Pairs with common factors
Pairs with common factors is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Generate the sequence where each term n is the count of the pairs (x,y) with 1 < x < y <= n, that have at least one common prime factor.
For instance, when n = 9, examine the pairs:
(2,3) (2,4) (2,5) (2,6) (2,7) (2,8) (2,9) (3,4) (3,5) (3,6) (3,7) (3,8) (3,9) (4,5) (4,6) (4,7) (4,8) (4,9) (5,6) (5,7) (5,8) (5,9) (6,7) (6,8) (6,9) (7,8) (7,9) (8,9)
Find all of the pairs that have at least one common prime factor:
(2,4) (2,6) (2,8) (3,6) (3,9) (4,6) (4,8) (6,8) (6,9)
and count them:
a(9) = 9
Terms may be found directly using the formula:
a(n) = n × (n-1) / 2 + 1 - 𝚺{i=1..n} 𝚽(i)
where 𝚽() is Phi; the Euler totient function.
For the term a(p), if p is prime, then a(p) is equal to the previous term.
- Task
- Find and display the first one hundred terms of the sequence.
- Find and display the one thousandth.
- Stretch
- Find and display the ten thousandth, one hundred thousandth, one millionth.
- See also
Factor
USING: formatting grouping io kernel math math.functions
math.primes.factors prettyprint ranges sequences
tools.memory.private ;
: totient-sum ( n -- sum )
[1..b] [ totient ] map-sum ;
: a ( n -- a(n) )
dup [ 1 - * 2 / ] keep totient-sum - ;
"Number of pairs with common factors - first 100 terms:" print
100 [1..b] [ a commas ] map 10 group simple-table. nl
7 <iota> [ dup 10^ a commas "Term #1e%d: %s\n" printf ] each
- Output:
Number of pairs with common factors - first 100 terms: 0 0 0 1 1 4 4 7 9 14 14 21 21 28 34 41 41 52 52 63 71 82 82 97 101 114 122 137 137 158 158 173 185 202 212 235 235 254 268 291 291 320 320 343 363 386 386 417 423 452 470 497 497 532 546 577 597 626 626 669 669 700 726 757 773 818 818 853 877 922 922 969 969 1,006 1,040 1,079 1,095 1,148 1,148 1,195 1,221 1,262 1,262 1,321 1,341 1,384 1,414 1,461 1,461 1,526 1,544 1,591 1,623 1,670 1,692 1,755 1,755 1,810 1,848 1,907 Term #1e0: 0 Term #1e1: 14 Term #1e2: 1,907 Term #1e3: 195,309 Term #1e4: 19,597,515 Term #1e5: 1,960,299,247 Term #1e6: 196,035,947,609
FreeBASIC
Function isPrime(n As Uinteger) As Boolean
If n Mod 2 = 0 Then Return false
For i As Uinteger = 3 To Int(Sqr(n))+1 Step 2
If n Mod i = 0 Then Return false
Next i
Return true
End Function
Function Totient(n As Uinteger) As Integer
Dim As Uinteger tot = n, i = 2
While i*i <= n
If n Mod i = 0 Then
Do
n /= i
Loop Until n Mod i <> 0
tot -= tot/i
End If
i += Iif(i = 2, 1, 2)
Wend
If n > 1 Then tot -= tot/n
Return tot
End Function
Dim As Uinteger n, limit = 1e6, sumPhi = 0
Dim As Uinteger a(limit)
For n = 1 To limit
sumPhi += Totient(n)
a(n) = Iif(isPrime(n), a(n-1), n * (n - 1) / 2 + 1 - sumPhi)
Next n
Print "Number of pairs with common factors - first one hundred terms:"
Dim As Uinteger j, count = 0
For j = 1 To 100
count += 1
Print Using " ##,###"; a(j);
If(count Mod 10) = 0 Then Print
Next j
Print !"\nThe 1st term: "; a(1)
Print "The 10th term: "; a(10)
Print "The 100th term: "; a(1e2)
Print "The 1,000th term: "; a(1e3)
Print "The 10,000th term: "; a(1e4)
Print "The 100,000th term: "; a(1e5)
Print "The 1,000,000th term: "; a(1e6)
Sleep- Output:
Number of pairs with common factors - first one hundred terms:
0 0 0 1 1 4 4 7 9 14
14 21 21 28 34 41 41 52 52 63
71 82 82 97 101 114 122 137 137 158
