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%I #57 Nov 29 2023 13:08:05
%S 1,2,3,4,6,7,8,10,11,12,14,16,17,18,20,22,23,24,26,27,28,30,32,34,35,
%T 36,38,39,40,42,44,46,47,48,50,51,52,54,56,57,58,60,62,63,64,66,68,70,
%U 71,72,74,75,76,78,80,81,82,84,86,87,88,90,92,94,95,96,98,99,100,102,104,105,106,108,110,111,112,114,116
%N Indices of terms in A354169 that have Hamming weight 1.
%C The terms of A354169 only have Hamming weights 0, 1, or 2.
%C Comment from _N. J. A. Sloane_, Jun 26 2022: (Start)
%C Taking first differences, then applying the RUNS transform twice gives [1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 13, 1, 1, 1, 13, 1, 1, 1, 29, 1, 1, 1, 29, 1, 1, 1, 61, 1, 1, 1, 61, 1, 1, 1, 125, 1, 1, 1, 125, 1, 1, 1, 253, 1, 1, 1, 253, 1, 1, 1, 509, 1, 1, 1,...].
%C If the initial terms 1, 1, 1, 1, 5 are replaced by a single 1, this has an obvious regular structure, which can then be analyzed to give a generating function for the sequence. See link below. (End)
%C For proofs of these statements see De Vlieger et al. (2022). - _N. J. A. Sloane_, Aug 29 2022
%H N. J. A. Sloane, <a href="/A354767/b354767.txt">Table of n, a(n) for n = 1..3296</a>
%H Michael De Vlieger, Thomas Scheuerle, Rémy Sigrist, N. J. A. Sloane, and Walter Trump, <a href="http://arxiv.org/abs/2209.04108">The Binary Two-Up Sequence</a>, arXiv:2209.04108 [math.CO], Sep 11 2022.
%H N. J. A. Sloane, <a href="/A354767/a354767.txt">A conjectured generating function for A354169</a>.
%F Conjecture from _N. J. A. Sloane_, Jun 30 2022, modified Jul 30 2022: (Start)
%F We first define a sequence {b(n)} as follows.
%F Define "fence posts" by F(0) = 1, F(2i+1) = 2^(i+5) - 3 for i >= 0, F(2i) = 3*2^(i+3) - 3 for i >= 1.
%F The F(i) sequence begins 1, 29, 45, 61, 93, 125, 189, 253, 381, 509, ... (cf. A136252 or A354788).
%F The value of b(n) at n = F(i) is V(i) = 0 if i = 0, V(i) =(F(i)-7)/2 if i >= 1.
%F The V(i) sequence begins 0, 11, 19, 27, 43, 59, 91, 123, 187, 251, ...
%F The first 28 terms are irregular, and we simply define b(n) for F(0) = 1 <= n <= 28 to be the n-th term of
%F [0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 4, 4, 4, 5, 6, 6, 6, 7, 7, 7, 8, 9, 10, 10, 10, 11, 11].
%F Assume now that n >= F(1) = 29, and define i and j by F(i) <= n < F(i+1), n = F(i) + j.
%F Then b(n) = V(i) + f(j), where f(0) ... f(5) are 0,1,2,3,3,3, and for j >= 6, f(j) = 2 + 2*floor((j-2)/4) + epsilon(j), where epsilon(j) is 1 if j==1 mod 4 and is otherwise 0.
%F The f(i), i >= 0, sequence (A354779) is independent of n (to find b(n) we use only an initial segment of f(n)), and begins:
%F 0, 1, 2, 3, 3, 3, 4, 4, 4, 5, 6, 6, 6, 7, 8, 8, 8, 9, 10, 10, 10, 11, 12, 12, 12, 13, 14, ...
%F With b(n) defined in this way, we conjecture that a(n) = b(n) + n. This has been checked for the first 3296 terms.
%F (End)
%F The conjecture is now known to be true. - _N. J. A. Sloane_, Aug 29 2022
%Y Cf. A354169, A354680, A354779, A354788, A354798.
%K nonn
%O 1,2
%A _N. J. A. Sloane_, Jun 21 2022
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