Abundant odd numbers: Difference between revisions

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// Abundant odd numbers. Nigel Galloway: August 1st., 2021

let fN g=Seq.initInfinite((+)1)|>Seq.takeWhile(fun n->n*n<=g)|>Seq.filter(fun n->g%n=0)|>Seq.sumBy(fun n->let i=g/n in n+(if i=n then 0 else i))

let fN g=Seq.initInfinite(int64>>(+)1L)|>Seq.takeWhile(fun n->n*n<=g)|>Seq.filter(fun n->g%n=0L)|>Seq.sumBy(fun n->let i=g/n in n+(if i=n then 0L else i))

let aon=Seq.initInfinite((*)2>>(+)1)|>Seq.map(fun g->(g,fN g))|>Seq.filter(fun(n,g)->2*n<g)

let aon n=Seq.initInfinite(int64>>(*)2L>>(+)n)|>Seq.map(fun g->(g,fN g))|>Seq.filter(fun(n,g)->2L*n<g)

aon|>Seq.take 25|>Seq.iter(fun(n,g)->printfn "The sum of the divisors of %d is %d" n g)

aon 1L|>Seq.take 25|>Seq.iter(fun(n,g)->printfn "The sum of the divisors of %d is %d" n g)

let n,g=aon|>Seq.item 999 in printfn "\nThe 1000th abundant odd number is %d. The sum of it's divisors is %d" n g

let n,g=aon 1L|>Seq.item 999 in printfn "\nThe 1000th abundant odd number is %d. The sum of it's divisors is %d" n g

let n,g=aon 1000000001L|>Seq.head in printfn "\nThe first abundant odd number greater than 1000000000 is %d. The sum of it's divisors is %d" n g

</lang>

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The 1000th abundant odd number is 492975. The sum of it's divisors is 1012336

The first abundant odd number greater than 1000000000 is 1000000575. The sum of it's divisors is 2083561584

</pre>

=={{header|Factor}}==

<lang factor>USING: arrays formatting io kernel lists lists.lazy math