158 173 185 202 212 235 235 254 268 291
291 320 320 343 363 386 386 417 423 452
470 497 497 532 546 577 597 626 626 669
669 700 726 757 773 818 818 853 877 922
922 969 969 1,006 1,040 1,079 1,095 1,148 1,148 1,195
1,221 1,262 1,262 1,321 1,341 1,384 1,414 1,461 1,461 1,526
1,544 1,591 1,623 1,670 1,692 1,755 1,755 1,810 1,848 1,907
The 1st term: 0
The 10th term: 14
The 100th term: 1907
The 1,000th term: 195309
The 10,000th term: 19597515
The 100,000th term: 1960299247
The 1,000,000th term: 196035947609
J
For this task, because of the summation of euler totient values, it's more efficient to generate the sequence with a slightly different routine than we would use to compute a single value. Thus:
(1 _1r2 1r2&p. - +/\@:(5&p:)) 1+i.1e2
0 0 0 1 1 4 4 7 9 14 14 21 21 28 34 41 41 52 52 63 71 82 82 97 101 114 122 137 137 158 158 173 185 202 212 235 235 254 268 291 291 320 320 343 363 386 386 417 423 452 470 497 497 532 546 577 597 626 626 669 669 700 726 757 773 818 818 853 877 922 922 969 969 1006 1040 1079 1095 1148 1148 1195 1221 1262 1262 1321 1341 1384 1414 1461 1461 1526 1544 1591 1623 1670 1692 1755 1755 1810 1848 1907
(1 _1r2 1r2&p.@{: - +/@:(5&p:)) 1+i.1e3
195309
(1 _1r2 1r2&p.@{: - +/@:(5&p:)) 1+i.1e4
19597515
(1 _1r2 1r2&p.@{: - +/@:(5&p:)) 1+i.1e5
1960299247
(1 _1r2 1r2&p.@{: - +/@:(5&p:)) 1+i.1e6
196035947609
Here, p. calculates a polynomial (1 + (-x)/2 + (x^2)/2 in this example), 5&p: is euler's totient function, @{: modifies the polynomial to only operate on the final element of a sequence, +/ is sum and +/\ is running sum, and 1+i.n is the sequence of numbers 1 through n.
Julia
using Formatting
using Primes
pcf(n) = n * (n - 1) ÷ 2 + 1 - sum(totient, 1:n)
foreach(p -> print(rpad(p[2], 5), p[1] % 20 == 0 ? "\n" : ""), pairs(map(pcf, 1:100)))
for expo in 1:6
println("The ", format(10^expo, commas = true), "th pair with common factors count is ",
format(pcf(10^expo), commas = true))
end
- Output:
0 0 0 1 1 4 4 7 9 14 14 21 21 28 34 41 41 52 52 63 71 82 82 97 101 114 122 137 137 158 158 173 185 202 212 235 235 254 268 291 291 320 320 343 363 386 386 417 423 452 470 497 497 532 546 577 597 626 626 669 669 700 726 757 773 818 818 853 877 922 922 969 969 1006 1040 1079 1095 1148 1148 1195 1221 1262 1262 1321 1341 1384 1414 1461 1461 1526 1544 1591 1623 1670 1692 1755 1755 1810 1848 1907 The 10th pair with common factors count is 14 The 100th pair with common factors count is 1,907 The 1,000th pair with common factors count is 195,309 The 10,000th pair with common factors count is 19,597,515 The 100,000th pair with common factors count is 1,960,299,247 The 1,000,000th pair with common factors count is 196,035,947,609
Phix
with javascript_semantics function totient(integer n) integer tot = n, i = 2 while i*i<=n do if mod(n,i)=0 then while true do n /= i if mod(n,i)!=0 then exit end if end while tot -= tot/i end if i += iff(i=2?1:2) end while if n>1 then tot -= tot/n end if return tot end function constant limit = 1e6 sequence a = repeat(0,limit) atom sumPhi = 0 for n=1 to limit do sumPhi += totient(n) if is_prime(n) then a[n] = a[n-1] else a[n] = n * (n - 1) / 2 + 1 - sumPhi end if end for string j = join_by(a[1..100],1,10,fmt:="%,5d") printf(1,"Number of pairs with common factors - first one hundred terms:\n%s\n",j) for l in {1, 10, 1e2, 1e3, 1e4, 1e5, 1e6} do printf(1,"%22s term: %,d\n", {proper(ordinal(l),"SENTENCE"), a[l]}) end for
