A335740 - OEIS

%I #20 Feb 10 2024 03:59:22

%S 3,5,6,7,9,10,11,13,14,15,17,18,19,21,22,23,25,26,27,29,30,31,33,34,

%T 35,36,37,38,39,41,42,43,45,46,47,49,50,51,53,54,55,57,58,59,61,62,63,

%U 65,66,67,69,70,71,72,73,74,75,77,78,79,81,82,83,85,86,87,89,90

%N Factorize each integer m >= 2 as the product of powers of nonunit squarefree numbers with distinct exponents that are powers of 2. The sequence lists m such that the factor with the largest exponent is not a power of 2.

%C Every missing number greater than 2 is a multiple of 4. Every power of 2 is missing. Every positive power of every squarefree number greater than 2 is present.

%C The defined factorization is unique. Every positive number is a product of at most one squarefree number (A005117), at most one square of a squarefree number (A062503), at most one 4th power of a squarefree number (A113849), at most one 8th power of a squarefree number, and so on.

%C Iteratively map m using A000188, until 1 is reached, as A000188^k(m), for some k >= 1. m is in the sequence if and only if the preceding number, A000188^(k-1)(m), is greater than 2. k can be shown to be A299090(m).

%C The asymptotic density of this sequence is 1 - Sum_{k>=0} (d(2^(k+1)) - d(2^k))/2^(2^(k+1)-1) = 0.78636642806078947931..., where d(k) = 2^(k-1)/((2^k-1)*zeta(k))) is the asymptotic density of odd k-free numbers for k >= 2, and d(1) = 0. - _Amiram Eldar_, Feb 10 2024

%H Amiram Eldar, <a href="/A335740/b335740.txt">Table of n, a(n) for n = 1..10000</a>

%F {a(n)} = {m : m >= 2 and A000188^(k-1)(m) > 2, where k = A299090(m)}.

%e 6 is a squarefree number, so its factorization for the definition (into powers of nonunit squarefree numbers with distinct exponents that are powers of 2) is the trivial "6^1". 6^1 is therefore the factor with the largest exponent, and is not a power of 2, so 6 is in the sequence.

%e 48 factorizes for the definition as 3^1 * 2^4. The factor with the largest exponent is 2^4, which is a power of 2, so 48 is not in the sequence.

%e 10^100 (a googol) factorizes in this way as 10^4 * 10^32 * 10^64. The factor with the largest exponent, 10^64, is a power of 10, not a power of 2, so 10^100 is in the sequence.

%t f[p_, e_] := p^Floor[e/2]; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; Select[Range[2, 100], FixedPointList[s, #] [[-3]] > 2 &] (* _Amiram Eldar_, Nov 27 2020 *)

%o (PARI) is(n) = {my(e = valuation(n, 2), o = n >> e); if(e == 0, n > 1, if(o == 1, e < 1, floor(logint(e, 2)) <= floor(logint(vecmax(factor(o)[,2]), 2))));} \\ _Amiram Eldar_, Feb 10 2024

%Y Complement within A020725 of A335738.

%Y A000188, A299090 are used in a formula defining this sequence.

%Y Powers of squarefree numbers: A005117(1), A144338(1), A062503(2), A113849(4).

%Y Subsequences: A042968\{1,2}, A182853, A268390.

%Y With {1}, numbers in the odd bisection of A336322.

%K nonn

%O 1,1

%A _Peter Munn_, Jun 20 2020