20200213, 08:13  #1 
Mar 2018
523 Posts 
Is there a powercounting function?
How many powers are there less than a given number x?
Is there a function that "counts" the powers less than x? For example powers less than 10 are three (4,8,9) 
20200214, 08:12  #2 
Romulan Interpreter
Jun 2011
Thailand
221E_{16} Posts 
This sounds like a honest question.
There are: (seriously, this would take milliseconds, for really huge numbers....) Edit: actually, not. I was a bit stupid, as this would count the powers of powers again, for example the powers of 4 or 9 or 36 will be counted again, they were already counted as powers of 2, 3, 6 respectively. But this is easy to avoid. Last fiddled with by LaurV on 20200214 at 08:52 Reason: forgot the lfoors and rfloors 
20200214, 09:06  #3 
Einyen
Dec 2003
Denmark
101101110101_{2} Posts 
Last fiddled with by ATH on 20200214 at 09:07 
20200214, 09:12  #4 
Romulan Interpreter
Jun 2011
Thailand
2×11×397 Posts 
Code:
gp > cnt=0;for(i=1,100000,if(ispower(i),print(cnt++": "i))) ..... <lot of lines removed> .... 359: 96721 360: 97336 361: 97344 362: 97969 363: 98596 364: 99225 365: 99856 366: 100000 gp > getnum(n)=s=0;for(i=2,log(n)/log(2),z=floor(n^(1/i)); s+=zgetnum(z)1);s %2 = (n)>s=0;for(i=2,log(n)/log(2),z=floor(n^(1/i));s+=zgetnum(z)1);s gp > getnum(10) %3 = 3 gp > getnum(16) %4 = 4 gp > getnum(10000) %5 = 124 gp > ## *** last result computed in 0 ms. gp > getnum(100000) %6 = 366 gp > ## *** last result computed in 0 ms. gp > getnum(10^50) time = 15 ms. %7 = 10000000046415898134516526 gp > ## *** last result computed in 15 ms. gp > Last fiddled with by LaurV on 20200214 at 09:13 
20200214, 09:49  #5 
Jun 2003
13×19^{2} Posts 
A faster version using inclusion/exclusion principle rather than recursion:
Code:
howmanyfactors(n)=#factor(n)[,1]; getnum(n)=my(s=0, z); for(i=2,log(n)/log(2), if(!issquarefree(i), next()); z=floor(n^(1/i))1; if(howmanyfactors(i)%2==0, s=z, s+=z)); s Last fiddled with by axn on 20200214 at 10:15 
20200214, 12:15  #6  
Nov 2016
2×1,117 Posts 
Quote:
e.g. powerpi(9453)=97+21+6+3+2+2=131 

20200214, 14:51  #7 
Feb 2017
Nowhere
2^{2}×5×173 Posts 
Note: The OP apparently forgot 1.
I worked the answer out for n = 10000. In the first routine, I handcalculated the limit on primes. The general bound is floor(log(n)/log(2)). For n = 10000 this is 13. Code:
? n=10000 %5 = 10000 ? j=0;for(i=1,n,if(ispower(i,2)ispower(i,3)ispower(i,5)ispower(i,7)ispower(i,11)ispower(i,13),j++));print(j) 125 Code:
? s=0;for(i=2,n,if(issquarefree(i),t=(floor(n^(1/i))1)*moebius(i);s+=t;print(i" "s" "t));if(t==0,break));s++ 2 99 99 3 119 20 5 124 5 6 121 3 7 123 2 10 122 1 11 123 1 13 124 1 14 124 0 %6 = 125 If n is very large, it is conceivable that there would be insufficient sig figs to calculate the roots to the nearest integer. My ancient version of PariGP has sqrtint(n), but doesn't have sqrtnint(n,i). Using sqrtnint(n,i)  1 in place of floor(n^(1/i))  1 may avoid this problem. Last fiddled with by Dr Sardonicus on 20200214 at 14:57 Reason: xignif ostpy 
20200215, 02:14  #8 
Romulan Interpreter
Jun 2011
Thailand
2×11×397 Posts 
Hihi, none of those are faster than the recursive version I posted, it takes longer than 15 ms to factor n in 10^50 range, as in my example. Well, there could be a "workaround" to always chose n extremely easy to factor (like 10^50 itself, which is trivial), because the result of the function won't change in a large range (around 10^50 there is no other "power" for quite a while! the precedent one is at 10^5019999999999999999999999999)**, but for a general random n, mine is faster by far***
