Ho do I convert pic related into a closed-form expression? I

>>8761093
>I need to compute values up to up to 2^56
but that will only give you 57 terms

>>8761141
What you mean?

>>8761272
6n + 0, 0, 1, 1, 3, 3, 4, 4, 7, 7, 8, 8, 10, 10, 11, 11, 15, 15, 16, 16, 18, 18, 19, 19, 22, 22, 23, 23, 25, 25, 26, 26, 31, 31, 32, 32, 34, 34, 35, 35, 38, 38, 39, 39, 41, 41, 42, 42, 46, 46, 47, 47, 49, 49, 50, 50, 53, 53, 54, 54, 56, 56, 57, 57, 63, 63, 64, 64, 66, 66, 67, 67, 70

>>8761362
I still have the problem to compute the correct element of your sequence given n.

Anyways, it looks easier. Thanks.

Hi folks, the closed result is obtained by looking at the chart to the right, which gives all the insight needed about the result, thanks OP for posting the function's results up to f(10) though, here goes :
[math]f(n)=\lfloor \frac{n+1}{4}\rfloor\left(\lfloor \frac{n+1}{4}\rfloor+1 \right)\frac{1}{2}+\lfloor \frac{n+1}{2}\rfloor+6n[/math]

>>8761701
I didn't actually check whether this is the "correct" sequence, I just substracted 6n from all the values you posted here >>8761093 and put it into this website: https://oeis.org/ (since your function is quite obviously some other function + 6n). You should check for yourself whether the first 30 values are actually equal

It says the function is "a(n) = n minus (number of 1's in binary expansion of n). Also highest power of 2 dividing n!."

A general formula apparently is: (I assume brackets means floor)
a(n) = a([n/2]) + [n/2] = [n/2] + [n/4] + [n/8] + [n/16] + ...

There are also some other formulas which might be even easier to compute

>>8761902
i fucked it up, my result has the [math]6n[/math] part one off, so it would then be :
[math]f(n+1)=\lfloor \frac{n+1}{4}\rfloor\left(\lfloor \frac{n+1}{4}\rfloor+1\right)+\lfloor \frac{n+1}{2}\rfloor+6(n+1)[/math]
This result, however, is right, take my word for it

>>8761927
Your function works for the first 10 values, after that it grows far faster than OPs function

OP, you should use this formula, and cache and reuse all a(n)s (because you're going to need them again for a(2*n)), this makes it possible to compute all function values in O(n)
>>8761904

I can't post the links because they are detected as spam. I will post a codes you can paste into google short links
>https://
>goo
>(dot)
>gl/

If I past my sequence into oeis.org I get
>Sorry, but the terms do not match anything in the table.
If you wanna check by yourself, here's the short links
Full sequence: oAQTuU (short link)
If I subtract 6n to each element: jEPXzM (short link)

>>8761902
>thanks OP for posting the function's results up to f(10)
There you go, up to 1000: tRWhuN (short link)

If you want more values, you just need to ask. If you want it in another format for an easier copy pasting, just tell me how should I print it.

>>8761904
>A general formula apparently is: (I assume brackets means floor)
>a(n) = a([n/2]) + [n/2] = [n/2] + [n/4] + [n/8] + [n/16] + ...
I didn't checked if it's correct, but computing something like this is as hard as computing the for loop I posted above (OP picture).
I need something much much faster. (if it's possible)

>>8761927
The first one is the sequence I need, the second one is what you get from your formula
> 0, 6, 13, 19, 27, 33, 40, 46, 55, 61, 68, 74, 82, 88, 95, 101, 111, 117, 124, 130, 138, 144, 151, 157
> 0, 6, 13, 19, 28, 34, 41, 47, 58, 64, 71, 77, 90, 96, 103, 109, 124, 130, 137, 143, 160, 166, 173, 179

If you have other ideas feel free to post them. I will check if they are correct.

>>8762136
so which one is it? the top or bottom sequence?

>>8762177
Top is what I want, bottom is what I get from your expression.
I need top, you gave me an expression for bottom.

File: numbers.png (2KB, 248x37px) Image search: [Google]

2KB, 248x37px

>>8762136
you fucked up your calculations apparently, geogebra says otherwise for me, pic related

>>8762201
>>8762136
okay i got it
forgot the half factor in the expression, i had it in the first one though, which was false, here is (again) the correct one :
[math]f(n+1)=\lfloor \frac{n+1}{4}\rfloor\left(\lfloor \frac{n+1}{4}\rfloor+1\right)\frac{1}{2}+\lfloor \frac{n+1}{2}\rfloor+6(n+1)[/math]

>>8762208
Top is what I want, bottom is what I get from your expression
> 0, 6, 13, 19, 27, 33, 40, 46, 55, 61, 68, 74, 82, 88, 95, 101, 111, 117, 124, 130, 138, 144, 151, 157
> 0, 6, 13, 19, 27, 33, 40, 46, 55, 61, 68, 74, 84, 90, 97, 103, 114, 120, 127, 133, 145, 151, 158, 164

>>8761093

import math

number = 23523452345

def bitCount(int_type):
count = 0
while(int_type):
int_type &= int_type - 1
count += 1
return(count)

print [ 7*n - bitCount(n) for n in range(1000)]

>>8762248
You're amazing, thanks! It seems to works really well.

>>8762248
how did you figure out this

>>8762269
Yeah its just the 6*n plus n minus the number of set bits in n.

The guy who posted the oasis thing should've looked down the page.

>>8762283
print 7*(2**256-1) - bin(2**256-1)[2:].count('1')
print [7*n - bin(n)[2:].count('1') for n in range(1000)]

this should be even faster according to stack overflow.