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Comparing the sequences OEIS A001951 $$a(n) = \lfloor n\sqrt{2}\rfloor = \lfloor \sqrt{2n^2 }\rfloor = \lfloor \sqrt{2*Square }\rfloor:\quad \{0, 1, 2, 4, 5, 7, 8, 9, 11, 12, 14, 15, 16, 18, 19, 21, ...\} \tag1$$ and OEIS A001953 $$a(n) = \left\lfloor\left(n - \frac12\right)\sqrt{2}\right\rfloor = \lfloor \sqrt{2(n^2-n) }\rfloor = \lfloor \sqrt{2*Oblong }\rfloor:\quad \{0, 0, 2, 3, 4, 6, 7, 9, 10, 12, 13, 14, 16, 17, 19, 20, ...\} \tag2$$ we realize that they have equal results at the indexes $n$ given by the sequence OEIS A059649 $$\{2, 7, 9, 12, 14, 19, 24, ...\}\tag{3}$$

Then, the first differences of $(3)$ are given by OEIS A059650: $$\{5, 2, 3, 2, 5, 5, 2, 5, 5, 2, 3, 2, 5, 2, ...\} \tag{4}$$

Questions reference to $(4)$:

  1. How do you prove that it is not possible to have any element other than 2, 3, or 5?
  2. Is this a cyclic sequence? How can you prove it?
  3. If it is, what is the string that always repeats?
  4. If it is not, what is its formation law?

For example:

A. The only element that repeats in two consecutive positions is 5.

B. The 3 always has two 2's isolating from the 5. That is, the 3 never succeeds or precedes the 5.

C. Thus, if we break all consecutive sequences of two 5's, we will have sequences of the numbers 525, 52325, and 523252325. Again the rules are repeated: 52325 never succeeds or precedes 523252325 which is the only element that repeats in two consecutive positions. The process of "breaking" the two 523252325's can also be repeated, and so on.

D. There seems to be an "almost"-cyclic when we allow one part to go clockwise and the other counterclockwise, to less than 1 element...

Some motivational considerations:

  1. The differences between the floor of the square root of two times a square number, and the floor of the square root of twice an oblong number is $\lfloor \sqrt{2*n^2}\rfloor-\lfloor \sqrt{2*(n^2-n)}\rfloor$ generate the sequence: ${1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, ...}$

  2. The number of $1$'s of the above differences are given by the sequence: ${1, 4, 1, 2, 1, 4, 4, 1, 4, 4, 1, 2, 1, 4, 1, 2, 1, 4, 4, 1, 2, 1, 4, 1, 2, 1, 4, 4, 1, 4, 4, 1, 2, 1, 4, 4, 1, 4, 4, 1, 2, 1, 4, 1, 2, 1, 4, 4, 1, 4, 4, 1, 2, 1, 4, 4, 1, 4, 4, 1, 2, 1, 4, 1, 2, 1, 4, 4, 1, 2, 1, 4, 1, 2, 1, 4, 4, 1, 4, 4, 1, 2, 1, 4, 1, 2, 1, 4, 4, 1, 2, 1, 4, 1, 2, 1, 4, 4, 1, 4, 4, 1, 2, 1, 4, 4, 1, 4, 4, 1, 2, 1, ...}$

There is an isomorphism between this sequence and OEIS A059650:

a. There is only the elements ${1,2,4}$.

b. The only element that repeats in two consecutive positions is 4.

c. The 2 never succeeds or precedes the 4.

d. The same process of "breaking" applies.

  1. This algorithm is also true for the equal results at the indexes $n$. The values when the floor of the square root of two times a square number is equal to the floor of the square root of the twice an oblong number are give by the sequence: ${2,9,12,16,19,26,33,36,43,50,53,57,60,67,70,74,77,84,91,94,98,101,108,111,115,118,125,132,135,142,149,152,156,159,166,173,176,183,190,193,197,200,207,210,214,217,224,231,234,241,248,251,255,258,265,272,275,282,289,292,296,299,306,309,313,316,323,330,333,337,340,347,350,354,357,364,371,374,381,388,391,395, ...}$

  2. The first differences are given by ${7, 3, 4, 3, 7, 7, 3, 7, 7, 3, 4, 3, 7, 3, 4, 3, 7, 7, 3, 4, 3, 7, 3, 4, 3, 7, 7, 3, 7, 7, 3, 4, 3, 7, 7, 3, 7, 7, 3, 4, 3, 7, 3, 4, 3, 7, 7, 3, 7, 7, 3, 4, 3, 7, 7, 3, 7, 7, 3, 4, 3, 7, 3, 4, 3, 7, 7, 3, 4, 3, 7, 3, 4, 3, 7, 7, 3, 7, 7, 3, 4, 3, 7, 3, 4, 3, 7, 7, 3, 4, 3, 7, 3, 4, 3, 7, 7, 3, 7, 7, 3, 4, 3, 7, 7, 3, 7, 7, 3, 4, 3, 7, 3, 4, 3, 7, 7, 3, 4, 3, 7, 3, 4, 3, 7, 7, 3, 7, 7, 3, 4, 3, 7, 3, 4, 3, 7, 7, 3, 4, 3, 7, 3, 4, 3, 7, 7, 3, 7, 7, 3, 4, 3, 7, 7, 3, 7, 7, 3, 4, 3, 7, 3, 4, 3, 7, 7, 3, 7, 7, 3, 4, 3, 7, 7, 3, 7, 7, 3, 4, 3, 7, 3, 4, 3, 7, 7, 3, 4, 3, 7, 3, 4, 3, 7, 7, 3, 7, 7, 3, 4, 3, 7, 7, 3, 7, 7, 3, 4, 3, 7, 3, 4, 3, 7, 7, 3, 7, 7, 3, 4, 3, 7, 7, 3, 7, 7, 3, 4, 3, 7, 3, 4, 3, 7, 7, 3, 4, 3, 7, 3, 4, 3, 7, 7, 3, 7, 7, 3, 4, 3, 7, ...}$

Again, there is an isomorphism between this sequence and OEIS A059650:

a. There is only the elements ${3,4,7}$.

b. The only element that repeats in two consecutive positions is 7.

c. The 4 never succeeds or precedes the 7.

d. The same process of "breaking" applies.

