What is the logic in this number sequence?
$begingroup$
I’ve got a calendar with usually fairly simple logic or math puzzles, but was unable to solve this. You are allowed to use multiplication, sum, minus and division to calculate the next number.
The answer was 58, but there was no explaining formula, does anyone have a clue?
Update: an example of another sequence that was much easier to solve is this:
9 - 17 - 14 - 22 - 19 - ?
Where the solution was simply alternating +8 and -3, so the answer here was 27
.
sequences-and-series contest-math puzzle
$endgroup$
|
show 3 more comments
$begingroup$
I’ve got a calendar with usually fairly simple logic or math puzzles, but was unable to solve this. You are allowed to use multiplication, sum, minus and division to calculate the next number.
The answer was 58, but there was no explaining formula, does anyone have a clue?
Update: an example of another sequence that was much easier to solve is this:
9 - 17 - 14 - 22 - 19 - ?
Where the solution was simply alternating +8 and -3, so the answer here was 27
.
sequences-and-series contest-math puzzle
$endgroup$
3
$begingroup$
You're expected to add 3, then 9, then 18, then 27; but why 27, I don't know. Those are all suspiciously three-ful, though.
$endgroup$
– Patrick Stevens
Feb 11 '18 at 9:35
$begingroup$
oeis.org/… has 12 sequences with $1,4,13,31$, but not one of them continues with 58.
$endgroup$
– Gerry Myerson
Feb 11 '18 at 9:38
$begingroup$
Not very positive reviews of the calendar bol.com/nl/p/2018-neurocampus-braintraining-scheurkalender/… (doesn't necesarilly mean anything, but thought it is interesting)
$endgroup$
– Sil
Feb 11 '18 at 9:55
$begingroup$
Maybe it has something to do with Dutch language?
$endgroup$
– Arnaud Mortier
Feb 11 '18 at 10:23
$begingroup$
Normally this is just a math puzzle. I didn't found tricked questions yet on the calendar, and those sequences were pretty easy before. That's why I was so surprised and wondering if I missed something obvious.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:34
|
show 3 more comments
$begingroup$
I’ve got a calendar with usually fairly simple logic or math puzzles, but was unable to solve this. You are allowed to use multiplication, sum, minus and division to calculate the next number.
The answer was 58, but there was no explaining formula, does anyone have a clue?
Update: an example of another sequence that was much easier to solve is this:
9 - 17 - 14 - 22 - 19 - ?
Where the solution was simply alternating +8 and -3, so the answer here was 27
.
sequences-and-series contest-math puzzle
$endgroup$
I’ve got a calendar with usually fairly simple logic or math puzzles, but was unable to solve this. You are allowed to use multiplication, sum, minus and division to calculate the next number.
The answer was 58, but there was no explaining formula, does anyone have a clue?
Update: an example of another sequence that was much easier to solve is this:
9 - 17 - 14 - 22 - 19 - ?
Where the solution was simply alternating +8 and -3, so the answer here was 27
.
sequences-and-series contest-math puzzle
sequences-and-series contest-math puzzle
edited Feb 11 '18 at 13:49
Justus Romijn
asked Feb 11 '18 at 9:23
Justus RomijnJustus Romijn
1294
1294
3
$begingroup$
You're expected to add 3, then 9, then 18, then 27; but why 27, I don't know. Those are all suspiciously three-ful, though.
$endgroup$
– Patrick Stevens
Feb 11 '18 at 9:35
$begingroup$
oeis.org/… has 12 sequences with $1,4,13,31$, but not one of them continues with 58.
$endgroup$
– Gerry Myerson
Feb 11 '18 at 9:38
$begingroup$
Not very positive reviews of the calendar bol.com/nl/p/2018-neurocampus-braintraining-scheurkalender/… (doesn't necesarilly mean anything, but thought it is interesting)
$endgroup$
– Sil
Feb 11 '18 at 9:55
$begingroup$
Maybe it has something to do with Dutch language?
$endgroup$
– Arnaud Mortier
Feb 11 '18 at 10:23
$begingroup$
Normally this is just a math puzzle. I didn't found tricked questions yet on the calendar, and those sequences were pretty easy before. That's why I was so surprised and wondering if I missed something obvious.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:34
|
show 3 more comments
3
$begingroup$
You're expected to add 3, then 9, then 18, then 27; but why 27, I don't know. Those are all suspiciously three-ful, though.
