[seqfan] Re: Is A290646 = A135517?
Vladimir Shevelev
shevelev at bgu.ac.il
Sat Aug 26 11:29:22 CEST 2017
Dear Seqfans,
The last arguments could be essentially
simplified. Recall, that I proved that
|e_k(n)|=binomial(n,k)*A002425((k+1)/2)/A006519(k+1).
Denote 2-adic order of n by r(n). So, denominator of
coefficients e*_k(n) in E_n(x)+E_n(0) is
2^(r(k+1)-r(binomial(n,k))-delta(n,k))=
2^(r((k+1)/binomial(n,k))-delta(n,k))=
2^(r((n+1)/binomial(n+1,k+1))-delta(n,k))=
2^(r(n+1)-r(binomial(n+1,k+1))-delta(n,k)).
So, for a(n) in A290646 we have
a(n)=2^(r(n+1)-min{odd k=1,...,n}(t(n+1,k+1)+delta(n,k))).
But min is 1, if n=2^m-1 and reaches for k=n, and is 0,
otherwise, and reaches for k=2^(v(n)-1)-1, where v(n) is
the number of (0,1)-digits in the binary expansion of n.
So, since r(n+1)=A007814(n+1), the required result
easily follows.
Best regards,
Vladimir
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Vladimir Shevelev [shevelev at exchange.bgu.ac.il]
Sent: 16 August 2017 16:23
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Is A290646 = A135517?
Dear Peter and SeqFans,
(I resend this message in connection with Neil's
letter "Bug in "email this user" program since Monday")
To complete the proof let us simplify my formula for a(n)
in A290646 in the previous message, using Kummer's
formula for A006519(binomial(n,k)).
Denote by t(n,k) the number of
carries which appear in addition
k and n-k in base 2. Then by Kummer's
result, A006519(binomial(n,k))=2^t(n,k).
So, in A290646 also
a(n) = 2^max(odd k=1..n)(A007814(k+1)-t(n,k) -delta(n,k)), (1)
where delta (n,k) is the Kronecker symbol: delta(i,j) is 1 if i=j,
otherwise it is 0.
Further, by my comment in A091090(n),
let binary expansion of n >= 1 end with m >= 0 1's. Then
A091090(n) = m if n = 2^m-1 and A091090(n) = m+1 if n > 2^m-1.
Let n be odd. In case n=2^m-1, we have t(n,k)=0 and in (1) we
obtain maximum, setting k=n, since A007814(n+1)=m takes the maximal
values, while t(n,k)=0. So since delta(n,n)=1, we have 2^(m-1)=
2^(A091090(n)-1)=A135517(n).
Let now n=1..10..01..1, where we have (from left to the right)
m_1 1's, r o's and m_2 1's. Set k=1..1 <n with m_1+r+m_2-1 1's
with the maximal possible A007814(k+1)=m_1+r+m_2 and the number
of carries in the subtracting k from n is r+m_1. Note that the same
number of carries we have for every number of the form k=1..1x..x1,
where we have m_1+r-1 1's, m_2-1 x's and the last 1. So our choice
gives the maximum for exponent in formula (1). So a(n) = 2^(m_1+r+m_2-
(r+m_1)) =2^m_2=2^(A091090(n)-1)=A135517(n).
Note that, if instead of k=1..1 <n with m_1+r+m_2-1 1's,
to take, say, k*=1..1 with m_1+r+m_2-s>=m_2+1 1's, A007814(k+1)
decreases on s-1 and t(n-k) decreases on exactly s-1, so we can have
several values of k for which the exponent in (1) reaches the maximum.
The odd numbers with more blocks of 1's and 0's are considered in the same
way.
Finally, if n is even, (1) remains true, and by the above arguments, we
see that a(n)=2^0 = 2^(A091090(n)-1) = A135517(n).
Best regards,
Vladimir
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Vladimir Shevelev [shevelev at exchange.bgu.ac.il]
Sent: 13 August 2017 18:43
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Is A290646 = A135517?
Dear Peter and SeqFans,
I proved that for odd n>=1
E(n,x) = x^n + Sum_{odd k=1..n} e_k(n)* x^(n-k),
where |e_k(n)| =
binomial(n,k)*A002425((k+1)/2)/A006519(k+1).
>From this in A290646 it easily follows that for odd n
a(n) = 2^max{odd k=1..n}(A007814(k+1) + A000120(n)-
A000120(n-k)-A000120(k)-delta(k,n)).
I think that anyone can prove that it is A135517(n)
(for odd n).
Best regards,
Vladimir
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Peter Luschny [peter.luschny at gmail.com]
Sent: 08 August 2017 19:18
To: seqfan at list.seqfan.eu
Subject: [seqfan] Is A290646 = A135517?
Thankful for any comment.
Peter
http://oeis.org/search?q=id:A290646|id:A135517
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