[seqfan] Re: An astonishing regularity with the participation of Thue-Morse sequence

[Vladimir Shevelev]

I submitted in A268866 a much more
precise formulation of my conjecture
("well posed problem is half-solved").
Conjecture 1) Let m(n) be the position
in A268865 corresponding to a(n). 
Then m(n)=(2/3)*(4^n-1), if n is even,
and m(n)=(2/3)*(4^(n-1)-1)+3*4^(n-1),
if n is odd. 
Conjecture 2) a(n)=2*m(n)+2, if n is even,
and a(n)=(7*m+12)/11, if n is odd. 

If Conjectures 1,2 are true, then we easily
have for even n>=0, m(n) == 0 (mod 10),
a(n)== 2 (mod 10), while a(1)= m(1)=3 and
for odd n>=3, m(n), a(n) == 8 (mod 10). 

Best regards,
Vladimir


________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Vladimir Shevelev [shevelev at exchange.bgu.ac.il]
Sent: 15 February 2016 15:38
To: seqfan at list.seqfan.eu
Subject: [seqfan] An astonishing regularity with the participation of Thue-Morse sequence

Dear SeqFans,

I have just submitted two new sequences A268865
and A268866.
Let b(n) be the parity of number of 1's in the
n-th term of A039724(n) {0,1,0,1,1,0,1,0,0,...}.
Let A_n be Thue-Morse sequence (A010060)
beginning with A010060(n). Then for A268865
a(n) is defined as the maximal number of the first
terms of {b(n)} coinciding with the corresponding
terms of A_n. A268865 begins (n>=0):
2, 0, 0, 3, 0, 1, 2, 0, 0, 1, 22
I supposed that a(10) =22 is somewhat "resonance".
The sequence A268866 lists the records of A268865.
And here, after Peter's calculations, I was surprised
by a nice regularity. Here a(n)=A268866.
Conjecture 1) To every a(n) corresponds a position
m=m(n) in A268865 such that either
a) m == 0 (mod 10) or b) m=3 or m == 8 (mod 10)
such that we have {m(n)}={0,3,10,58,170,938,2730,
15018,43690,...};
2) In case a) a(n)=2*m+2; in case b) a(n)=(7*m+12)/11.
It is a rare event when for records of a non-simple sequence
there appear so remarkable regularities!

All comments are welcome.

Best regards,
Vladimir

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