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A319952
Let M = A022342(n) be the n-th number whose Zeckendorf representation is even; then a(n) = A129761(M).
1
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
2,2
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).
FORMULA
If the Zeckendorf representation of M ends with exactly k zeros, ...10^k, then a(n) = ceiling(2^k/3).
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;
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved