[Maxima] Sums of digits and other tricks and factor()

Richard Hennessy rich.hennessy at verizon.net
Thu Dec 10 16:29:47 CST 2009

Hi List,

I have been wondering why factor() is slow on factoring big exact numbers that are in the form y/x where x is a big even 
integer and y is an integer.  It is easy to tell that factor 8788797887776565343256785434546789796543568097876592 is 
factorable by 2 just by looking at the last digit.

I have a function of two variables defined as a power series f(x,y,terms) and another home made function factorsum().

factorsum(f(x,y,48),y)$  /* this is my factorsum() function not Maxima's.*/
Evaluation took 2.5900 seconds (2.5900 elapsed)


factorsum(f(1/2,y,48), y)$
Evaluation took 58.5300 seconds (58.5300 elapsed)

takes a lot longer.  I have determined that the result is slower because of the call to factor() inside factorsum() and 
the time it takes to factor y^n/(2^48 * . . .) + . . . .  There are a lot of tricks I know for decimal numbers that can 
tell quickly that a number is factorable by 2,3 5,7,11.  I imagine there are similar tricks that can be used in binary 
and for larger primes.

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