cp's OEIS Frontend

This sequence is a permutation of the positive integers.

[Proof from N. J. A. Sloane, Apr 20 2022: a(n) always exists, so the sequence is infinite. Every time n is a power of 2, n-reversed is 1, and a(n) is the smallest missing number. Since there are infinitely many powers of 2, every number will eventually appear.]

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.