- Output:
Number of pairs with common factors - first one hundred terms:
0 0 0 1 1 4 4 7 9 14
14 21 21 28 34 41 41 52 52 63
71 82 82 97 101 114 122 137 137 158
158 173 185 202 212 235 235 254 268 291
291 320 320 343 363 386 386 417 423 452
470 497 497 532 546 577 597 626 626 669
669 700 726 757 773 818 818 853 877 922
922 969 969 1,006 1,040 1,079 1,095 1,148 1,148 1,195
1,221 1,262 1,262 1,321 1,341 1,384 1,414 1,461 1,461 1,526
1,544 1,591 1,623 1,670 1,692 1,755 1,755 1,810 1,848 1,907
First term: 0
Tenth term: 14
One hundredth term: 1,907
One thousandth term: 195,309
Ten thousandth term: 19,597,515
One hundred thousandth term: 1,960,299,247
One millionth term: 196,035,947,609
Raku
use Prime::Factor;
use Lingua::EN::Numbers;
my \𝜑 = 0, |(1..*).hyper.map: -> \t { t × [×] t.&prime-factors.unique.map: { 1 - 1/$_ } }
sub pair-count (\n) { n × (n - 1) / 2 + 1 - sum 𝜑[1..n] }
say "Number of pairs with common factors - first one hundred terms:\n"
~ (1..100).map(&pair-count).batch(10)».&comma».fmt("%6s").join("\n") ~ "\n";
for ^7 { say (my $i = 10 ** $_).&ordinal.tc.fmt("%22s term: ") ~ $i.&pair-count.&comma }
- Output:
Number of pairs with common factors - first one hundred terms:
0 0 0 1 1 4 4 7 9 14
14 21 21 28 34 41 41 52 52 63
71 82 82 97 101 114 122 137 137 158
158 173 185 202 212 235 235 254 268 291
291 320 320 343 363 386 386 417 423 452
470 497 497 532 546 577 597 626 626 669
669 700 726 757 773 818 818 853 877 922
922 969 969 1,006 1,040 1,079 1,095 1,148 1,148 1,195
1,221 1,262 1,262 1,321 1,341 1,384 1,414 1,461 1,461 1,526
1,544 1,591 1,623 1,670 1,692 1,755 1,755 1,810 1,848 1,907
First term: 0
Tenth term: 14
One hundredth term: 1,907
One thousandth term: 195,309
Ten thousandth term: 19,597,515
One hundred thousandth term: 1,960,299,247
One millionth term: 196,035,947,609
Wren
import "./math" for Int
import "./fmt" for Fmt
var totient = Fn.new { |n|
var tot = n
var i = 2
while (i*i <= n) {
if (n%i == 0) {
while(n%i == 0) n = (n/i).floor
tot = tot - (tot/i).floor
}
if (i == 2) i = 1
i = i + 2
}
if (n > 1) tot = tot - (tot/n).floor
return tot
}
var max = 1e6
var a = List.filled(max, 0)
var sumPhi = 0
for (n in 1..max) {
sumPhi = sumPhi + totient.call(n)
if (Int.isPrime(n)) {
a[n-1] = a[n-2]
} else {
a[n-1] = n * (n - 1) / 2 + 1 - sumPhi
}
}
System.print("Number of pairs with common factors - first one hundred terms:")
Fmt.tprint("$,5d ", a[0..99], 10)
System.print()
var limits = [1, 10, 1e2, 1e3, 1e4, 1e5, 1e6]
for (limit in limits) {
Fmt.print("The $,r term: $,d", limit, a[limit-1])
}- Output:
Number of pairs with common factors - first one hundred terms:
0 0 0 1 1 4 4 7 9 14
14 21 21 28 34 41 41 52 52 63
71 82 82 97 101 114 122 137 137 158
158 173 185 202 212 235 235 254 268 291
291 320 320 343 363 386 386 417 423 452
470 497 497 532 546 577 597 626 626 669
669 700 726 757 773 818 818 853 877 922
922 969 969 1,006 1,040 1,079 1,095 1,148 1,148 1,195
1,221 1,262 1,262 1,321 1,341 1,384 1,414 1,461 1,461 1,526
1,544 1,591 1,623 1,670 1,692 1,755 1,755 1,810 1,848 1,907
The 1st term: 0
The 10th term: 14
The 100th term: 1,907
The 1,000th term: 195,309
The 10,000th term: 19,597,515
The 100,000th term: 1,960,299,247
The 1,000,000th term: 196,035,947,609