20180302, 09:12  #1 
Mar 2018
17×31 Posts 
A suggestion for factordb.
I would suggest that factordb can detect the number of digits of a number , like PFGW can it with the lencommand.
Last fiddled with by enzocreti on 20180302 at 09:35 
20201219, 12:59  #2 
Nov 2016
2,819 Posts 
Other suggestions:
Operator "" for integer division (i.e. a\b = floor(a/b)) (FactorDB already has "%" for the modulo operator) Operator "n!!" for double factorial of n Operator "n!!!" for triple factorial of n Operator "An" for n!  (n1)! + (n2)!  ... 1! (A005165(n)) Operator "Kn" for 1! + 2! + 3! + ... + n! Operator "Bn" for Bell(n) Operator "En" for Euler(n) Operator "Cn" for Catalan(n) Operator "W(m,n)" (Wolstenholme numbers) for the numerator of 1 + 1/(2^m) + 1/(3^m) + ... + 1/(n^m) Operator "I(m,n)" for the nth mFibonacci number (1, 2, 3, 4, 5, 6, ...) Operator "J(m,n)" for the nth mstep Fibonacci number (2, 3, 4, 5, 6, ...) Operator "P(n)" for the nth Perrin number Operator "Phi(m,n)" for the mth cyclotomic polynomial at n Operator "R(m,n)" for the repunit in base m with length n Operator "S(m,n)" for the concatenate the first n integers in base m Also for the first n digits of (pi, e, sqrt(2), golden ratio, ln(2), gamma, ...) in base m Last fiddled with by sweety439 on 20201219 at 13:06 
20201223, 07:53  #3 
Romulan Interpreter
Jun 2011
Thailand
5·1,889 Posts 
The guy was either trolling or brain farting (FDB displays the length of the number always, as a small index after the number, unless the number is too small).
But some of your suggestions are good (some other are futile, or step over the existent notations). 
20201223, 18:00  #4 
"Daniel Jackson"
May 2011
14285714285714285714
2·5^{2}·13 Posts 
That would save a lot of typing:
(10^911)\(3^911) is shorter than (10^911((10^911)%(3^911)))/3^911. 64!! would probably end up as 126886932185884164103433389335161480802865516174545192198801894375214704230400000000000000!, a number that's way too large for the DB. This is due to how the DB parses expressions. A better way of representing it would be n!2 for n!!, n!3 for n!!!, etc. A109 is much shorter than 109!108!+107!...24+62+1. K109 is much shorter than 109!+108!+107!+...+24+6+2+1. The Bell, Euler, Catalan, etc. functions would make it easier to add those numbers, if anyone wanted to do so. Phi(1001,2) is much shorter than M1001/M143*M13/M91*M7*M11/M77. R(314159,919) would definitely be easier to type than (314159^9191)/314158 S(36,35) is definitely shorter than (6^721261)/1225 (See http://www.factordb.com/index.php?sh...685782&base=36 for proof of concatenation). 
20201223, 18:15  #5  
Nov 2016
2,819 Posts 
Quote:


20201223, 18:23  #6 
Nov 2016
2819_{10} Posts 
There are also many types of numbers to factorize, which cannot be simply typed in factorDB, unlike the formula (a*b^n+c)/d (nearrepdigit numbers and quasirepdigit numbers in base b)
I think that we can use Q(n) for part(n) (since P(n) is already used for Perrin numbers), and use D(n) for distinctpart(n) Also, currently factorDB only supports Fibonacci numbers I(n) and Lucas numbers L(n), but not support the Fibonacci polynomial (we may use I(m,n) for the nth Fibonacci polynomial evaluated at m, which is the same as the nth mFibonacci number, e.g. for m=2, this is the nth Pell number, and for m=3, this is A006190(n)) (I think that we can even use U(P,Q,n) and V(P,Q,n) for the nth number in Lucas sequences), factorDB also not supports tribonacci numbers, tetracci numbers, etc. Last fiddled with by sweety439 on 20201223 at 18:30 
20210105, 10:33  #7  
Romulan Interpreter
Jun 2011
Thailand
5·1,889 Posts 
Quote:
64!! = 1130138339199322632554990773529330319360000000 You can get it in pari with "prod(x=1,32,2*x)". 

20210105, 14:45  #8  
Feb 2017
Nowhere
11BE_{16} Posts 
Quote:
For iterated factorials, use parentheses. The factorial of 64! is (64!)!. It's too big for me to want to calculate exactly. I'll content myself with its natural logarithm. Even for the logarithm, I have to bump up the real precision from its default of 38. Code:
? n=64! %1 = 126886932185884164103433389335161480802865516174545192198801894375214704230400000000000000 ? default(realprecision,150) %2 = 150 ? lngamma(n+1) %3 = 25906276482267964036342506230950052158110911566157592844228132441739611236380625962625795073.4807460213107849024090809743801213248377167429072675193492 

20210105, 15:59  #9 
Nov 2016
2,819 Posts 
The workers in factordb can only:
* Factoring numbers <70 digits * Proving probable primes <300 digits * Checking the smallest number with status unknown <20000 digits For the number factoring, it can also factor numbers <100 digits but not immediately (I do not know about it, maybe somebody factored these numbers, like that maybe somebody assign the first numbers >=20000 digits to worker to test their (probable)primality) However, in 2010, they can prove probable primes <1000 digits (see https://web.archive.org/web/20100921...com/status.php) I hope that factordb can: * Factoring numbers <100 digits * Proving probable primes <2500 digits * Checking the smallest number with status unknown <50000 digits Is this possible? 
20210105, 17:57  #10 
"Curtis"
Feb 2005
Riverside, CA
7·683 Posts 
Go ahead and do it you know it's users doing those factoring jobs and proving those prps, right?
Become that user. Download composites, factor them, upload the factors. Or run Primo proofs for any prps you wish were proven. Yes, really. You. 
20210105, 19:49  #11 
"Nuri, the dragon :P"
Jul 2016
Good old Germany
2·13·31 Posts 

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