A325240 Numbers whose minimum prime exponent is 2.

4, 9, 25, 36, 49, 72, 100, 108, 121, 144, 169, 196, 200, 225, 288, 289, 324, 361, 392, 400, 441, 484, 500, 529, 576, 675, 676, 784, 800, 841, 900, 961, 968, 972, 1089, 1125, 1152, 1156, 1225, 1323, 1352, 1369, 1372, 1444, 1521, 1568, 1600, 1681, 1764, 1800

Offset: 1

Author

Gus Wiseman, Apr 15 2019

Keywords

• nonn

Comments

Or barely powerful numbers, a subset of powerful numbers A001694.

The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), so these are Heinz numbers of integer partitions whose minimum multiplicity is 2 (counted by A244515).

Powerful numbers (A001694) that are not cubefull (A036966). - Amiram Eldar, Jan 30 2023

Examples

			The sequence of terms together with their prime indices begins:
    4: {1,1}
    9: {2,2}
   25: {3,3}
   36: {1,1,2,2}
   49: {4,4}
   72: {1,1,1,2,2}
  100: {1,1,3,3}
  108: {1,1,2,2,2}
  121: {5,5}
  144: {1,1,1,1,2,2}
  169: {6,6}
  196: {1,1,4,4}
  200: {1,1,1,3,3}
  225: {2,2,3,3}
  288: {1,1,1,1,1,2,2}
  289: {7,7}
  324: {1,1,2,2,2,2}
  361: {8,8}
  392: {1,1,1,4,4}
  400: {1,1,1,1,3,3}
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Positions of 2's in A051904.

Maximum instead of minimum gives A067259.

Cf. A001221, A001222, A001358, A001694, A007774, A036966, A051903, A052485, A118914, A244515, A325241.

Cf. A065483, A082695.

Programs

Mathematica

Select[Range[1000],Min@@FactorInteger[#][[All,2]]==2&]

PARI

is(n)={my(e=factor(n)[,2]); n>1 && vecmin(e) == 2; } \\ Amiram Eldar, Jan 30 2023

Python

from math import isqrt, gcd
from sympy import integer_nthroot, factorint, mobius
def A325240(n):
    def squarefreepi(n): return int(sum(mobius(k)*(n//k**2) for k in range(1, isqrt(n)+1)))
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x):
        c, l = n+x, 0
        j = isqrt(x)
        while j>1:
            k2 = integer_nthroot(x//j**2,3)[0]+1
            w = squarefreepi(k2-1)
            c -= j*(w-l)
            l, j = w, isqrt(x//k2**3)
        c -= squarefreepi(integer_nthroot(x,3)[0])-l
        for w in range(1,integer_nthroot(x,5)[0]+1):
            if all(d<=1 for d in factorint(w).values()):
                for y in range(1,integer_nthroot(z:=x//w**5,4)[0]+1):
                    if gcd(w,y)==1 and all(d<=1 for d in factorint(y).values()):
                        c += integer_nthroot(z//y**4,3)[0]
        return c
    return bisection(f,n,n**2) # Chai Wah Wu, Oct 02 2024

Formula

Sum_{n>=1} 1/a(n) = zeta(2)*zeta(3)/zeta(6) - Product_{p prime} (1 + 1/(p^2*(p-1))) = A082695 - A065483 = 0.6038122832... . - Amiram Eldar, Jan 30 2023