1)</li><li>5 (4^1 + 4^0)</li><li>16 (4^2)</li><li>17 (4^2 + 4^0)</li><li>20 (4^2 + 4^1)</li><li>21 (4^2 + 4^1 + 4^0)</li><li>64 (4^3)</li><li>65 (4^3 + 4^0)</li></ul>Since nstart=2 and nend=6 in this example, you take the second through the sixth terms of this sequence. The correct output would be 4+5+16+17+20, or 62. Notice that the number 8 does not occur in this pattern. While 8 is a multiple of 4, 8=4^1+4^1. Because there are two 4^1 terms in the sum, 8 does not qualify as a sum of distinct powers of 4. You can assume that all three will be integers, base&gt;1, and that nstart&lt;nend. Good luck!

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Last Solution submitted on Mar 30, 2023

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@bmtran (Bryant Tran) on 17 Oct 2017

I'm having some trouble understanding the problem. Could you provide an example for test 7?

base=7;nstart=1;nend=10;y_correct=1265;

My understanding of the problem leads me to believe the answer should be 1600: 1 + 7 + 8 + 49 + 50 + 56 + 343 + 344 + 350 + 392

James on 18 Oct 2017

You've almost got it, bmtran. You are missing 7^2 + 7^1 + 7^0 = 57 in your example. Put that in, remove the 392 (since that's term #11 in the sequence) and you have the correct answer.

@bmtran (Bryant Tran) on 18 Oct 2017

Thanks, that makes sense. I was under the impression that it could only be the sum of 2 distinct powers based on the example.

James on 18 Oct 2017

I've changed the description to make it a bit clearer that you can use any number of terms in the sequence. Hope this helps.

Rafael S.T. Vieira on 8 Jul 2020

The problem wants all unique combinations of distinct powers of a given base up to a point. If we forget some factors it will overflow, if we repeat sequences, their sum will not match the test cases.

that means
0
1
2
3
1 0
2 0
2 1
3 0
3 1
3 2
2 1 0
3 1 0
3 2 0
3 2 1
and so on (and resulting numbers must be sorted if we are not following their natural order)

Rafael S.T. Vieira on 8 Jul 2020

It is not clear at all, James, that we must sum all possible unique combinations of exponents for a base. However, it is a good problem.

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