Conjectures 111-113 from "100 Conjectures from the OEIS". Correction.

[Antti Karttunen]

Antti Karttunen wrote:

>
> %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.)

> 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
>>>
>>>
>>
>
>