A319952 Let M = A022342(n) be the n-th number whose Zeckendorf representation is even; then a(n) = A129761(M).

1, 2, 3, 1, 6, 1, 2, 11, 1, 2, 3, 1, 22, 1, 2, 3, 1, 6, 1, 2, 43, 1, 2, 3, 1, 6, 1, 2, 11, 1, 2, 3, 1, 86, 1, 2, 3, 1, 6, 1, 2, 11, 1, 2, 3, 1, 22, 1, 2, 3, 1, 6, 1, 2, 171, 1, 2, 3, 1, 6, 1, 2, 11, 1, 2, 3, 1, 22, 1, 2, 3, 1, 6, 1

Offset: 2

Author

Jeffrey Shallit and N. J. A. Sloane, Oct 03 2018

Keywords

• nonn

Comments

The Zeckendorf representations of numbers are given in A014417. The even ones are specified by A022342.

The offset here is 2 (because A129761 should really have had offset 1 not 0).

Crossrefs

Cf. A003714, A014417, A022342, A072649, A129761, A130234.

Programs

Maple

with(combinat): F:=fibonacci:
A130234 := proc(n)
    local i;
    for i from 0 do
        if F(i) >= n then
            return i;
        end if;
    end do:
end proc:
A014417 := proc(n)
    local nshi, Z, i ;
    if n <= 1 then
        return n;
    end if;
    nshi := n ;
    Z := [] ;
    for i from A130234(n) to 2 by -1 do
        if nshi >= F(i) and nshi > 0 then
            Z := [1, op(Z)] ;
            nshi := nshi-F(i) ;
        else
            Z := [0, op(Z)] ;
        end if;
    end do:
    add( op(i, Z)*10^(i-1), i=1..nops(Z)) ;
end proc:
A072649:= proc(n) local j; global F; for j from ilog[(1+sqrt(5))/2](n)
       while F(j+1)<=n do od; (j-1); end proc:
A003714 := proc(n) global F; option remember; if(n < 3) then RETURN(n); else RETURN((2^(A072649(n)-1))+A003714(n-F(1+A072649(n)))); fi; end proc:
A129761 := n -> A003714(n+1)-A003714(n):
a:=[];
for n from 1 to 120 do
   if (A014417(n) mod 2) = 0 then a:=[op(a), A129761(n-1)]; fi;
od;
a;

Formula

If the Zeckendorf representation of M ends with exactly k zeros, ...10^k, then a(n) = ceiling(2^k/3).