A372405 Exponentially powerful numbers whose prime factorization exponents are all powerful numbers > 1.
1, 16, 81, 256, 512, 625, 1296, 2401, 6561, 10000, 14641, 19683, 20736, 28561, 38416, 41472, 50625, 65536, 83521, 104976, 130321, 160000, 194481, 234256, 279841, 314928, 320000, 390625, 456976, 614656, 707281, 810000, 923521, 1185921, 1229312, 1336336, 1500625, 1679616
Offset: 1
Keywords
Examples
16 = 2^4 and 4 = A001694(2) is a powerful number. a(7) = 1296 = 2^4*3^4. a(12) = 19683 = 3^9 (9 = A001694(4) is a powerful number).
Links
- David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1580 terms from Michael De Vlieger)
Programs
-
Mathematica
nn = 2^21; f[x_] := f[x] = Times @@ FactorInteger[x][[All, 1]]; Select[Union@ Flatten@ Table[a^7*b^6*c^5*d^4, {d, Surd[nn, 4]}, {c, Surd[nn/d^4, 5]}, {b, Surd[nn/(c^5*d^4), 6]}, {a, Surd[nn/(b^6*c^5*d^4), 7]}], AllTrue[FactorInteger[#][[All, -1]], Divisible[#, f[#]^2] &] &] (* Michael De Vlieger, Apr 29 2024 *)
-
PARI
isok(k) = if (ispowerful(k), my(f=factor(k)[,2]); #select(ispowerful, f) == #f); \\ Michel Marcus, Apr 30 2024
Formula
Sum_{n>=1} 1/a(n) = Product_{p prime} (1 + Sum_{k>=2} 1/p^A001694(k)) = 1.08410926642148594327... . - Amiram Eldar, May 12 2024
Extensions
More terms from Michael De Vlieger, Apr 29 2024
Comments