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When 1 Merit is given it generate .5 sMerits. I found many times user trying to calculate like from 23045 will create 11522 and then 5761  and so on.
But they are not able to reach the number at end (after all the sum of halving merits).

So I decided to calculate it Mathematically. Since number of smerits will halved every time and 23045 is not perfectly divisible by 2.
We need to represent 23045 as sum of numbers that can be easily divisible by 2 at every stage till we reach Merit 1 .

In other words, first we need to put 23045 as sum of numbers in power of 2

23045 = 16384+4096+2048+512+4+1
23045= 2¹⁴ + 2¹² + 2¹¹ + 2⁹ +2² +2⁰

Now we start with  2¹⁴ i.e  16384 and check how many merits it can generate.

16384 + 8192   + 4096 + 2048 +1024 + 512 + 256 + 128 + 64 +32 + 16 + 8 + 4 + 2 + 1 =  32767
  
Similarly, for other terms
4096 + 2048 +1024 + 512 + 256 + 128 + 64 +32 + 16 + 8 + 4 + 2 + 1 =                              8191
2048 +1024 + 512 + 256 + 128 + 64 +32 + 16 + 8 + 4 + 2 + 1 =                                          4095
512 + 256 + 128 + 64 +32 + 16 + 8 + 4 + 2 + 1 =                                                                1023
4 + 2 + 1 =                                                                                                                                7
1=                                                                                                                                              1

So sum of these number will be  32767+8191+4095+1023+7+1=46084
Since we have 6 number that make up to the sum of 23045 , so  we get 6/2 =3. and further 3/2 =1
You might be wondering why I did not use same formula for 6 (2²+2¹ ) because 6 is converted to merit ,it is not smerit anymore.

So in total 46084+3+1 = 46088 Merit will be generated Mathematically.

This all calculation is derived by using the fundamentals of Geometric Progression and sequence, their n^th term and sum till the n^th term.

Some links to understand Geometric Progression are here (http://www.mathguide.com/lessons/SequenceGeometric.html)  and  here (https://en.wikipedia.org/wiki/Geometric_progression) and here (https://www.mathsisfun.com/algebra/sequences-sums-geometric.html)

Tabular Data of the Calculation
https://i.imgur.com/Boq5Yfz.png


Hope you all enjoyed the calculation.

PS: 46088  is mathematically feasible and maximum limit that can be generated if we used all source merit but actual number can be less or more because many users will left with residual smerits but this residual smerit can be used next time when they receive a single merit or odd number of merits.

All these calculations to come up with 46088 instead of 46090?

So what you are saying is that a 46088 merits can be generated every month? ( the maximum )

Thanks for sharing this to be honest, somewhat useful info :)

 

I did manual calculation for 50 sMerits and I come to result of 98. So I am not doubting the mathematics and my formula.

Just to point out that there is some difference if you calculate from sMerits vs Merits that I pointed in my OP

Your topic reminds me of the old days in high school. Smerits are generated like radioactive decay and above sum is a geometric progression
I still remember that there's a general method to calculate this sum:

Assume that S =           16384 + 8192   + 4096 + 2048 +1024 + 512 + 256 + 128 + 64 +32 + 16 + 8 + 4 + 2 + 1
                      =           2¹⁴ + 2¹³ + 2¹² + ... + 2¹ +2⁰ (*)
then          2S  = 2¹⁵ + 2¹⁴ + 2¹³ + ... +2² +2¹           (**)

(**) - (*) side by side, then S = 2¹⁵-1 =  32767

Btw, it's an interesting discovery OP.

 I know, I can directly use the formula, but I just wanted it to be in laymen terms, where everybody can visualize each divison and the sum.