$endgroup$
– Patrick Stevens
Feb 11 '18 at 9:35
$begingroup$
oeis.org/… has 12 sequences with $1,4,13,31$, but not one of them continues with 58.
$endgroup$
– Gerry Myerson
Feb 11 '18 at 9:38
$begingroup$
Not very positive reviews of the calendar bol.com/nl/p/2018-neurocampus-braintraining-scheurkalender/… (doesn't necesarilly mean anything, but thought it is interesting)
$endgroup$
– Sil
Feb 11 '18 at 9:55
$begingroup$
Maybe it has something to do with Dutch language?
$endgroup$
– Arnaud Mortier
Feb 11 '18 at 10:23
$begingroup$
Normally this is just a math puzzle. I didn't found tricked questions yet on the calendar, and those sequences were pretty easy before. That's why I was so surprised and wondering if I missed something obvious.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:34
3
3
$begingroup$
You're expected to add 3, then 9, then 18, then 27; but why 27, I don't know. Those are all suspiciously three-ful, though.
$endgroup$
– Patrick Stevens
Feb 11 '18 at 9:35
$begingroup$
You're expected to add 3, then 9, then 18, then 27; but why 27, I don't know. Those are all suspiciously three-ful, though.
$endgroup$
– Patrick Stevens
Feb 11 '18 at 9:35
$begingroup$
oeis.org/… has 12 sequences with $1,4,13,31$, but not one of them continues with 58.
$endgroup$
– Gerry Myerson
Feb 11 '18 at 9:38
$begingroup$
oeis.org/… has 12 sequences with $1,4,13,31$, but not one of them continues with 58.
$endgroup$
– Gerry Myerson
Feb 11 '18 at 9:38
$begingroup$
Not very positive reviews of the calendar bol.com/nl/p/2018-neurocampus-braintraining-scheurkalender/… (doesn't necesarilly mean anything, but thought it is interesting)
$endgroup$
– Sil
Feb 11 '18 at 9:55
$begingroup$
Not very positive reviews of the calendar bol.com/nl/p/2018-neurocampus-braintraining-scheurkalender/… (doesn't necesarilly mean anything, but thought it is interesting)
$endgroup$
– Sil
Feb 11 '18 at 9:55
$begingroup$
Maybe it has something to do with Dutch language?
$endgroup$
– Arnaud Mortier
Feb 11 '18 at 10:23
$begingroup$
Maybe it has something to do with Dutch language?
$endgroup$
– Arnaud Mortier
Feb 11 '18 at 10:23
$begingroup$
Normally this is just a math puzzle. I didn't found tricked questions yet on the calendar, and those sequences were pretty easy before. That's why I was so surprised and wondering if I missed something obvious.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:34
$begingroup$
Normally this is just a math puzzle. I didn't found tricked questions yet on the calendar, and those sequences were pretty easy before. That's why I was so surprised and wondering if I missed something obvious.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:34
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
With $$a_n=1+frac{n(n+1)(n-1)}2,$$
the next term would be 61 instead.
Seriously, number sequences are highly arbitrary. Who would guess that the next number after $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59$ is $60$ - because it lists the orders of non-trivial simple groups?
$endgroup$
$begingroup$
@quasi well it means that you can create a formula using those operands. So something like (N * 3 + 1) which works for 1,4,13 but doesn't for 31. Those sequences take usually the current number as input for a formula to calculate the next one.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:33
$begingroup$
@hagen-von-eitzen I agree that number sequences can be super arbitrary, but in this case it involves a calendar that usually has fairly straightforward puzzles, so I was surprised that I could not figure out the formula behind this one.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:40
$begingroup$
@hagen-von-eitzen: Apparently I misinterpreted the problem.
$endgroup$
– quasi
Feb 11 '18 at 13:52
add a comment |
$begingroup$
Got the answer from the creators of the calendar:
(translated from Dutch)
The number sequence is 1, 4, 13, 31, 58. You get to this in the following way:
Between 1 and 4 is 3, between 4 and 13 is 9, between 13 en 31 is 18, between 31 and 58 27.
This related to the table of 3's in this way:
1 X 3 = 3
3 X 3 = 9
6 X 3 = 18
9 X 3 = 27
This how you get to the next number.
