A062503 Squarefree numbers squared.
1, 4, 9, 25, 36, 49, 100, 121, 169, 196, 225, 289, 361, 441, 484, 529, 676, 841, 900, 961, 1089, 1156, 1225, 1369, 1444, 1521, 1681, 1764, 1849, 2116, 2209, 2601, 2809, 3025, 3249, 3364, 3481, 3721, 3844, 4225, 4356, 4489, 4761, 4900, 5041, 5329, 5476
Offset: 1
Keywords
Links
- Harry J. Smith, Table of n, a(n) for n = 1..1000
Crossrefs
Programs
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Haskell
a062503 = a000290 . a005117 -- Reinhard Zumkeller, Jul 07 2013
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Mathematica
Select[Range[100], SquareFreeQ]^2
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PARI
je=[]; for(n=1,200, if(issquarefree(n),je=concat(je,n^2),)); je
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PARI
n=0; for (m=1, 10^5, if(issquarefree(m), write("b062503.txt", n++, " ", m^2); if (n==1000, break))) \\ Harry J. Smith, Aug 08 2009
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PARI
is(n)=issquare(n,&n) && issquarefree(n) \\ Charles R Greathouse IV, Sep 18 2015
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Python
from math import isqrt from sympy import mobius def A062503(n): def f(x): return n-1+x-sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1)) kmin, kmax = 1,2 while f(kmax) >= kmax: kmax <<= 1 while True: kmid = kmax+kmin>>1 if f(kmid) < kmid: kmax = kmid else: kmin = kmid if kmax-kmin <= 1: break return kmax**2 # Chai Wah Wu, Aug 19 2024
Formula
Numbers k such that Sum_{d|k} mu(d)*mu(k/d) = 1. - Benoit Cloitre, Mar 03 2004
a(n) ~ (Pi^4/36) * n^2. - Charles R Greathouse IV, Nov 24 2015
a(n) = A046692(a(n))^2. - Torlach Rush, Jan 05 2019
For all k in the sequence, Omega(k) = 2*omega(k). - Wesley Ivan Hurt, Apr 30 2020
Sum_{n>=1} 1/a(n) = zeta(2)/zeta(4) = 15/Pi^2 (A082020). - Amiram Eldar, May 22 2020
Comments