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%I #17 Feb 23 2020 17:09:48
%S 3,5,15,23,29,34,39,57,58,60,90,92,95,102,111,125,126,144,147,149,159,
%T 165,178,183,207,237,243,249,267,335,343,390,399,413,414,432,435,437,
%U 447,467,469,474,495,500,503,612,619,621,633,634,636,667,670,686,700
%N Numbers having twice as many 1's in their binary expansion as terms in their Zeckendorf expansion.
%C Numbers k such that A000120(k) = 2 * A007895(k).
%e The binary expansion of 15, "1111", contains four 1's, and the Zeckendorf expansion contains two terms: 15 = 13 + 2. There are twice as many 1's in the binary expansion, so 15 is in the sequence.
%t Position[DigitCount[(v = Select[Range[10^4], BitAnd[#, 2#] == 0 &]), 2, 1] / DigitCount[Range @ Length[v], 2, 1], _?(# == 1/2 &)]//Flatten (* _Amiram Eldar_, Jan 12 2020 after _Jean-François Alcover_ at A007895 *)
%Y Cf. A000045, A000120, A007895, A220116, A329853.
%K nonn,base
%O 1,1
%A _Alex Ratushnyak_, Nov 22 2019
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