How would you find the formula for the following numbers in a list?
0
$begingroup$
177 (starting number, i don't know the previous numbers)
188 + 12 (188 + 12 = 200 and so forth)
200 + 12
212 + 12
224 + 13
237 + 13
250 + 13
263 + 14
277 + 14
291 + 15
306 + 15
321 + 15
336 + 15
351 + 16
367 + 16
383 + 17
400 + 17
417 + 17
434 + 18
452 + 18
470 + 18
488 + 19
507 + 19
526 + 19
545 + 20
565 + 20
585 + 20
605 + 21
626 + 21
647 + 22
669 + 22
691 + 22
713 + 22
735 + 23
758 + 23
781 + 24
805 + 24
829 + 24
853 + 25
878 + 25
903 + 25
928 + 26
954 (ending number, I do not know the next numbers)
I've been trying to figure out the formula to these numbers, but without success. Wondering if anyone could give it an attempt to figure it out?
I want to be able to predict the coming numbers, after 954. This is hard.
probability
asked Dec 16 '18 at 0:18
Jeffrey AndersonJeffrey Anderson
11
$endgroup$
add a comment |
0
$begingroup$
177 (starting number, i don't know the previous numbers)
188 + 12 (188 + 12 = 200 and so forth)
200 + 12
212 + 12
224 + 13
237 + 13
250 + 13
263 + 14
277 + 14
291 + 15
306 + 15
321 + 15
336 + 15
351 + 16
367 + 16
383 + 17
400 + 17
417 + 17
434 + 18
452 + 18
470 + 18
488 + 19
507 + 19
526 + 19
545 + 20
565 + 20
585 + 20
605 + 21
626 + 21
647 + 22
669 + 22
691 + 22
713 + 22
735 + 23
758 + 23
781 + 24
805 + 24
829 + 24
853 + 25
878 + 25
903 + 25
928 + 26
954 (ending number, I do not know the next numbers)
I've been trying to figure out the formula to these numbers, but without success. Wondering if anyone could give it an attempt to figure it out?
I want to be able to predict the coming numbers, after 954. This is hard.
probability
asked Dec 16 '18 at 0:18
Jeffrey AndersonJeffrey Anderson
11
$endgroup$
$begingroup$
Why is it $224+12$ then? Is it $+13$ instead?
$endgroup$
– Mindlack
Dec 16 '18 at 0:22
$begingroup$
Correct. Updated the question.
$endgroup$
– Jeffrey Anderson
Dec 16 '18 at 0:27
$begingroup$
If any simple pattern is likely, it's "3324233". The second cycle was just completed at 25. So you'll want +26 3 times, +27 3 times, +28 2 times etc.
$endgroup$
– David Peterson
Dec 16 '18 at 0:33
$begingroup$
@DavidPeterson is there a formula to use for this though?
$endgroup$
– Jeffrey Anderson
Dec 16 '18 at 0:37
add a comment |
0
0
0
$begingroup$
177 (starting number, i don't know the previous numbers)
188 + 12 (188 + 12 = 200 and so forth)
200 + 12
212 + 12
224 + 13
237 + 13
250 + 13
263 + 14
277 + 14
291 + 15
306 + 15
321 + 15
336 + 15
351 + 16
367 + 16
383 + 17
400 + 17
417 + 17
434 + 18
452 + 18
470 + 18
488 + 19
507 + 19
526 + 19
545 + 20
565 + 20
585 + 20
605 + 21
626 + 21
647 + 22
669 + 22
691 + 22
713 + 22
735 + 23
758 + 23
781 + 24
805 + 24
829 + 24
853 + 25
878 + 25
903 + 25
928 + 26
954 (ending number, I do not know the next numbers)
I've been trying to figure out the formula to these numbers, but without success. Wondering if anyone could give it an attempt to figure it out?
I want to be able to predict the coming numbers, after 954. This is hard.
probability
asked Dec 16 '18 at 0:18
Jeffrey AndersonJeffrey Anderson
11
$endgroup$
177 (starting number, i don't know the previous numbers)
188 + 12 (188 + 12 = 200 and so forth)
200 + 12
212 + 12
224 + 13
237 + 13
250 + 13
263 + 14
277 + 14
291 + 15
306 + 15
321 + 15
336 + 15
351 + 16
367 + 16
383 + 17
400 + 17
417 + 17
434 + 18
452 + 18
470 + 18
488 + 19
507 + 19
526 + 19
545 + 20
565 + 20
585 + 20
605 + 21
626 + 21
647 + 22
669 + 22
691 + 22
713 + 22
735 + 23
758 + 23
781 + 24
805 + 24
829 + 24
853 + 25
878 + 25
903 + 25
928 + 26
954 (ending number, I do not know the next numbers)
I've been trying to figure out the formula to these numbers, but without success. Wondering if anyone could give it an attempt to figure it out?
