3
$\begingroup$

Find the correct pattern and continue the following sequence

4, 3, 4, 7, 6, 5, 2, 6, 8, 7, 3, ...

You should be able to find at least one hint in the title


EDIT

Just realized I made a little mistake. It is still a valid sequence, but I need to give a hint, because it would be too arbitrary without this one.

The sequence will never contain a number greater than 8


Hint 1

Further explanation of the initial hint, you will need mod 9

Hint 2

switch (i mod x) {...}

Hint 3

In 1748 you were able to continue this sequence to the 23th 24th number

Hint 4

The title might be a hint on (not only) a important number used in this puzzle, even it's not there in its true form

Hint 5

The sequence is based on a number

Hint 6

A hint on how you could create the sequence with some kind of pseudocode (addition to hint 2)

for (i = 0; i < basenumber.length; i++)
     switch (i mod x) {...}

Hint 7

$2 * 2 = 4, 7 * 3 = 21, 1 * 4 = 4$

$\endgroup$
  • $\begingroup$ Ah, cr*p. I was about to ask exactly what you already changed about of 1748. $\endgroup$ – Racso Sep 27 '18 at 22:35
  • $\begingroup$ @Racso yeah I was a bit in a rush when I wrote this hint :D $\endgroup$ – Ian Fako Sep 28 '18 at 6:58
  • $\begingroup$ I think I got it. $\endgroup$ – Racso Sep 28 '18 at 12:30
2
+50
$\begingroup$

The sequence continues:

$2,0,0,7$

This is because:

$\text{e}=2.71\;828\;182\;845\;904\dots$.

Your sequence is $2a_0, 3a_1, 4a_2, 2a_3, 3a_4, 4a_5, \dots \pmod 9$.

So $4,21,4,\;16,6,32,\;2,24,8,\;16,12,20,\;18,0,16 \dots \pmod9$
$ \to 4,3,4,\;7,6,5,\;2,6,8,\;7,3,2,\;0,0,7\dots$

$\endgroup$
  • 1
    $\begingroup$ @user477343 actually he was the first to suggest this answer, so he will be awarded :D $\endgroup$ – Ian Fako Sep 29 '18 at 11:43
  • $\begingroup$ @IanFako oh, well, okey then. I'll delete my comment. $(+1)$ :) $\endgroup$ – Feeds Sep 29 '18 at 11:44
4
$\begingroup$

I might have the answer, although it could be a coincidence

After reading your edit:

I suspected the answer might be found using modular arithmetic, specifically mod 9

No lightbulb yet, but I thought:

Perhaps I can combine the digits into 2-digit numbers and the remainder after mod 9 might be of interest

Then I noticed:

Skipping the first number, the next two form 34, then 76, 52, 68, 73
I found that:
34 = 4 * 9 + 8
76 = 8 * 9 + 4

Surely there is a mathematical term for it, but let's call them 'inverse modulo's' for now :)

This also holds for the next two numbers:
52 = 5 * 9 + 7
68 = 7 * 9 + 5

That just leaves
73 = 8 * 9 + 1
and our answer
1 * 9 + 8 = 17
Meaning the next two numbers are 1, 7

$\endgroup$
  • 1
    $\begingroup$ Interesting, that this works up to this point. I checked if it holds for the next couple of numbers but it seems to stop after the last 3 in the sequence. Anyway +1 for the effort $\endgroup$ – Ian Fako Sep 13 '18 at 6:47
  • $\begingroup$ Thanks for checking the remainder of the sequence. Could have been an artefact of the actual sequence, but now I know to keep searching $\endgroup$ – P1storius Sep 13 '18 at 7:13
3
$\begingroup$

Answer: If I'm correct, the first digits of the sequence are one of the following:

43476526873290716319

43476526873200716310

The second one takes into account the "mod 9" rule.

Here's the algorithm I used:

Take each digit of e = 271828182845904523536 (decimal separator removed) and its position i = 0, 1, 2... That means (2,0), (7,1), (1,2), (8,3), ...

