A319952 - OEIS

%I #15 Oct 03 2018 15:27:34

%S 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,

%T 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,

%U 1,2,3,1,22,1,2,3,1,6,1

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

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

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

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

%p with(combinat): F:=fibonacci:

%p A130234 := proc(n)

%p     local i;

%p     for i from 0 do

%p         if F(i) >= n then

%p             return i;

%p         end if;

%p     end do:

%p end proc:

%p A014417 := proc(n)

%p     local nshi, Z, i ;

%p     if n <= 1 then

%p         return n;

%p     end if;

%p     nshi := n ;

%p     Z := [] ;

%p     for i from A130234(n) to 2 by -1 do

%p         if nshi >= F(i) and nshi > 0 then

%p             Z := [1, op(Z)] ;

%p             nshi := nshi-F(i) ;

%p         else

%p             Z := [0, op(Z)] ;

%p         end if;

%p     end do:

%p     add( op(i, Z)*10^(i-1), i=1..nops(Z)) ;

%p end proc:

%p A072649:= proc(n) local j; global F; for j from ilog[(1+sqrt(5))/2](n)

%p        while F(j+1)<=n do od; (j-1); end proc:

%p 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:

%p A129761 := n -> A003714(n+1)-A003714(n):

%p a:=[];

%p for n from 1 to 120 do

%p    if (A014417(n) mod 2) = 0 then a:=[op(a), A129761(n-1)]; fi;

%p od;

%p a;

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

%K nonn

%O 2,2

%A _Jeffrey Shallit_ and _N. J. A. Sloane_, Oct 03 2018