  1. Note that during the breaking process, the number of digits that are formed in the first break is the same in all of the above sequences. The number of digits form the sequence ${5, 3, 9, 9, 3, 5, 3, 9, 3, 5, 3, 9, 9, 3, 9, 9, 3, 5, 3, 9, 9, 3, 9, 9, 3, 5, 3, 9, 3, 5, 3, 9, 9, 3, 5, 3, 9, 3, 5, 3, 9, 9, 3, 9, 9, 3, 5, 3, 9, 3, 5, 3, 9, 9, 3, 5, 3, 9, 3, 5, 3, 9, 9, 3, 9, 9, 3, 5, 3, 9, 9, 3, 9, 9, 3, 5, 3, 9, 3, 5, 3, 9, 9, 3, 9, 9, 3, 5, 3, 9, 9, 3, 9, 9, 3, 5, 3, 9, 3, 5, 3, 9, 9, 3, 5, 3, 9, 3, 5, 3, 9, 9, 3, 9, 9, 3, 5, 3, 9, 9, 3, 9, 9, 3, 5, 3, 9, 3, 5, 3, 9, 9, 3, 9, 9, 3, 5, 3, 9, 9, 3, 9, 9, 3, 5, 3, 9, 3, 5, 3, 9, 9, 3, 5, 3, 9, 3, 5, 3, 9, 9, 3, 9, 9, 3, 5, 3, 9, 3, 5, 3, 9, 9, 3, 5, 3, 9, 3, 5, 3, 9, 9, 3, 9, 9, 3, 5, 3, 9, 9, 3, 9, 9, 3, 5, 3, 9, 3, 5, 3, 9, 9, 3, 5, 3, 9, 3, 5, 3, 9, 9, 3, 9, 9, 3, 5, 3, 9, 3, 5, 3, 9, 9, 3, 5, 3, 9, 3, 5, 3, 9, 9, 3, 9, 9, 3, 5, 3, 9, ...}$

And this last sequence also has properties equal to the sequence OEIS A059650.

  1. If we can ask one more question, note that we used in the initial sequences above, before the "break", the trios (1,2,4); (3,4,7); and (3,5,9). Can anyone find an initial sequence with 6 and/or 8 in a trio?
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    $\begingroup$ Please make questions as self-contained as possible. In this case, rather than requiring the reader to visit an external site to determine what you're even talking about, include in your question the description of the sequence of interest. (Since the sequence is defined as the first differences of another sequence, which is in turn defined as positions of $1$s in a third sequence, excluding this information places quite a context-seeking burden on the reader.) You might also want to say something about why you find this particular sequence interesting, to help motivate answerers. $\endgroup$
    – Blue
    Jun 22, 2021 at 3:39
  • $\begingroup$ The motivation is based on the fact that the sequences oeis.org/A001951 and oeis.org/A001953 have intrinsic relation with oeis.org/A059648. At the end, all these sequences are related to the fact that each prime number is located between a square number and its following oblong number (oeis.org/A307508) or an oblong number and its following square number (oeis.org/A334163). $\endgroup$
    – user925112
    Jun 22, 2021 at 4:49
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    $\begingroup$ I appreciate the update. (You should use MathJax to format your mathematical expressions.) Since comments are easily overlooked and may be hidden, make sure to include all context and clarifications in the body of the question itself. Cheers! :) $\endgroup$
    – Blue
    Jun 22, 2021 at 4:52

1 Answer 1

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Disclaimer: I only have an answer to the first question so far.


The two sequences will have equal results when $$\{ n\sqrt{2} \} > \frac{\sqrt{2}}{2}\approx 0.707$$

For the first part, given that $\frac{\sqrt{2}}{2} < \{n\sqrt{2}\} < 1$ (i.e. given that $n$ "works"), we want to show that the next value of $m$ such that $\{m\sqrt{2}\} > \sqrt{2}/2$ is in $\{n+2, n+3, n+5\}$. The range of $\{(n+1)\sqrt{2}\}$ is $(-2+\frac{3}{2}\sqrt{2},-1+\sqrt{2}) \approx (0.121, 0.414)$, so it's impossible for $n+1$ to work. The range of $\{(n+2)\sqrt{2}\}$ is $(-3+\frac{5}{2}\sqrt{2},-2+2\sqrt{2}) \approx (0.536, 0.828)$, so when $\{n\sqrt{2}\}$ is sufficiently high, the next term would be $n+2$.

Specifically, if $$\{n\sqrt{2}\} > 3-\frac{3}{2}\sqrt{2} \approx 0.879$$

the next term will be $n+2$. Continuing in the same vein, if $$\{n\sqrt{2}\} < 5-3\sqrt{2} \approx 0.757$$ the next term will be $n+3$. And if $$\{ n\sqrt{2} \} < 8-5\sqrt{2} \approx 0.929$$ the next term would be $n+5$, except for the fact that for much of this interval, the next term is already $n+2$ or $n+3$. So the next term is $n+5$ if and only if $$5-3\sqrt{2} < \{n\sqrt{2}\} < 3-\frac{3}{2}\sqrt{2}$$

These conditions cover the whole interval for $\{n\sqrt{2}\} > \frac{\sqrt{2}}{2}$, so no terms in the sequence other than $2, 3$, and $5$ are possible.

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