Next in Sequence: What Comes After 61?

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Andre
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Just an idea.

which number is the next in sequence?

21, 34, 57, 61,...
 
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24, 35??

Just a thought. :biggrin:
 
Nice try, but no, the next one is perfectly logical and explainable, but there is a twist.
 
How about; 93, 88, 129, 115?
 
You're guessing :-p

You really need to find the logic, but think about the twist
 
Ambiguous twists actually :rolleyes:
 
Amazing. But you know you are right :smile:

You defeated your heuristic bias. Great job.

Can anybody else explain why?
 
We're still looking for the explanation, I guess. Defeating heuristic bias is obviously thinking out of the box. And the box here is obviously that a row of numbers usually means figuring out, which calculations are consistent with those numbers.
 
Andre said:
Just an idea.

which number is the next in sequence?

21, 34, 57, 61,...


There is no unique answer to these. It's a number based on your rule,
or the creator of the sequence (if not you).

You could be asking if we can figure out what rule/process you
happened to use.
 
The rule is...
that at least one of the lines of the left character/symbol is curved, twisted: 2,3,5,6,8,9,0 (the hint of the twist -used double-, ambigeously) while the right symbol is made of straight lines 1,4,7.

I wonder which other hidden -unintended- rules could exist. Can you give an example?
 
Last edited:
Okay, out of an infinite number of possibilities for the rules, I chose this one:


After starting out with 23,

it could be for the subsequent tens digits that a person can
add 1, add 2, add 1, add 2, etc. to the previous tens digit.


And for the subsequent units digits, they could (take turns)
cycling as 1, 4, 7, 1, 4, 7, etc.


[tex]Hence, \ \ \ 21, \ 34, \ 57, \ 61, \ \boxed{84}, \ ...[/tex]





As an aside, using my method, if I were to figure the next number
following 84, I would add 1 to 8 to get 9 for the tens digit,
and the units digit would be a 7 from the cycling as mentioned above.


[tex]\ ... , \ 84, \ 97, \ ...[/tex]
 
I got a better rule. :biggrin:

21 → [2+1=3] And [21/7=3]
+13
34 → [3+4=7]
+23
57 → [5+7=12→1+2=3] And [57/3=19]
+4
61 → [3+4=7]
+23
84 → [8+4=12→1+2=3] And [84/7=12]
+13
97 → [3+4=7]
+23
120 → [1+2+0=3] And [120/3=40] [Hello checkitagain. :biggrin:]
+4
124 → [3+4=7]
+23
147 → [1+4+7=12→1+2=3] And [147/7=21]
+13
160 → [3+4=7]
+23
183
...


Rule: if the sum of all characters [digits] of the given number, results in 7 [keep adding the resultant digits until you end up with only one digit] then you add up 23 to the number to get the next in sequence. The other resultant number you may get other than 7, is 3. Here you'll have to add up 4 or 13 to the number. If that same number accepts the division by seven [it should result in an integer quotient] then you'd have to add 13 to get the following number. But if it accepts the division by three, then you add up 4 to the number. As you can see, you don't have to do this calculations all through, just figure the first next in sequence, and then add up 4 or 13 alternatively, of course with 23 in between.


You just can't do this intuitively. :biggrin:
 
-1, 2, 3, 8, 15, 24, 35, ...

Any takers?
 
I like Serena said:
That's... very twisted! As requested by andre. :biggrin:

Well, couldn't come up with any convincing method other than this.

So, what's yours??
 
Gad said:
Well, couldn't come up with any convincing method other than this.

So, what's yours??

Well, I was thinking of doing the twist.
So I was thinking left-right-left-right.
So I made 2 sequences from 21, 34, 57, 61.
21-57 and 34-61
First digit goes 3 up, second digit goes 6 up.
And since 4+6=10, summing those digits gives 1.

Next would be 57 + 36 -> (8)(7+6) -> (8)(13) -> 84!

To my surprise, andre was sure I had the right answer.

"Come on let's twist again like we did last summer" :cool:
 
QuarkCharmer said:
-1, 2, 3, 8, 15, 24, 35, ...

Any takers?


QuarkCharmer,

I would bet a lot that you have a typo, and that your
sequence should instead be:


-1, 0, 3, 8, 15, 24, 35, ...


In that case, the differences between consecutive
entries are consecutive odd integers.


So 35 + 13 =
[tex]\boxed{48}[/tex]