A060687 Numbers k such that there exist exactly 2 Abelian groups of order k, i.e., A000688(k) = 2.
4, 9, 12, 18, 20, 25, 28, 44, 45, 49, 50, 52, 60, 63, 68, 75, 76, 84, 90, 92, 98, 99, 116, 117, 121, 124, 126, 132, 140, 147, 148, 150, 153, 156, 164, 169, 171, 172, 175, 188, 198, 204, 207, 212, 220, 228, 234, 236, 242, 244, 245, 260, 261, 268, 275, 276, 279
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..5000 from Enrique Pérez Herrero)
- Eckford Cohen, Arithmetical notes. VIII. An asymptotic formula of Rényi, Proc. Amer. Math. Soc. 13 (1962), pp. 536-539.
- Alfred Rényi, On the density of certain sequences of integers, Publications de l'Institut Mathématique, Vol. 8 (1955), pp. 157-162.
- Index entries for sequences related to groups.
Programs
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Haskell
a060687 n = a060687_list !! (n-1) a060687_list = filter ((== 1) . a046660) [1..] -- Reinhard Zumkeller, Nov 29 2015
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Mathematica
Select[Range[500], PrimeOmega[#] - PrimeNu[#] == 1 &] (* Harvey P. Dale, Sep 08 2011 *)
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PARI
for(n=1,279,if(bigomega(n)-omega(n)==1,print1(n,",")))
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PARI
is(n)=factorback(factor(n)[,2])==2 \\ Charles R Greathouse IV, Sep 18 2015
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PARI
list(lim)=my(s=lim\4,v=List(),u=vectorsmall(s,i,1),t,x); forprime(k=2,sqrtint(s), t=k^2; forstep(i=t,s,t, u[i]=0)); forprime(k=2,sqrtint(lim\1), t=k^2; for(i=1,#u, if(u[i] && gcd(k,i)==1, x=t*i; if(x>lim, break); listput(v,x)))); Set(v) \\ Charles R Greathouse IV, Aug 02 2016
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Python
from math import isqrt from sympy import mobius, primerange def A060687(n): def bisection(f,kmin=0,kmax=1): while f(kmax) > kmax: kmax <<= 1 kmin = kmax >> 1 while kmax-kmin > 1: kmid = kmax+kmin>>1 if f(kmid) <= kmid: kmax = kmid else: kmin = kmid return kmax def g(x): return sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1)) def f(x): return int(n+x+sum(sum(g(x//p**j) if j&1 else -g(x//p**j) for j in range(2,x.bit_length())) for p in primerange(isqrt(x)+1))) return bisection(f,n,n) # Chai Wah Wu, Feb 24 2025
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Python
from sympy import factorint def is_A060687(n): return sum(v := factorint(n).values()) == len(v) + 1 # David Radcliffe, Jul 28 2025
Formula
Extensions
Corrected and extended by Vladeta Jovovic, Jul 05 2001
Comments