Abundant odd numbers: Difference between revisions

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<lang julia>using Primes
 
function factorsofpropfact(n)
f = [one(n)]
for (p, x) in factor(n)
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end
 
isabundant(n) = sum(factorsofpropfact(n)) > n
prettyprintfactors(n) = (a = factorsofpropfact(n); println("$n has proper divisors $a, these sum to $(sum(a))."))
 
function oddabundantsfrom(startingint, needed, nprint=0, targetf=(x -> false))
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