A057828 Number of perfect squares, k^2, where k^2 <= n and gcd(k,n) = 1.
1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 3, 1, 3, 2, 2, 2, 4, 1, 4, 2, 3, 2, 4, 1, 4, 3, 4, 3, 5, 1, 5, 3, 4, 3, 4, 2, 6, 3, 4, 2, 6, 2, 6, 3, 3, 3, 6, 2, 6, 3, 5, 4, 7, 3, 6, 3, 5, 4, 7, 2, 7, 4, 4, 4, 7, 3, 8, 4, 6, 2, 8, 3, 8, 4, 5, 4, 7, 3, 8, 3, 6, 5, 9, 2, 8, 5, 6, 5, 9, 2, 8, 5, 6, 5, 8, 3, 9, 4, 6, 4, 10, 3, 10, 5
Offset: 1
Examples
Only 2 squares, 1 and 9, are <= 14 and relatively prime to 14. So a(14) = 2.
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
- Eric Weisstein's World of Mathematics, Totative.
Programs
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Haskell
a057828 x = length $ filter ((== 1) . (gcd x)) $ takeWhile (<= x) $ tail a000290_list -- Reinhard Zumkeller, Jul 22 2012
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Maple
with(numtheory): seq(add(mobius(d)*floor(sqrt(n)/d), d in divisors(n)), n=1..100); # Ridouane Oudra, Jan 26 2025
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Mathematica
Table[Count[Range[Sqrt@ n]^2, _?(CoprimeQ[#, n] &)], {n, 104}]
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PARI
a(n) = sumdiv(n, d, moebius(d)*(sqrt(n)\d)); \\ Michel Marcus, Jan 27 2025
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PARI
a(n,f=factor(n))=my(g=f,d); g[,2]=vectorv(#f~,i,1); d=divisors(g,1); sum(i=1,#d, moebius(d[i][2])*sqrtint(n\d[i][1]^2)) \\ Charles R Greathouse IV, Mar 26 2025
Formula
a(n) = Sum_{d|n} mu(d)*floor(sqrt(n)/d). - Ridouane Oudra, Jan 26 2025
a(n) = Sum_{k=1..floor(sqrt(n))} A054521(n,k). - Ridouane Oudra, Mar 25 2025
Comments