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n= 1,2,3,4,5,6,...n c= 1,3,6,10,,15,21,...n(n+1)/2 m= 0,1,2,4,6,9...

Am trying to find out if there a formulae that could be generated to fill in the sequence for m. I know the pattern as follows: m1+1=m2 => 0+1=1 m2+1=m3 => 1+1=2 m3+2=m4 => 2+2=4 m4+2=m5 => 4+2=6 m5+3=m6 => 6+3=9

The ratio keeps adding by 1 after every two values. any help towards figuring out a formula would be much appreciated.

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  • $\begingroup$ Not sure I have the pattern right, but how about $m_i=\big\lfloor \frac {i^2}2 \big \rfloor$? $\endgroup$ – lulu Oct 19 '18 at 19:06
  • $\begingroup$ To be clear...you appear to have strung a whole bunch of sequences together, with no clear connections between them. I am guessing that you are just asking about the last one and I am further guessing that I have the right pattern for that one. My sequence goes $0,1,2,4,6,9,12,16,20,25,30,36,\cdots$. $\endgroup$ – lulu Oct 19 '18 at 19:09
  • $\begingroup$ Can you clarify your question? Does the sequence I proposed answer your question? Note: What I wrote contains a typo. I meant to write $m_i=\big \lfloor \frac {i^2}4\big \rfloor$. $\endgroup$ – lulu Oct 19 '18 at 19:26
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Found this on the encyclopedia of the integer sequences

The formula is the following:

$$a_n=\sum_{j=1}^{n} \left\lfloor \sum_{k=1}^{j}k\sqrt3-\lfloor k\sqrt3\rfloor\right \rfloor $$

I wrote a small script on python:

Try it here

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