OFFSET
1,1
COMMENTS
Numbers of the form (2*k+1) * 2^e where e >= 1, k+e is even. In other words, union of {(4*m+1) * 2^(2t)} and {(4*m+3) * 2^(2t-1)}, where m >= 0, t > 0.
Numbers whose quaternary (base-4) expansion ends in 100...00 or 1200..00 or 3200..00. At least one trailing zero is required in the first case but not in the latter two cases.
There are precisely 2^(N-2) terms <= 2^N for every N >= 2.
Also even indices of 1 in A209615. - Jianing Song, Apr 24 2021
Complement of A343500 with respect to the even numbers. - Jianing Song, Apr 26 2021
LINKS
Jianing Song, Table of n, a(n) for n = 1..16384 (all terms <= 2^16).
FORMULA
a(n) = 2*A338691(n). - Hugo Pfoertner, Apr 26 2021
EXAMPLE
MATHEMATICA
okQ[n_] := If[OddQ[n], False, Module[{e = IntegerExponent[n, 2], k}, k = (n/2^e - 1)/2; EvenQ[k + e]]];
Select[Range[1000], okQ] (* Jean-François Alcover, Apr 19 2021, after PARI *)
PROG
(PARI) isA343501(n) = if(n%2, 0, my(e=valuation(n, 2), k=bittest(n, e+1)); !((k+e)%2))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Jianing Song, Apr 17 2021
STATUS
approved