A268335 Exponentially odd numbers.
1, 2, 3, 5, 6, 7, 8, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 24, 26, 27, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 46, 47, 51, 53, 54, 55, 56, 57, 58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 82, 83, 85, 86, 87, 88, 89, 91, 93, 94, 95, 96, 97
Offset: 1
Keywords
Links
- Peter J. C. Moses, Table of n, a(n) for n = 1..2000
- Eckford Cohen, Arithmetical functions associated with the unitary divisors of an integer, Mathematische Zeitschrift, Vol. 74 (1960), pp. 66-80.
- Eckford Cohen, Some sets of integers related to the k-free integers, Acta Sci. Math. (Szeged), Vol. 22, No. 3-4 (1961), pp. 223-233.
- Vladimir Shevelev, Exponentially S-numbers, arXiv:1510.05914 [math.NT], 2015.
- Vladimir Shevelev, Set of all densities of exponentially S-numbers, arXiv preprint arXiv:1511.03860 [math.NT], 2015.
- Vladimir Shevelev, S-exponential numbers, Acta Arithmetica, Vol. 175(2016), 385-395.
- D. Suryanarayana and R. Sita Rama Chandra Rao, Distribution of unitarily k-free integers, Journal of the Australian Mathematical Society, Vol. 20 , No. 2 (1975), pp. 129-141.
Crossrefs
Programs
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Mathematica
Select[Range@ 100, AllTrue[Last /@ FactorInteger@ #, OddQ] &] (* Version 10, or *) Select[Range@ 100, Times @@ Boole[OddQ /@ Last /@ FactorInteger@ #] == 1 &] (* Michael De Vlieger, Feb 02 2016 *)
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PARI
isok(n)=my(f = factor(n)); for (k=1, #f~, if (!(f[k,2] % 2), return (0))); 1; \\ Michel Marcus, Feb 02 2016
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Python
from itertools import count, islice from sympy import factorint def A268335_gen(startvalue=1): # generator of terms >= startvalue return filter(lambda n:all(e&1 for e in factorint(n).values()),count(max(startvalue,1))) A268335_list = list(islice(A268335_gen(),20)) # Chai Wah Wu, Jun 22 2023
Formula
Sum_{a(n)<=x} 1 = C*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 C = Product_{prime p} (1 - 1/p*(p + 1)) = 0.7044422009991... (A065463).
Sum_{n>=1} 1/a(n)^s = zeta(2*s) * Product_{p prime} (1 + 1/p^s - 1/p^(2*s)), s>1. - Amiram Eldar, Sep 26 2023
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