OFFSET
0,2
COMMENTS
Local minima of A119435.
Let U(n,j) be the j-th smallest missing number in A119435(1..n-1). Example: for A119435(1..11), U(12,j) begins {6, 8, 10, 11, 14, ...}. Therefore we may alternatively define A119435(n) = U(n, A030101(n)).
Theorem: A119435(2^k) represents a local minimum. Proof: Observe that A030101(2^k) = 1. 2^k expressed in binary is 1 followed by zeros. When we reverse this number, the leading zeros are trivial and we read the number 1 in the 2^0 place. Therefore we select U(2^k, 1), which is the smallest missing number in A119435(1..n-1). Hence, a(n) = A119435(2^n).
Also positions of 2^n in A119436.
LINKS
Rémy Sigrist, PARI program
MATHEMATICA
a = {1}; nn = 2^14; Do[AppendTo[a, Complement[Range[i + 2 nn], a][[IntegerReverse[i, 2]] ]], {i, 2, nn}]; Array[a[[2^#]] &, Floor@ Log2@ Length@ a - 1, 0]]
PROG
(PARI) See Links section.
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Michael De Vlieger, Apr 18 2022
EXTENSIONS
More terms from Rémy Sigrist, Apr 19 2022
STATUS
approved