%I #16 Jan 22 2015 00:47:10
%S 1,2,4,9,18,36,72,145,290,580,1161,2322,4644,9288,18576,37153,74306,
%T 148612,297225,594450,1188900,2377800,4755601,9511202,19022404,
%U 38044809,76089618,152179236,304358472,608716944,1217433888,2434867777,4869735554,9739471108
%N Numbers whose binary expansion equals the first n digits of the binary sequence A252488 whose run lengths are given by A001511 (the ruler function).
%C The binary sequence with run lengths given by A001511 (1,2,1,3,1,2,1,4,1, ...) begins 1001000100100001.... Truncated to the first n digits and expressed as decimal numbers, this yields:
%C 1 1
%C 10 2
%C 100 4
%C 1001 9
%C 10010 18
%C 100100 36
%C 1001000 72
%C 10010001 145
%C 100100010 290
%C 1001000100 580
%C 10010001001 1161
%C 100100010010 2322
%C 1001000100100 4644
%C 10010001001000 9288
%C 100100010010000 18576
%C 1001000100100001 37153
%C This is a superincreasing sequence (every element of the sequence is greater than the sum of all previous elements in the sequence).
%C The binary sequence appears to match the parity of A170849.
%H Jeremy Gardiner, <a href="/A253585/b253585.txt">Table of n, a(n) for n = 1..40</a>
%o (PARI) a001511(n) = valuation(n, 2) + 1;
%o lista(nn) = {a = 0; for (n=1, nn, for (j=1, a001511(n), a *= 2; if (n % 2, a += 1); print1(a, ", ");););} \\ _Michel Marcus_, Jan 11 2015
%Y Cf. A001511, A170849, A252488.
%K nonn,base
%O 1,2
%A _Jeremy Gardiner_, Jan 04 2015