The first step being 1 x 3
really messes up the sequence in my opinion, but yeah I guess that is just the problem with arbitrary number sequences.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2645609%2fwhat-is-the-logic-in-this-number-sequence%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
With $$a_n=1+frac{n(n+1)(n-1)}2,$$
the next term would be 61 instead.
Seriously, number sequences are highly arbitrary. Who would guess that the next number after $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59$ is $60$ - because it lists the orders of non-trivial simple groups?
$endgroup$
$begingroup$
@quasi well it means that you can create a formula using those operands. So something like (N * 3 + 1) which works for 1,4,13 but doesn't for 31. Those sequences take usually the current number as input for a formula to calculate the next one.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:33
$begingroup$
@hagen-von-eitzen I agree that number sequences can be super arbitrary, but in this case it involves a calendar that usually has fairly straightforward puzzles, so I was surprised that I could not figure out the formula behind this one.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:40
$begingroup$
@hagen-von-eitzen: Apparently I misinterpreted the problem.
$endgroup$
– quasi
Feb 11 '18 at 13:52
add a comment |
$begingroup$
With $$a_n=1+frac{n(n+1)(n-1)}2,$$
the next term would be 61 instead.
Seriously, number sequences are highly arbitrary. Who would guess that the next number after $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59$ is $60$ - because it lists the orders of non-trivial simple groups?
$endgroup$
$begingroup$
@quasi well it means that you can create a formula using those operands. So something like (N * 3 + 1) which works for 1,4,13 but doesn't for 31. Those sequences take usually the current number as input for a formula to calculate the next one.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:33
$begingroup$
@hagen-von-eitzen I agree that number sequences can be super arbitrary, but in this case it involves a calendar that usually has fairly straightforward puzzles, so I was surprised that I could not figure out the formula behind this one.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:40
$begingroup$
@hagen-von-eitzen: Apparently I misinterpreted the problem.
$endgroup$
– quasi
Feb 11 '18 at 13:52
add a comment |
$begingroup$
With $$a_n=1+frac{n(n+1)(n-1)}2,$$
the next term would be 61 instead.
Seriously, number sequences are highly arbitrary. Who would guess that the next number after $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59$ is $60$ - because it lists the orders of non-trivial simple groups?
$endgroup$
With $$a_n=1+frac{n(n+1)(n-1)}2,$$
the next term would be 61 instead.
Seriously, number sequences are highly arbitrary. Who would guess that the next number after $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59$ is $60$ - because it lists the orders of non-trivial simple groups?
edited Feb 12 '18 at 1:43
Gerry Myerson
147k8149302
147k8149302
answered Feb 11 '18 at 9:35


Hagen von EitzenHagen von Eitzen
282k23272505
282k23272505
$begingroup$
@quasi well it means that you can create a formula using those operands. So something like (N * 3 + 1) which works for 1,4,13 but doesn't for 31. Those sequences take usually the current number as input for a formula to calculate the next one.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:33
$begingroup$
@hagen-von-eitzen I agree that number sequences can be super arbitrary, but in this case it involves a calendar that usually has fairly straightforward puzzles, so I was surprised that I could not figure out the formula behind this one.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:40
$begingroup$
@hagen-von-eitzen: Apparently I misinterpreted the problem.
$endgroup$
– quasi
Feb 11 '18 at 13:52
add a comment |
$begingroup$
@quasi well it means that you can create a formula using those operands. So something like (N * 3 + 1) which works for 1,4,13 but doesn't for 31. Those sequences take usually the current number as input for a formula to calculate the next one.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:33
$begingroup$
@hagen-von-eitzen I agree that number sequences can be super arbitrary, but in this case it involves a calendar that usually has fairly straightforward puzzles, so I was surprised that I could not figure out the formula behind this one.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:40
$begingroup$
@hagen-von-eitzen: Apparently I misinterpreted the problem.
$endgroup$
– quasi
Feb 11 '18 at 13:52
$begingroup$
@quasi well it means that you can create a formula using those operands. So something like (N * 3 + 1) which works for 1,4,13 but doesn't for 31. Those sequences take usually the current number as input for a formula to calculate the next one.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:33
$begingroup$
@quasi well it means that you can create a formula using those operands. So something like (N * 3 + 1) which works for 1,4,13 but doesn't for 31. Those sequences take usually the current number as input for a formula to calculate the next one.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:33
$begingroup$
@hagen-von-eitzen I agree that number sequences can be super arbitrary, but in this case it involves a calendar that usually has fairly straightforward puzzles, so I was surprised that I could not figure out the formula behind this one.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:40
$begingroup$
@hagen-von-eitzen I agree that number sequences can be super arbitrary, but in this case it involves a calendar that usually has fairly straightforward puzzles, so I was surprised that I could not figure out the formula behind this one.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:40
$begingroup$
@hagen-von-eitzen: Apparently I misinterpreted the problem.
