A383394 Perfect powers of Achilles numbers.
5184, 11664, 40000, 82944, 153664, 186624, 250000, 373248, 419904, 455625, 640000, 746496, 937024, 944784, 1259712, 1265625, 1327104, 1750329, 1827904, 1882384, 2458624, 3240000, 3779136, 4000000, 5345344, 6718464, 7290000, 8000000, 8340544, 9529569, 10240000
Offset: 1
Examples
Table of n, a(n) for n = 1..12: n a(n) -------------------------------- 1 5184 = 72^2 = 2^6 * 3^4 2 11664 = 108^2 = 2^4 * 3^6 3 40000 = 200^2 = 2^6 * 5^4 4 82944 = 288^2 = 2^10 * 3^4 5 153664 = 392^2 = 2^6 * 7^4 6 186624 = 432^2 = 2^8 * 3^6 7 250000 = 500^2 = 2^4 * 5^6 8 373248 = 72^3 = 2^9 * 3^6 9 419904 = 648^2 = 2^6 * 3^8 10 455625 = 675^2 = 3^6 * 5^4 11 640000 = 800^2 = 2^10 * 5^4 12 746496 = 864^2 = 2^10 * 3^6
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
nn = 2^24; mm = Sqrt[nn]; i = 1; k = 2; MapIndexed[Set[S[First[#2]], #1] &, Rest@ Select[Union@ Flatten@ Table[a^2*b^3, {b, Surd[mm, 3]}, {a, Sqrt[mm/b^3]}], GCD @@ FactorInteger[#][[;; , -1]] == 1 &]]; Union@ Reap[While[j = 2; While[S[i]^j < nn, Sow[S[i]^j]; j++]; j > 2, k++; i++] ][[-1, 1]] -
Python
from math import isqrt from sympy import integer_nthroot, mobius def A383394(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 f(kmin) < kmin: kmin >>= 1 kmin = max(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): c, l = squarefreepi(integer_nthroot(x,3)[0])+sum(mobius(k)*(integer_nthroot(x, k)[0]-1) for k in range(2, x.bit_length()))-1, 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) return c-l def f(x): return n+x-sum(g(integer_nthroot(x, k)[0]) for k in range(2, x.bit_length())) return bisection(f,n,n) # Chai Wah Wu, Aug 11 2025
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