This is a sequence puzzle I found online which I'm stuck on:

452801, 773924, 102410, 471056, ?

I figured out how to get from the first number to the second (+3, +2, +1, +1, +2, +3) for the digits, but this rule doesn't work for the rest. Any ideas?

  • $\begingroup$ 050321 might be next number in sequence $\endgroup$ – CR241 Oct 3 '18 at 23:54
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    $\begingroup$ I checked, it's not @CR241 $\endgroup$ – Duck Oct 25 '18 at 23:35
  • $\begingroup$ I haven't had the time yet, but seems like it might be a determined value being added each time, but the sum is capped as would a variable in software be (example, if you do addition with an integer 16-bit value and the answer is above the cap, it literally cuts off the highest value digits to keep within the integer 16-bit value range) $\endgroup$ – nine9 Mar 1 at 8:28

The next number in the sequence is:


I got it by creating an algorithm that tested positive integers until a result was generated indicating that the question was answered correctly. The algorithm submitted an answer one at a time and waited for a response before submitting the next answer to make sure that the website wasn't overwhelmed.

For the curious - it took about 5 hours.

Even with the answer, I currently haven't found a strong correlation between the numbers in the sequence however @Mr Pie's strange observation is correct!

  • 1
    $\begingroup$ Hahah thank you! I thought about doing it myself but never did it. Thanks for doing it for me. I upvoted but haven't accepted your answer yet since like you said you haven't found a pattern yet. $\endgroup$ – simonzack May 3 at 10:57
  • $\begingroup$ Oh, what do you know! :P $\endgroup$ – Mr Pie May 5 at 2:58

One thing all 4 numbers have in common is this :

Rule: They are all the last 6 digits of the integer part of the quarter-square of some integer.


The quarter square of a number $x$ is defined as $(\frac{x}{2})^2$ or $\frac{x^2}{4}$. A list of quarter-squares upto 100,000 can be found here.

For example

The quarter-square of the number 216,371 is :


The last 6 digits of the integer part are 102,410.

Going up to 2 million,

which is a 7-digit number, and also a period for the above rule,

we can find a list of numbers that adhere to the above rule :

452801 : [69698, 492802, 507198, 930302, 1069698, 1492802, 1507198, 1930302]

773924 : [54764, 179764, 320236, 445236, 554764, 679764, 820236, 945236, 1054764, 1179764, 1320236, 1445236, 1554764, 1679764, 1820236, 1945236]

102410 : [216371, 627379, 1372621, 1783629]

471056 : [53832, 76065, 94815, 196168, 303832, 305185, 323935, 446168, 476065, 494815, 553832, 696168, 705185, 723935, 803832, 876065, 894815, 946168, 1053832, 1105185, 1123935, 1196168, 1276065, 1294815, 1303832, 1446168, 1505185, 1523935, 1553832, 1676065, 1694815, 1696168, 1803832, 1905185, 1923935, 1946168]

The above lists were calculated using

this code written in Swift.

But I still can't find a relationship between the numbers in those lists.

I hope this is helpful!

  • 1
    $\begingroup$ Wow nice a lot of work I upvoted cause of that, still not sure about the answer. $\endgroup$ – simonzack Mar 30 at 13:58
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    $\begingroup$ How did you find that?! :o $\endgroup$ – Mr Pie Apr 8 at 3:36

This is not exactly a pattern, but there is one odd thing these numbers have in common:

If you get each term and divide it by $\sqrt{e^\pi}$ then it is close to an integer.

Proof (correct to $5$ decimal places):

$452801\div\sqrt{e^\pi}=94128.08005\approx 94128$

$773924\div\sqrt{e^\pi}=160822.99325\approx 160823$

$102410\div\sqrt{e^\pi}=21288.94741\approx 21289$

$471056\div\sqrt{e^\pi}=97922.92172\approx 97922$

It additionally appears that

The answers are becoming less nearer to an integer the further down the sequence we continue, thus perhaps we have to add or subtract a constant to each term that correspondingly gets bigger, in order to make the answers become closer to an integer.

This might have to do with the overall pattern.

Hope this helps!

  • 2
    $\begingroup$ Yea in Adam's answer we have $216101\div\sqrt{e^\pi}=44922.984328976$. $\endgroup$ – simonzack May 6 at 6:21

I am not quite sure but was able to make some quite good relation

1. 452801=a , 773924=b , 102410=c , 471056=d
2. Now let us take b-a=321123=A , d-c=368646=B , b-c=671514=C , d-a=18255=D
3. Adding A+B=689769
4. Adding C+D=689769
5. From above A+B=C+D
6. Considering the same pattern the answer might be 157715
7. Let us take x as required number
8. 773924-x + 102410-x = 671514+47105-x
9. X=876334-718619=157715

  • 2
    $\begingroup$ hate to burst your bubble, but A+B=b+d-a-c=C+D $\endgroup$ – JonMark Perry Mar 1 at 10:30
  • $\begingroup$ Oh let me cut that first sentence $\endgroup$ – user56760 Mar 1 at 10:41
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    $\begingroup$ you can type your guess into the website mentioned in OQ (you don't need to enter all answers) and it marks it for you. for example, the answer to (3) is 25,1 (previously solved on PSE) and press submit at the bottom of the page. $\endgroup$ – JonMark Perry Mar 1 at 10:54

protected by JonMark Perry Apr 8 at 19:27

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