Calculate d[i](i%3+2). That means:
2
(0%3+2) = 2*2 = 4
7*(1%3+2) = 7*3 = 21
1*(2%3+2) = 1*4 = 4
8*(3%3+2) = 8*2 = 16

Add the digits of each answer together. That leads us to 4, 3, 4, 7, ...

TL; DR:

Multiply each digit of e (including the whole part) by 2, 3, 4, 2, 3, 4, 2, 3, 4, ... and add the digits of the result.

$\endgroup$
  • 1
    $\begingroup$ I'd say you're pretty close ;) $\endgroup$ – Ian Fako Sep 27 '18 at 16:10
  • $\begingroup$ correct now but JonMark Perry was faster, +1 anyway $\endgroup$ – Ian Fako Sep 29 '18 at 11:44
2
$\begingroup$

I would assume this is part of the pattern:

Six consecutive numbers: A,B,C,D,E,F

A = some number
B = A + x
C = A + B
D = C - 1
E = D - x
F = E - A
(next A = F)

I just don't know the rule for x..
But I think for the solution I don't need to know x.
Applying my pattern to your given numbers I would get 1 as the next number.

$\endgroup$
  • $\begingroup$ A can't see how your solution matches even the first few numbers. Can you please illustrate your algorithm with the specific numbers from the puzzle? $\endgroup$ – rhsquared Sep 10 '18 at 13:14
  • $\begingroup$ I just assumed the given numbers are in the following order: E,F/A,B,C,D,E,F/A,B,C,D,E,... Therefore the last given number (3) would be E, and the next numbers would be F/A,B,C,D,... (keep in mind A and F label the same position in my solution) $\endgroup$ – npkllr Sep 10 '18 at 13:24
  • $\begingroup$ In order to convince me that this is correct, you're going to need to define the heuristics for x. You should also explain the hint provided in the title and its relation to your solution. $\endgroup$ – Ian MacDonald Sep 10 '18 at 13:29
  • $\begingroup$ @IanMacDonald That's why I used x: I just don't know the heuristics for x. I also don't understand the hint in the title. I assume with the hint it should be possible to define the rules for x. $\endgroup$ – npkllr Sep 10 '18 at 13:37
  • $\begingroup$ Sorry you're on the wrong track ;) $\endgroup$ – Ian Fako Sep 10 '18 at 14:23
2
$\begingroup$

I think I got the hint in the title.

The capital letters in the title are E, E, E and + 1 which when read is 'three E plus 1', which is equal to 4. I think it gives a hint as to how the first term is derived for the sequence.

$\endgroup$
  • $\begingroup$ Not bad, this is partially correct $\endgroup$ – Ian Fako Sep 13 '18 at 6:33
2
$\begingroup$

Partial Answer

So clearly going by the hints and the title the sequence is based off the number

e

Because

The title has E capitalized multiple times. Euler calculated e to the 23rd digit back in 1748.

This is where I'm stuck...

$\endgroup$
  • $\begingroup$ So far so good ;) $\endgroup$ – Ian Fako Sep 21 '18 at 11:05
1
$\begingroup$

Just notes to myself.

I have a feeling that the the title could mean either 15 or 16 (since E=5 in alphabet). Otherwise, it might mean 1555 (as in hint 3 you suggested a year). Another find from the title is that the words have 8 3 8 letters, and the sequence starts with 4 3 4. Probably does not have to do with anything though.

From hint 2:

Switch..case should change logic based on module of previous number, could be mod 2, mod 3 or something else. Checked a couple of ideas, no luck so far.

From hint 3:

I suspect that 2018 is somehow of use here.

Finally,

Tried combining numbers, since it seemed promising: 43 34 47 76 65... Results of division by 9 were 4(+7) 3(+7) 5(+2) 8(+4) 7(+2). For a moment I thought that next number is created by the division and mod, but didn't work out for now. Maybe there is something here.

Will continue gnawing on this puzzle.

$\endgroup$
  • $\begingroup$ I like the way you try to solve this. Not quite there yet but not bad. And a little reminder, you can solve all my puzzles without hints (unlikely but not impossible :D) $\endgroup$ – Ian Fako Sep 18 '18 at 17:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.