Conjectures 111-113 from "100 Conjectures from the OEIS". Second part.

[Antti Karttunen]

And for the latter part of the conjecture:
"a(n) = 0 <=> n=3k with k not in A036556."

The ==> direction holds, as a(n) is 0 only for multiples of 3,
but for the odious multiples of 3 the A065359 must give a non-zero integer.

For <== direction, n must be 3k for some k, that a(n) were zero.
Is there k, such that 3k is one of the evil numbers, A001969
for which A065359(3k) is not zero?
At least 2839893 (1010110101010101010101 in binary) =  946631 * 3
is, thus giving a counter-example for <== direction of the conjecture.
(A065359(2839893) gives 9-3 = 6, while there are 9+3 = 12 1-bits).

Rest is absurd monologue where I wrote, and wrote and contradicted myself.
Correction was wrong, the corrected version was the correct. But the 
beans were good!

>> %S A036556 
>> 7,14,23,27,28,29,31,39,46,54,56,57,58,62,71,78,87,91,92,93,95,103,107,
>> %T A036556 
>> 108,109,111,112,113,114,115,116,117,119,123,124,125,127,135,142,151,
>> %U A036556 155,156,157,159,167,174,182,184,185,186,190,199,206,214
>> %N A036556 Multiples of 3 with an odd number of one bits in base 2.
>>
>> Shouldn't it be: "Integers, which when multiplied by 3 yield an 
>> odious number (A000069)".
>> (Or "intersection of A000069 and A008585 (multiples of 3), divided by 
>> 3.") ?
>>
>> And the first half of Ralph's conjecture in 
>> http://www.research.att.com/~njas/sequences/A065359
>>
>>  a(n) = 3 or -3 iff n in 3*A036556; a(n) = 0 iff n=3k with k not in 
>> A036556.
>>
>> then would mean that A065359(n) = 3 or -3 <==> n is an odious number 
>> which is multiple of 3.
>>
>> I think the first counter-example to <== direction of the conjecture 
>> is 87381 = (3*29127) = 10101010101010101 in binary,
>> whose alternating sum is 9, although it is an odious multiple of 3.
>
>
> So far so good, but the part below went wrong.
>
>> However, ==> direction holds always, because first, of the 
>> "11"-divisibility algorithm
>> (three is the "eleven" of the binary system), the alternating sum
>> of 3's multiples is always a multiple of 3,  <--- THIS IS NOT TRUE IN 
>> BINARY!
>
>
> (Instead, I recall that it is always eiher -1, 0 or +1. Have to check.)

How could it be only -1, 0 or +1, when we just saw cases where it is +3, 
and +9?
(No wonder that I almost floundered my courses of the elementary logic.)
But, 3 is also "nine" of the base-4. So taking successive blocks of bits 
i and i+1
and summing those bit-pairs should yield a three's multiple if the 
number is three's multiple.
And from that we should deduce that my original remebrance was correct,
that the alternating sum (in binary system) of 3's multiples is always a 
multiple of 3.

>
>> and secondly, if the alternating sum is 3 or -3,
>> then the number of 1's in odd (or even) positions
>> is three more than the number of 1's in even (or respectively: odd)
>> positions. So the total number of 1's is 2*k + 3, that is, odd.
>> QED.
>>
> And I leave the latter part of the conjecture: "a(n) = 0 <=> n=3k with 
> k not in A036556."
> open, until I have maintained my brain's glucose levels, or somebody 
> gets there before me.
>
> -- Same.
>
>>
>> Yours,
>>
>> Antti Karttunen
>>
>>
>>
>>>
>>> Conjecture 113 seems to be given in a wrong direction in your paper.
>>> You ask to prove that if a(3k)=0 then k belongs to A006288. But it is
>>> opposite to proving that a(3*A006288) = 0.
>>>
>>> Max
>>>
>>> On 1/4/07, Ralf Stephan <ralf at ark.in-berlin.de> wrote:
>>>
>>>> Max, short answer first.
>>>> > In Conjecture 111:
>>>> > Let n=21 (=10101 in binary).
>>>> > Then a_{21}=3 but 21 does not belong to the set { m | m=3k & k=3i &
>>>> > e_1(k)=1 mod 2 } (simply because all elements of the set are 
>>>> multiples
>>>> > of 9 while 21 is not).
>>>> > Is n=21 a counterexample to Conjecture 111?
>>>> >
>>>> > In Conjecture 112:
>>>> > Let n=63 (=111111 in binary).
>>>> > Then a_{63}=0 and m=n/3=21. But 21 belongs to the set { k | k=3i &
>>>> > e_1(k)=1 mod 2 }.
>>>> > Is n=63 a counterexample to Conjecture 112?
>>>>
>>>> These two refer to the following OEIS entry
>>>>
>>>> %N A065359 Alternating bit sum for n: replace 2^k with (-1)^k in 
>>>> binary expansion of n.
>>>> %C A065359 Conjectures: a(n) = 3 or -3 iff n in 3*A036556; a(n) = 0 
>>>> iff n=3k with k not in A036556. - Ralf Stephan 
>>>> (ralf(AT)ark.in-berlin.de), Mar 07 2003
>>>>
>>>>
>>>> > In Conjecture 113:
>>>> > Let k=18. Then a_{3k}=a_{54}=0:
>>>> > a_{54} = 1-a_{27} = 1+a_{13} = 1-a_6 = a_3 = -a_1 = a_0 = 0.
>>>> > But in the base-4 the last digit of 18 must be different from 
>>>> -1,0,1.
>>>> > Is k=18 a counterexample to Conjecture 113?
>>>>
>>>> This refers to:
>>>>
>>>> %C A083905 Conjecture: a(3*A006288) = 0.
>>>>
>>>>
>>>> ralf
>>>>
>>>>
>>>
>>
>>
>
>