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A217586
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a(1) = 1 and, for all n >= 1, if a(n) = 0 then a(2*n) = 1 and a(2*n+1) = 0 whereas if a(n) = 1 then a(2*n) = 0 and a(2*n+1) = 0.
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1
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1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0
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OFFSET
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1
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COMMENTS
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The indices where 1 occurs in {a(n)} are {1, 4, 6, 10, 14, 16, ...} which appear to correspond precisely to the terms of A171945. (That is, if true, then this would be a characteristic function of A171945.) If this were proved then A171945 could be easily extended.
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LINKS
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FORMULA
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Here is a proof of Velin Yanev's conjecture.
There is an offset problem. We will give the period-doubling sequence offset 1, and will also consider the characteristic function of powers of 2, A209229, instead of the Fredholm-Rueppel sequence A036987.
Let alpha be the period doubling morphism
alpha: 0 -> 01, 1 -> 00.
Let PD = 01000101... be the fixed point of alpha, with offset 1, so PD(n+1) = A096268(n).
Let beta be the time reversal of alpha
beta: 0 -> 10, 1 -> 00.
By definition, (a(n)) satisfies (a(n)) = 1 beta((a(n))), implying that
(*) (a(n)) = 1 beta(1) beta^2(1) beta^3(1) ....
Note that Yanev's conjecture can be reformulated as
/ PD(n) if n is not a power of 2,
a(n) = -|
\ 1-PD(n) if n is a power of 2.
It is easy to see that for all n
beta^n(0) = beta^n(1), EXCEPT for the first letter:
beta^n(0) = delta(n)R, beta^n(1) = (1-delta(n))R, where
delta(2n) = 0, delta(2n+1) = 1.
We will use below that, similarly, alpha^n(0) = alpha^n(1), EXCEPT for the LAST letter.
From (*) we see now that the conjecture will be proved if one shows that
(**) alpha^n(0) = 0 beta(0)beta^2(0)...beta^{n-1}(0) delta(n),
since the first letters of the beta^(k) in this concatenation occur exactly at positions 2^k.
The formula (**) has a companion
(***) beta^n(0) = delta(n) 0 beta(0)beta^2(0)...beta^{n-1}(0).
We will prove both formula's together with induction on n.
For n=2 (**) and (***) one has to check
alpha^2(0)=0beta(0)delta(2)=0100, beta^2(0)=delta(2)0beta(0)=0010,
which are true. Suppose (**), (***) hold for some n>1. Then
alpha^{n+1}(0) = alpha^n(0) alpha^n(1) =
0 beta(0)beta^2(0)...beta^{n-1}(0) delta(n)0
beta(0)beta^2(0)...beta^{n-1}(0) (1-delta(n)) =
0 beta(0)beta^2(0)...beta^n(0) (1-delta(n)) =
0 beta(0)beta^2(0)...beta^n(0)delta(n+1).
The formula for (***) with n+1 follows in a similar way.
(End)
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MATHEMATICA
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t = {1}; Do[If[EvenQ[n], a = 1 - t[[n/2]], a = 0]; AppendTo[t, a], {n, 2, 100}]; t (* T. D. Noe, Oct 10 2012 *)
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PROG
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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