$endgroup$
– quasi
Feb 11 '18 at 13:52
$begingroup$
@hagen-von-eitzen: Apparently I misinterpreted the problem.
$endgroup$
– quasi
Feb 11 '18 at 13:52
add a comment |
$begingroup$
Got the answer from the creators of the calendar:
(translated from Dutch)
The number sequence is 1, 4, 13, 31, 58. You get to this in the following way:
Between 1 and 4 is 3, between 4 and 13 is 9, between 13 en 31 is 18, between 31 and 58 27.
This related to the table of 3's in this way:
1 X 3 = 3
3 X 3 = 9
6 X 3 = 18
9 X 3 = 27
This how you get to the next number.
The first step being 1 x 3
really messes up the sequence in my opinion, but yeah I guess that is just the problem with arbitrary number sequences.
$endgroup$
add a comment |
$begingroup$
Got the answer from the creators of the calendar:
(translated from Dutch)
The number sequence is 1, 4, 13, 31, 58. You get to this in the following way:
Between 1 and 4 is 3, between 4 and 13 is 9, between 13 en 31 is 18, between 31 and 58 27.
This related to the table of 3's in this way:
1 X 3 = 3
3 X 3 = 9
6 X 3 = 18
9 X 3 = 27
This how you get to the next number.
The first step being 1 x 3
really messes up the sequence in my opinion, but yeah I guess that is just the problem with arbitrary number sequences.
$endgroup$
add a comment |
$begingroup$
Got the answer from the creators of the calendar:
(translated from Dutch)
The number sequence is 1, 4, 13, 31, 58. You get to this in the following way:
Between 1 and 4 is 3, between 4 and 13 is 9, between 13 en 31 is 18, between 31 and 58 27.
This related to the table of 3's in this way:
1 X 3 = 3
3 X 3 = 9
6 X 3 = 18
9 X 3 = 27
This how you get to the next number.
The first step being 1 x 3
really messes up the sequence in my opinion, but yeah I guess that is just the problem with arbitrary number sequences.
$endgroup$
Got the answer from the creators of the calendar:
(translated from Dutch)
The number sequence is 1, 4, 13, 31, 58. You get to this in the following way:
Between 1 and 4 is 3, between 4 and 13 is 9, between 13 en 31 is 18, between 31 and 58 27.
This related to the table of 3's in this way:
1 X 3 = 3
3 X 3 = 9
6 X 3 = 18
9 X 3 = 27
This how you get to the next number.
The first step being 1 x 3
really messes up the sequence in my opinion, but yeah I guess that is just the problem with arbitrary number sequences.
answered Feb 15 '18 at 9:58
Justus RomijnJustus Romijn
1294
1294
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2645609%2fwhat-is-the-logic-in-this-number-sequence%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
$begingroup$
You're expected to add 3, then 9, then 18, then 27; but why 27, I don't know. Those are all suspiciously three-ful, though.
$endgroup$
– Patrick Stevens
Feb 11 '18 at 9:35
$begingroup$
oeis.org/… has 12 sequences with $1,4,13,31$, but not one of them continues with 58.
$endgroup$
– Gerry Myerson
Feb 11 '18 at 9:38
$begingroup$
Not very positive reviews of the calendar bol.com/nl/p/2018-neurocampus-braintraining-scheurkalender/… (doesn't necesarilly mean anything, but thought it is interesting)
$endgroup$
– Sil
Feb 11 '18 at 9:55
$begingroup$
Maybe it has something to do with Dutch language?
$endgroup$
– Arnaud Mortier
Feb 11 '18 at 10:23
$begingroup$
Normally this is just a math puzzle. I didn't found tricked questions yet on the calendar, and those sequences were pretty easy before. That's why I was so surprised and wondering if I missed something obvious.
$endgroup$
– Justus Romijn
Feb 11 '18 at 10:34