I want to be able to predict the coming numbers, after 954. This is hard.
probability
probability
asked Dec 16 '18 at 0:18
Jeffrey AndersonJeffrey Anderson
11
asked Dec 16 '18 at 0:18
Jeffrey AndersonJeffrey Anderson
11
edited Dec 16 '18 at 0:26
Jeffrey Anderson
asked Dec 16 '18 at 0:18
Jeffrey AndersonJeffrey Anderson
11
asked Dec 16 '18 at 0:18
Jeffrey AndersonJeffrey Anderson
11
asked Dec 16 '18 at 0:18
Jeffrey AndersonJeffrey Anderson
11
11
$begingroup$
Why is it $224+12$ then? Is it $+13$ instead?
$endgroup$
– Mindlack
Dec 16 '18 at 0:22
$begingroup$
Correct. Updated the question.
$endgroup$
– Jeffrey Anderson
Dec 16 '18 at 0:27
$begingroup$
If any simple pattern is likely, it's "3324233". The second cycle was just completed at 25. So you'll want +26 3 times, +27 3 times, +28 2 times etc.
$endgroup$
– David Peterson
Dec 16 '18 at 0:33
$begingroup$
@DavidPeterson is there a formula to use for this though?
$endgroup$
– Jeffrey Anderson
Dec 16 '18 at 0:37
add a comment |
$begingroup$
Why is it $224+12$ then? Is it $+13$ instead?
$endgroup$
– Mindlack
Dec 16 '18 at 0:22
$begingroup$
Correct. Updated the question.
$endgroup$
– Jeffrey Anderson
Dec 16 '18 at 0:27
$begingroup$
If any simple pattern is likely, it's "3324233". The second cycle was just completed at 25. So you'll want +26 3 times, +27 3 times, +28 2 times etc.
$endgroup$
– David Peterson
Dec 16 '18 at 0:33
$begingroup$
@DavidPeterson is there a formula to use for this though?
$endgroup$
– Jeffrey Anderson
Dec 16 '18 at 0:37
$begingroup$
Why is it $224+12$ then? Is it $+13$ instead?
$endgroup$
– Mindlack
Dec 16 '18 at 0:22
$begingroup$
Why is it $224+12$ then? Is it $+13$ instead?
$endgroup$
– Mindlack
Dec 16 '18 at 0:22
$begingroup$
Correct. Updated the question.
$endgroup$
– Jeffrey Anderson
Dec 16 '18 at 0:27
$begingroup$
Correct. Updated the question.
$endgroup$
– Jeffrey Anderson
Dec 16 '18 at 0:27
$begingroup$
If any simple pattern is likely, it's "3324233". The second cycle was just completed at 25. So you'll want +26 3 times, +27 3 times, +28 2 times etc.
$endgroup$
– David Peterson
Dec 16 '18 at 0:33
$begingroup$
If any simple pattern is likely, it's "3324233". The second cycle was just completed at 25. So you'll want +26 3 times, +27 3 times, +28 2 times etc.
$endgroup$
– David Peterson
Dec 16 '18 at 0:33
$begingroup$
@DavidPeterson is there a formula to use for this though?
$endgroup$
– Jeffrey Anderson
Dec 16 '18 at 0:37
$begingroup$
@DavidPeterson is there a formula to use for this though?
$endgroup$
– Jeffrey Anderson
Dec 16 '18 at 0:37
add a comment |
0
active
oldest
votes
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
});
}
});
draft saved
draft discarded
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%2f3042101%2fhow-would-you-find-the-formula-for-the-following-numbers-in-a-list%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3042101%2fhow-would-you-find-the-formula-for-the-following-numbers-in-a-list%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
$begingroup$
Why is it $224+12$ then? Is it $+13$ instead?
$endgroup$
– Mindlack
Dec 16 '18 at 0:22
$begingroup$
Correct. Updated the question.
$endgroup$
– Jeffrey Anderson
Dec 16 '18 at 0:27
$begingroup$
If any simple pattern is likely, it's "3324233". The second cycle was just completed at 25. So you'll want +26 3 times, +27 3 times, +28 2 times etc.
$endgroup$
– David Peterson
Dec 16 '18 at 0:33
$begingroup$
@DavidPeterson is there a formula to use for this though?
$endgroup$
– Jeffrey Anderson
Dec 16 '18 at 0:37