Timeline for The largest digital sum of the square of an n-digit number
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Apr 18 at 3:51 | comment | added | Gerry Myerson | Related question recently asked on mathstack, math.stackexchange.com/questions/4900379/… | |
| Apr 16 at 5:49 | answer | added | Mrexcel | timeline score: 3 | |
| S Nov 19, 2021 at 19:51 | history | suggested | Freddy Barrera | CC BY-SA 4.0 |
Add OEIS link.
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| Nov 19, 2021 at 18:44 | review | Suggested edits | |||
| S Nov 19, 2021 at 19:51 | |||||
| Oct 12, 2021 at 14:52 | comment | added | markvs | @TimothyChow: "you further implied" is purely your imagination. Clearly you do not need all squares. For example you do not need squares of numbers which are divisible by 10. There are many others which are not needed. $10^9$ is a small number for many similar questions. The OP gave first 10 values. But even the brute force (using some slow CAS) would give you $n=11$ in a few minutes. | |
| Oct 12, 2021 at 14:41 | comment | added | Timothy Chow | @markvs Your question was whether the OP checked $n< 10^9$ and I do not expect you to know whether the OP did in fact check $n < 10^9$. But saying that $n \approx 10^9$ is "small" implies that checking $n < 10^9$ is a feasible computation. When Yaakov Baruch asked for confirmation and pointed out that $n \approx 10^9$ does not appear to be a feasible computation, you affirmed that you did mean $n < 10^9$, and by saying, "You do not need squares of all numbers" you further implied that you had in mind some nontrivial way of making the computation feasible. I am just asking you for more details. | |
| Oct 12, 2021 at 14:15 | comment | added | markvs | @TimothyChow: I asked a question to the OP. Why do you think I should know the answer? | |
| Oct 12, 2021 at 12:28 | comment | added | Timothy Chow | @markvs Can you please elaborate on how exactly one might computationally check the case $n=10^9$ in a reasonable amount of time? | |
| Oct 11, 2021 at 19:36 | comment | added | markvs | @YaakovBaruch: You do not need squares of all numbers. I did mean $n\lt 10^9$. But if the OP can check it for other small $n$, it is fine. By the way, it is obvious that the sequence is not decreasing. The only question is whether $a_m$ can be equal to $a_{m+1}$. | |
| Oct 11, 2021 at 19:21 | comment | added | Yaakov Baruch | @markvs: What do you mean exactly? For $n=10^9$ one would have to compute the squares of about $10^{10^9-1}$ numbers. If you meant $n=9$ then the OP did do that. | |
| Oct 11, 2021 at 19:12 | history | edited | Bernardo Recamán Santos |
edited tags
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| Oct 11, 2021 at 8:52 | comment | added | Moritz Firsching | for the analogues sequence in base 2, namely $1, 2, 3, 5, 6, 8, 9, 13, 13, 15, 16, 18, 20, 22, 24, 25, 27, ...$ it is not strictly increasing, since $13, 13$ appears. | |
| Oct 11, 2021 at 1:51 | history | edited | Bernardo Recamán Santos | CC BY-SA 4.0 |
clarification
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| Oct 11, 2021 at 0:56 | comment | added | Bernardo Recamán Santos | @markvs No, I only have those values for the sequence. | |
| Oct 11, 2021 at 0:14 | comment | added | markvs | Did you check that for small $n$, say, $n<10^9$? | |
| Oct 10, 2021 at 22:47 | history | asked | Bernardo Recamán Santos | CC BY-SA 4.0 |