\\ Kevin Ryde, November 2021.
\\
\\ Usage: gp <a349238.gp
\\
\\ This is some straightforward code for calculating A349238 Zeckendorf
\\ base digit reversal, and the related A349239 reverse add, and A349240
\\ reverse and subtract.

default(recover,0);
default(strictargs,1);


A349238_Zeckendorf_reverse(n) =
{
  if(n==0,return(0));
  my(k=logint(n,2)*3>>1 + 3,  \\ over-estimate of log_phi(n)
     f=fibonacci(k),
     g=fibonacci(k-1));

  \\ reduce f,g to where f is the largest Zeckendorf term in n
  while(f>n,
        [f,g]=[g,f-g]);  \\ Fibonacci step down

  my(rev_f=1, rev_g=1,   \\ pair of Fibonacci, working upwards
     ret=0);
  while(n,
        if(n>=f, n-=f; ret+=rev_f);   \\ Zeckendorf 1 digit
        [f,g] = [g,f-g];              \\ step down
        [rev_f, rev_g] = [rev_f+rev_g, rev_f]);  \\ step up
  ret;
}

A349239_Zeckendorf_reverse_add(n) = n + A349238_Zeckendorf_reverse(n);
A349240_Zeckendorf_reverse_sub(n) = n - A349238_Zeckendorf_reverse(n);

{
  print("A349238_Zeckendorf_reverse(n) for n>=0");
  print1("  = ");
  for(n=0,21,
     print1(A349238_Zeckendorf_reverse(n),","));
  print("...");
}
{
  print("A349239_Zeckendorf_reverse_add(n) for n>=0");
  print1("  = ");
  for(n=0,17,
     print1(A349239_Zeckendorf_reverse_add(n),","));
  print("...");
}
{
  print("A349240_Zeckendorf_reverse_sub(n) for n>=0");
  print1("  = ");
  for(n=0,20,
     print1(A349240_Zeckendorf_reverse_sub(n),","));
  print("...");
}


\\ In A349238_Zeckendorf_reverse(), the Zeckendorf reprsentation is found by
\\ a logarithm for an over-estimate of the greatest Fibonacci number
\\ required, then successive comparing and subtract for the terms (A035516)
\\ in the required "greedy" way.
\\
\\ Each Fibonacci term at the most significant end reverses to a Fibonacci
\\ term at the least significant end.
\\
\\ The initial f,g could be calculated together by polmod powering, but that
\\ measures no speedup.  (Perhaps in the future fibonacc() will have an
\\ option to get a pair of values.)
\\
\\ A349239_Zeckendorf_reverse_add() would be as easy as starting "ret = n"
\\ within the reversal, but doing a separate addition is enough.  Similarly
\\ A349240_Zeckendorf_reverse_sub() could be ret = n and then subtract the
\\ low Fibonacci terms but again a separate subtract is enough.

print("end");