I like the analogy of sMerits vs radioactive decay but only difference is that they do not produce any unwanted product (like radiations)

That may be my fault. I had a couple of rusty merits in the bottom of my bag. I've given them to you, so that we can keep the record straight. :)

It is interesting to see the potential maximum that can be derived from a single month’s Merit Source total aggregate. When I looked at it some time ago, I used a simpler approach:

-   For a given integer number (not too small), the aggregate of itself and all its halvings is roughly double the initial number (i.e. 23.045 sMerits generate roughly a potential maximum of 46.090 sMerits aggregating all the halvings).

-   The number of times the halving can be performed is CEILING(LOG(Number;2);1)-1 (base 2 logarithm).

-   The result of each division generates =FLOOR(Number/2;1) sMerits, where Number is the number of sMerits that resulted from the previous division.

The quick way of seeing it, with a small margin of error, is simply as double the initial figure.

Note: I’m not quite sure why you aggregate the extra 4 sMerits to the 46.084 based on the number of decomposed initial numbers (I mean the reason behind).

i still need to do more to really understand this calculations. hope i find more useful information

The geometric sum formula assumes an infinite series of descending terms, which is not the case as the minimum merit is 1. That should explain discrepancies IMO.

23045= 2¹⁴ + 2¹² + 2¹¹ + 2⁹ +2² +2⁰

23045 is used as sum of 6 numbers and each number (or series) ended at 1. Practically 1 Merit is not useful to generate any Merit but when 6 series (each ended with 1) then you got extra 6 Merits, These 6 Merits can produce 4 full merits leaving the redundant smerits.


Thanks Paxmao, making it more clear.

TeQuiero also explained how the sum in GP done and I already given links to the GP tutorials in OP. 46090  is no way correct.

Any way , after calculating with few more values, I got the generalized formula  for Merit generation from sMerits.

X (smerits)=> 2X - 1 Merits  (X >0, and X can be expressed as 2^y like 2,4,8, 16 and so on.)

X(smerits)=> 2X-2   Merits  (X >0, and X can not be expressed as 2^y  like 6,7 ,12 etc..)





If 23045 sMerits generate 46090 Merits. And these 46090 Merits will again able to  generate 23045  sMerits. Then cycle can simply go on.
We know already, this type of cycle is not going on ,so even without using any mathematics I can say by practical experience, it is absolutely incorrect.

That’s what I figured. But let’s say that you gave each of your numbers (the six you split the original into) to a different person, and each person sMerits another non-intersecting person in a linear manner. As you say, there will be six people with one Merit at the leaf of each linear chain. What we cannot do is then scoop those six units together, generate a batch of 6 sMerits, and generate four more after two steps. What I mean by this is that the last part of the algorithm does not seem to apply in practice to sMerits, because we cannot scoop the ones together...

I agree, I think we will end up in much larger leaf Nodes, Everybody who has earned odd number of Merit has  potential .5 sMerit in unusable form.
46088 just define max potential that can be achieved by using all ideal conditions.

I think , if Theymos decide to decay the sMerits in future, instead of decaying it , it should be re assigned to Merit Source.

But why all those merits were not shared every month? even half of them are not shared yet per month because we still have lot of airdropped merits under circulation so merit sources need to be more active at finding the posts? or there are not enough worthy posts to get merits. ???

 You can refer to Merit Awarded Data (https://bitcointalk.org/index.php?topic=2818350.msg46065728#msg46065728) here.
 I think you are wrong in claiming that Merit Source are not distributing Merits. Majority of air dropped merits were exhausted in a month.

To get the ideal value , Only Merit source are not responsible. If you got Merits and you did not merited others, you just stopped a part of chain and further generation of merits.

My point is: just double the total amount of sMerit, and you're (almost) at the right number. You end up 2 sMerits lower. But it might just as well be send to users who already had 0.5 sMerit, meaning a few more sMerit becomes available.
Just doubling it makes the calculations much easier, and the rounding depends on various circumstances anyway.

I got your sentiments. For a  big number "X",  2X-2 ,2X-1 and 2X  all have approximately same values so you can continue with 2X for sake of simplicity.