 **challenge: the fastest "nextpower()" and "precpower()" functions? ***you may however, need to redefine realprecision to match the range in this case Last fiddled with by LaurV on 20200215 at 03:37 
20200215, 04:09  #9  
Jun 2003
13·19^{2} Posts 
Quote:
Code:
? \p 50 realprecision = 57 significant digits (50 digits displayed) getnumlaurv(n)=my(s=0, z); for(i=2,log(n)/log(2),z=floor(n^(1/i)); s+=zgetnumlaurv(z)1); s howmanyfactors(n)=#factor(n)[,1]; getnumaxn(n)=my(s=0, z); for(i=2,log(n)/log(2), if(!issquarefree(i), next()); z=floor(n^(1/i))1; if(howmanyfactors(i)%2==0, s=z, s+=z)); s getnumdrs(n)=my(s=0, t); for(i=2,n,if(issquarefree(i),t=(floor(n^(1/i))1)*moebius(i);s+=t);if(t==0,break)); s++ ? # timer = 1 (on) ? r=10^50+random(2^64) %5 = 100000000000000000000000000000013282407956253574712 ? ? for(i=1, 1000, getnumlaurv(r)) time = 8,918 ms. ? for(i=1, 1000, getnumaxn(r)) time = 437 ms. ? for(i=1, 1000, getnumdrs(r)) time = 426 ms. ? 

20200215, 07:38  #10 
Romulan Interpreter
Jun 2011
Thailand
2×11×397 Posts 
Whoops...
fast reading, you confused me with those "n" and "i" mixture Of course, you are right. (I would however, test to see if pari is not caching the result, hihi, don't want to give up so easy, which is possible for your function, but not for mine, due to recursion  you know, like when reading files in windows, if you read the same file many times, subsequent accesses are much faster, hihi) (just kidding) Last fiddled with by LaurV on 20200215 at 07:51 
20200215, 13:23  #11 
Feb 2017
Nowhere
2^{2}×5×173 Posts 
My clunky PariGP script (including a print() statement for each term in the sum, and an inevitable superfluous zero term in the sum), my ancient version of PariGP, and doddering old computer still did the count for 10^50 in 2 milliseconds. Since I count 1 as a power, my answer is 1 greater than yours.
If you take out the print() statement, replace floor(n^(1/i)) with sqrtnint(n,i), and (if you want) take out the s++ statement at the end if you want to ignore the power 1, and run it on your machine, you'll probably do 10^50 in less than a millisecond, and also be able to do arbitrary n. With the default precision, my script craps out at around 10^77 due to insufficient precision. Code:
n=10^50; ? s=0;for(i=2,n,if(issquarefree(i),t=(floor(n^(1/i))1)*moebius(i);s+=t;print(i" "s" "t));if(t==0,break));s++ 2 9999999999999999999999999 9999999999999999999999999 3 10000000046415888336127786 46415888336127787 5 10000000046415898336127784 9999999998 6 10000000046415898120684316 215443468 7 10000000046415898134579269 13894953 10 10000000046415898134479271 99998 11 10000000046415898134514381 35110 13 10000000046415898134521397 7016 14 10000000046415898134517671 3726 15 10000000046415898134515518 2153 17 10000000046415898134516390 872 19 10000000046415898134516817 427 21 10000000046415898134516578 239 22 10000000046415898134516392 186 Code:
163 10000000046415898134516526 1 165 10000000046415898134516527 1 166 10000000046415898134516526 1 167 10000000046415898134516526 0 time = 2 ms. %2 = 10000000046415898134516527 
Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
New confirmed pi(10^27), pi(10^28) prime counting function records  kwalisch  Computer Science & Computational Number Theory  32  20200831 16:58 
Prime counting function  Steve One  Miscellaneous Math  20  20180303 22:44 
Prime counting function records  D. B. Staple  Computer Science & Computational Number Theory  43  20161225 22:15 
Legendre's prime counting function  pbewig  Information & Answers  0  20110714 00:47 
Counting on Fingers  Orgasmic Troll  Lounge  3  20051231 22:22 