A077642 Number of squarefree integers in the closed interval [10^n, -1 + 2*10^n], i.e., among 10^n consecutive integers beginning with 10^n.

1, 7, 61, 607, 6077, 60787, 607951, 6079284, 60792732, 607927092, 6079270913, 60792710227, 607927101577, 6079271018873, 60792710185938, 607927101853650, 6079271018542500, 60792710185398417, 607927101854027370, 6079271018540264581, 60792710185402679735, 607927101854026683706

Offset: 0

Author

Labos Elemer, Nov 14 2002

Keywords

• nonn

Comments

What a(n)/n is converging to?

Limit_{n->oo} a(n)/10^n = 1/zeta(2). [Max Alekseyev, Oct 18 2008]

Examples

			n=10: among numbers {10,...,19} seven are squarefree [10,11,13,14,15,17,19], so a(1)=7.
		

Links

• Lucas A. Brown, Python program.

Crossrefs

Cf. A077641, A077643.

Programs

Maple

with(numtheory): for n from 0 to 5 do ct:=0: for k from 10^n to 2*10^n-1 do if abs(mobius(k))>0 then ct:=ct+1 else ct:=ct: fi: od: a[n]:=ct: od: seq(a[n],n=0..5); # Emeric Deutsch, Mar 28 2005

Mathematica

Table[Apply[Plus, Table[Abs[MoebiusMu[10^w+j]], {j, 0, -1+10^(w-1)}]], {w, 0, 6}]

PARI

{ a(n) = sum(m=1,sqrtint(2*10^n-1), moebius(m) * ((2*10^n-1)\m^2 - (10^n-1)\m^2) ) } \\ Max Alekseyev, Oct 18 2008

Formula

a(n) = Sum_{j=0..-1+10^n} abs(mu(10^n + j)).

Extensions

6079284 from Emeric Deutsch, Mar 28 2005

a(8)-a(15) from Max Alekseyev, Oct 18 2008

a(16)-a(21) from Lucas A. Brown, Feb 25 2024