A262675 Exponentially evil numbers.

1, 8, 27, 32, 64, 125, 216, 243, 343, 512, 729, 864, 1000, 1024, 1331, 1728, 1944, 2197, 2744, 3125, 3375, 4000, 4096, 4913, 5832, 6859, 7776, 8000, 9261, 10648, 10976, 12167, 13824, 15552, 15625, 16807, 17576, 19683, 21952, 23328, 24389, 25000, 27000, 27648, 29791

Offset: 1

Author

Vladimir Shevelev, Sep 27 2015

Keywords

• nonn
• base

Comments

Or the numbers whose prime power factorization contains primes only in evil exponents (A001969): 0, 3, 5, 6, 9, 10, 12, ...

If n is in the sequence, then n^2 is also in the sequence.

A268385 maps each term of this sequence to a unique nonzero square (A000290), and vice versa. - Antti Karttunen, May 26 2016

Examples

			864 = 2^5*3^3; since 5 and 3 are evil numbers, 864 is in the sequence.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Reinhard Zumkeller)

Crossrefs

Subsequence of A036966.

Apart from 1, a subsequence of A270421.

Indices of ones in A270418.

Sequence A270437 sorted into ascending order.

Cf. A001969, A209061, A138302, A197680, A000578, A000584, A001014, A001017, A008456, A010803, A010805, A010806, A010808, A010811, A010812, A001221, A010059, A124010, A268385, A270428.

Programs

Haskell

a262675 n = a262675_list !! (n-1)
a262675_list = filter
   (all (== 1) . map (a010059 . fromIntegral) . a124010_row) [1..]
-- Reinhard Zumkeller, Oct 25 2015

Mathematica

{1}~Join~Select[Range@ 30000, AllTrue[Last /@ FactorInteger[#], EvenQ@ First@ DigitCount[#, 2] &] &] (* Michael De Vlieger, Sep 27 2015, Version 10 *)
expEvilQ[n_] := n == 1 || AllTrue[FactorInteger[n][[;; , 2]], EvenQ[DigitCount[#, 2, 1]] &]; With[{max = 30000}, Select[Union[Flatten[Table[i^2*j^3, {j, Surd[max, 3]}, {i, Sqrt[max/j^3]}]]], expEvilQ]] (* Amiram Eldar, Dec 01 2023 *)

PARI

isok(n) = {my(f = factor(n)); for (i=1, #f~, if (hammingweight(f[i,2]) % 2, return (0));); return (1);} \\ Michel Marcus, Sep 27 2015

Perl

use ntheory ":all"; sub isok { my @f = factor_exp($[0]); return scalar(grep { !(hammingweight($->[1]) % 2) } @f) == @f; } # Dana Jacobsen, Oct 26 2015

Formula

Product_{k=1..A001221(n)} A010059(A124010(n,k)) = 1. - Reinhard Zumkeller, Oct 25 2015

Sum_{n>=1} 1/a(n) = Product_{p prime} (1 + Sum_{k>=2} 1/p^A001969(k)) = Product_{p prime} f(1/p) = 1.2413599378..., where f(x) = (1/(1-x) + Product_{k>=0} (1 - x^(2^k)))/2. - Amiram Eldar, May 18 2023, Dec 01 2023

Extensions

More terms from Michel Marcus, Sep 27 2015