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A197680 Numbers whose exponents in their prime power factorization are squares.

%I A197680 #77 Apr 29 2025 05:33:42
%S A197680 1,2,3,5,6,7,10,11,13,14,15,16,17,19,21,22,23,26,29,30,31,33,34,35,37,
%T A197680 38,39,41,42,43,46,47,48,51,53,55,57,58,59,61,62,65,66,67,69,70,71,73,
%U A197680 74,77,78,79,80,81,82,83,85,86,87,89,91,93,94,95,97,101
%N A197680 Numbers whose exponents in their prime power factorization are squares.
%C A197680 Numbers whose prime factorization has the form Product_i p_i^e_i where the e_i are all squares.
%C A197680 All squarefree numbers (A005117) are in the sequence. - _Vladimir Shevelev_, Nov 16 2015
%C A197680 Let h_k be the density of the subsequence of A197680 of numbers whose prime power factorization (PPF) has the form Product_i p_i^e_i where the e_i all squares <= k^2. Then for every k>1 there exists eps_k>0 such that for any x from the interval (h_k-eps_k, h_k) there is no sequence S of positive integers such that x is the density of numbers whose PPF has the form Product_i p_i^e_i where the e_i are all in S. - For a proof, see [Shevelev], second link. - _Vladimir Shevelev_, Nov 17 2015
%C A197680 Numbers with an odd number of exponential divisors (A049419). - _Amiram Eldar_, Nov 05 2021
%H A197680 Alois P. Heinz, <a href="/A197680/b197680.txt">Table of n, a(n) for n = 1..10000</a>
%H A197680 Math StackExchange, <a href="http://math.stackexchange.com/questions/73354/two-dirichlets-series-related-to-the-divisor-summatory-function-and-to-the-riem">Question 73354</a>, 2011.
%H A197680 Vladimir Shevelev, <a href="http://arxiv.org/abs/1510.05914">Exponentially S-numbers</a>, arXiv:1510.05914 [math.NT], 2015.
%H A197680 Vladimir Shevelev, <a href="http://arxiv.org/abs/1511.03860">Set of all densities of exponentially S-numbers</a>, arXiv:1511.03860 [math.NT], 2015.
%H A197680 Vladimir Shevelev, <a href="http://dx.doi.org/10.4064/aa8395-5-2016">S-exponential numbers</a>, Acta Arithmetica, Vol. 175 (2016), pp. 385-395.
%F A197680 Sum_{i<=x, i is in A197680} 1 = h*x + O(sqrt(x)*log x*e^(c*sqrt(log x)/(log(log x))), where c=4*sqrt(2.4/log 2)=7.44308... and h=Product_{prime p} (1+Sum_{i>=2} (u(i)-u(i-1))/p^i)=0.641115... where u(n) is the characteristic function of sequence A000290. The calculations of h in the formula were done independently by _Juan Arias-de-Reyna_ and _Peter J. C. Moses_. For a proof of the formula, see the first Shevelev link. - _Vladimir Shevelev_, Nov 17 2015
%p A197680 a:= proc(n) option remember; local k; for k from 1+
%p A197680       `if`(n=1, 0, a(n-1)) while 0=mul(`if`(issqr(
%p A197680        i[2]), 1, 0), i=ifactors(k)[2]) do od; k
%p A197680     end:
%p A197680 seq(a(n), n=1..80);  # _Alois P. Heinz_, Jun 30 2016
%t A197680 Select[Range[100], Union[IntegerQ /@ Sqrt[Transpose[FactorInteger[#]][[2]]]][[1]] &] (* _T. D. Noe_, Oct 18 2011 *)
%o A197680 (PARI) isok(n) = {my(f = factor(n)[,2]); #select(x->issquare(x), f) == #f; } \\ _Michel Marcus_, Oct 23 2015
%Y A197680 Cf. A000290, A049419, A005117, A115063, A209061.
%K A197680 nonn
%O A197680 1,2
%A A197680 _A. Neves_, Oct 17 2011
%E A197680 Reformulation of the name by _Vladimir Shevelev_, Oct 14 2015