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%I #31 Feb 14 2022 01:24:05
%S 63,95,111,119,123,125,126,159,175,183,187,189,190,207,215,219,221,
%T 222,231,235,237,238,243,245,246,249,250,252,287,303,311,315,317,318,
%U 335,343,347,349,350,359,363,365,366,371,373
%N Numbers with exactly 6 ones in binary expansion.
%C Sequence appears to include all numbers m such that 8^5 is the highest power of 2 dividing A005148(m). General conjecture: numbers k such that 8^j is the highest power of 2 dividing A005148(k) is the same sequence as numbers having exactly (j+1) 1's in their binary representation. - _Benoit Cloitre_, Jun 22 2002
%H Ivan Neretin, <a href="/A023688/b023688.txt">Table of n, a(n) for n = 1..10000</a>
%H Robert Baillie, <a href="http://arxiv.org/abs/0806.4410">Summing the curious series of Kempner and Irwin</a>, arXiv:0806.4410 [math.CA], 2008-2015. See p. 18 for Mathematica code irwinSums.m.
%F a(n+1) = A057168(a(n)). - _M. F. Hasler_, Aug 27 2014
%F Sum_{n>=1} 1/a(n) = 1.387753111935705074750004158584017188750706394077047633137401652680870607884... (calculated using Baillie's irwinSums.m, see Links). - _Amiram Eldar_, Feb 14 2022
%t Select[ Range[ 63, 380 ], (Count[ IntegerDigits[ #, 2 ], 1 ]==6)& ]
%o (PARI) is_A023688(n)=hammingweight(n)==6 \\ _M. F. Hasler_, Aug 27 2014
%o (PARI) print1(t=2^6-1); for(i=2, 50, print1(", "t=A057168(t))) \\ _M. F. Hasler_, Aug 27 2014
%Y Cf. A000079, A018900, A014311, A014312, A014313, A023689, A023690, A023691 (Hamming weight = 1..9).
%Y Cf. A005148, A057168.
%K nonn,base,easy
%O 1,1
%A _Olivier Gérard_
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