1
$\begingroup$

Let $a(n)$ be A290254, the viabin numbers of the self-conjugate integer partitions, also defined as $\left\lbrace 0 \right\rbrace$ union fixed points of A059894, self-inverse permutation defined as follows:

Complement and reverse the order of all but the most significant bit in binary expansion of $n$. $n = 1ab\cdots yz$ -> $1ZY\cdots BA = A059894(n)$, where $A = 1-a, B = 1-b, \cdots$

I conjecture that $$a(n) = \begin{cases} n-1,&\text{if $n < 3$}\\ a(n-1) + 2\cdot 4^{f(n-1) - 1} + 3\cdot 2^{f(n-1) - 1} - 1,&\text{if $n = 2^k + 1, k > 0$}\\ a(n-1) + (2^{g(n-1) + 2} - 3)\cdot 2^{f(h(n-2))},&\text{otherwise} \end{cases}$$ where

  • $f(n)$ is A000523, $f(n) = \left\lfloor\log_2(n)\right\rfloor$
  • $g(n)$ is A007814, exponent of highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.
  • $h(n)$ is A025480, $h(2n) = n, h(2n+1) = h(n)$

Is there a way to prove it?

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ This should follow straightforwardly from the characterisation that $a(2^k + m)$ with $1 \le m \le 2^k$ is given by $4^k + 2^k (m-1) + \operatorname{revinv}_k(m-1)$ where $\operatorname{revinv}_k$ reverses and inverts the bits of a $k$-bit number. $\endgroup$
    – Peter Taylor
    Commented Sep 24, 2021 at 12:08

0

Reset to default

You must log in to answer this question.