Abundant, deficient and perfect number classifications: Difference between revisions
Abundant, deficient and perfect number classifications (view source)
Revision as of 09:03, 12 July 2015
, 8 years ago→{{header|Javascript}}
Line 442:
=={{header|Javascript}}==
<lang Javascript>for (var dpa=[1,0,0], n=2; n<=20000; n+=1) {
for (var ds=0, d=1, e=n/2+1; d<e; d+=1) if (n%d==0) ds+=d
dpa[ds<n ? 0 : ds==n ? 1 : 2]+=1
}
document.write('Deficient:',dpa[0], ', Perfect:',dpa[1], ', Abundant:',dpa[2], '<br>' )</lang>
'''Or:'''
<lang Javascript>for (var dpa=[1,0,0], n=2; n<=20000; n+=1) {
for (var ds=1, d=2, e=Math.sqrt(n); d<e; d+=1) if (n%d==0) ds+=d+n/d
Line 448 ⟶ 454:
}
document.write('Deficient:',dpa[0], ', Perfect:',dpa[1], ', Abundant:',dpa[2], '<br>' )</lang>
'''Or:'''
<lang Javascript>function primes(t) {
var ps = {2:true, 3:true}
next: for (var n=5, i=2; n<=t; n+=i, i=6-i) {
var s = Math.sqrt( n )
for ( var p in ps ) {
if ( p > s ) break
if ( n % p ) continue
continue next
}
ps[n] = true
}
return ps
}
function factorize(f, t) {
var cs = {}, ps = primes(t)
for (var n=f; n<=t; n++) if (!ps[n]) cs[n] = factors(n)
return cs
function factors(n) {
for ( var p in ps ) if ( n % p == 0 ) break
var ts = {}
ts[p] = 1
if ( ps[n /= p] ) {
if ( !ts[n] ) ts[n]=1; else ts[n]+=1
}
else {
var fs = cs[n]
if ( !fs ) cs[n] = fs = factors(n)
for ( var e in fs ) ts[e] = fs[e] + (e==p ? 1 : 0)
}
return ts
}
}
function pContrib(p, e) {
for (var pc=1, n=1, i=1; i<=e; i+=1) pc+=n*=p;
return pc
}
for (var dpa=[1,0,0], t=20000, cs=factorize(2,t), n=2; n<=t; n+=1) {
var ds=1, fs=cs[n]
if (fs) {
for (var p in fs) ds *= pContrib(p, fs[p])
ds -= n
}
dpa[ds<n ? 0 : ds==n ? 1 : 2]+=1
}
document.write('Deficient:',dpa[0], ', Perfect:',dpa[1], ', Abundant:',dpa[2], '<br>' )</lang>
{{output}}
<pre>Deficient:15043, Perfect:4, Abundant:4953</pre>
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