Let $T(n,k)$ be A035506, i.e., Stolarsky array read by antidiagonals. Here we consider that $T(n,k)=0$ for $n<1, k<1$.
Let $a(n)$ be A200714, i.e., Stolarsky representation interpreted as binary to decimal integers.
Let $\operatorname{tr}(n)$ be A007814, i.e., the number of trailing zeros in the binary representation of $n$.
Let $b(n)$ be the sequence of positive integers such that $$b(n)=T(b(\left\lfloor\frac{n}{2^{\operatorname{tr}(n)+1}}\right\rfloor+1),\operatorname{tr}(n)+1), b(1)=1$$
I conjecture that $b(n)$ is a permutation of natural numbers. I also conjecture that $$a(b(n))=n-1$$
To verify this conjecture one may use this PARI prog:
{Stolarsky(r, c)= tau=(1+sqrt(5))/2; a=floor(r*(1+tau)-tau/2); b=round(a*tau); if(c==1, a, if(c==2, b, for(i=1, c-2, d=a+b; a=b; b=d; ); d))} \\ from A035506
a(n) = {if (n == 1, return (0)); tau = (1 + sqrt(5))/2; mn = 0; while ((m = round(mn*tau)) < n, mn++; ); if (m == n, return (2*a(mn)+1)); mn = 0; while ((m = floor(mn*(1+tau)-tau/2)) < n, mn++; ); if (m == n, return (2*a(mn))); error("neither A nor B !!"); } \\ from A200714
b(n)=if(n==1, 1, Stolarsky(b(n\2^(valuation(n, 2) + 1) + 1), valuation(n, 2) + 1))
test(n)=a(b(n))==(n - 1)
Is there a way to prove it?
