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  • Let $a(n)$ be an integer sequence with binary values (that is, $a(n) \in \{0,1\}$) such that there at least one $0$ after each $1$.
  • Let $b(n)$ be an integer sequence of numbers $k$ such that $a(k)=1$. In other words, $a(n)$ is characteristic function of $b(n)$.
  • Let $c(n)$ be an integer sequence with generating function $C(x)$ such that $$ C(x) = \sum\limits_{k=1}^{\infty} \frac{x^{b(k)}}{(1-x)^k}. $$
  • Start with vector $\nu$ of fixed length $m$ with elements $\nu_i=1$ (that is, $\nu = \{1,1,\dotsc,1\}$) and for $i$ from $1$ to $m-1$, for $j$ from $i+1$ to $m$ consecutively apply $$ \nu_j := \nu_j + \nu_{j-1}. $$ if $a(j-i)=1$.

I conjecture that after the whole tranform we have $$ \nu_{n+1} - \nu_{n} = c(n). $$

Here is the PARI/GP program to check it numerically:

a(n) = issquare(n)
upto1(n) = my(A, v1); A = sum(k=1, n, x^(k^2)/(1-x)^k) + x*O(x^n); v1 = vector(n, i, polcoeff(A,i,x))
upto2(n) = my(v1); v1 = vector(n+1, i, 1); v2 = vector(n, i, a(i)); for(i=1, n, for(j=i+1, n+1, if(v2[j-i], v1[j] += v1[j-1]))); v1 = vector(n, i, v1[i+1] - v1[i])
test(n) = upto1(n) == upto2(n)

Is there a way to prove it?

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    $\begingroup$ Not sure if I understand the definition of b(n): if k=b(n) then a(k)=0, but also conversely?, if a(k)=0, there is n s.t. k=b(n) ? $\endgroup$ Commented Mar 4 at 21:45
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    $\begingroup$ @PietroMajer, thank you for comment! For example, if $a(1) = 1, a(2) = 0, a(3) = 1, a(4) = 0, a(5) = 0, a(6) = 1$, then we have $b(1) = 1, b(2) = 3, b(3) = 6$. $\endgroup$ Commented Mar 5 at 4:51

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