[seqfan] Re: A conjecture for sums of binomials

Andrew N W Hone A.N.W.Hone at kent.ac.uk
Thu Jun 22 11:04:39 CEST 2017


Dear Vladimir, 

I'm afraid I misread your inequality. You had cos(2Pi/N), and I intended to write cos(Pi/N)!
 
In order for this to be an asymptotic expansion, in Landau notation we have 

R(n)  = o ((2*cos(Pi/N))^n). 

In what I wrote, the correct ratio (tending to zero as n->infinity) should have been  

R(n)  / (2*cos(Pi/N))^n = 2/N ( Sum_{ j=2..floor((N-1)/2) } (cos(Pi*j/N) / cos(Pi/N))^n*cos(Pi*j*n/N) ). 

The ratio you want is 

R(n)  / (2*cos(2Pi/N))^n = 2/N * cos(2*Pi*n/N) + 2/N ( Sum_{ j=3..floor((N-1)/2) } (cos(Pi*j/N) / cos(2Pi/N))^n*cos(Pi*j*n/N) ). 

So the leading term in the asymptotics of R(n)  / (2*cos(2Pi/N))^n is  2/N * cos(2*Pi*n/N), which indeed does not tend to zero as n->infinity, as you say, but all the correction terms do. 

Best wishes,
Andy 
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Vladimir Shevelev [shevelev at bgu.ac.il]
Sent: 21 June 2017 15:13
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: A conjecture for sums of binomials

Dear Andy,

Here you are not right.

Indeed, after the two main terms, the remainder
term is R_N(n)=
2/N((2cos(2*Pi/N))^n*cos(2*Pi*n/N)) +
(2cos(3*Pi/N))^n*cos(3*Pi*n/N))+...)
with a finite sum (j<=floor((N-1)/2).
Then abs(R_N(n))<=2/N((2cos(2*Pi/N))^n+
(2cos(3*Pi/N))^n+...) and since for N>=6,
cos(2*Pi/N)>cos(3*Pi/N)>..., then
abs(R_N(n))<=2/N*(2cos(2*Pi/N))^n)*((N-1)/2-1)
<(2cos(2*Pi/N))^n that I wrote.
But if R_N(n)/(2cos(2*Pi/N)^n-> 0 as n -> infinity,
then cos(2*Pi*n/N)-> 0 as n -> infinity, which
is impossible.

Best regards,
Vladimir
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Andrew N W Hone [A.N.W.Hone at kent.ac.uk]
Sent: 21 June 2017 12:04
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: A conjecture for sums of binomials

Dear Vladimir,

Since this is an asymptotic formula in n, a stronger statement holds:

R(n)  / (2*cos(2Pi/N))^n -> 0 as n -> infinity.

All the best,
Andy

________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Vladimir Shevelev [shevelev at bgu.ac.il]
Sent: 20 June 2017 10:46
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: A conjecture for sums of binomials

Dear Andy,

Thank you for the proof of my conjecture!
Thus for N>=6 we have
 S_N(n)=2/N*(2^(n-1)+
(2cos(Pi/N))^n*cos(Pi*n/N))+R(n)
with R(n) <(2*cos(2Pi/N))^n.

Best,
Vladimir
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Andrew N W Hone [A.N.W.Hone at kent.ac.uk]
Sent: 20 June 2017 00:42
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: A conjecture for sums of binomials

Dear Seqfans,

By combining suitable Nth roots of unity, the exact formula is

S_N(n)=2/N*( 2^(n-1) + Sum_{ j=1..floor((N-1)/2) } (2*cos(Pi*j/N))^n*cos(Pi*j*n/N) ),

which confirms the two leading order terms in the asymptotics conjectured below, by just taking the j=1 term in the sum.

Best wishes,
Andy
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Vladimir Shevelev [shevelev at bgu.ac.il]
Sent: 19 June 2017 14:43
To: seqfan at list.seqfan.eu
Subject: [seqfan] A conjecture for sums of binomials

Dear SeqFans,

Let S_N(n)=Sum_{t>=0}C(n,N*t), n>=0,
where C(a,b) denotes Binomial(a,b).
Using formula S_N(n)=1/N*Sum_{j=1..N}
(omega_N^j+1)^n, where omega_N=
exp(2*Pi*i)/N),  it is easy to show that
S_3(n)=(2/3)*(2^(n-1)+cos(Pi*n/3)),
S_4(n)=(2/4)*(2^(n-1)+sqrt(2)^n*cos(Pi*n/4))
and recently (not so simply) I obtain also that
S_5(n)=round((2/5)*(2^(n-1)+phi^n*cos(Pi*n/5),
where phi is the golden ratio. Since phi=2cos(Pi/5),
the general term of the sequence 1,sqrt(2),phi,...
should be 2cos(Pi/N). I conjecture that the two
main terms in asymptotic of S_N(n) is (2/N)*
(2^(n-1)+(2cos(Pi/N))^n*cos(Pi*n/N)).
If it is true, then I proved that S_{N,r)(n)=
Sum_{t>=0}C(n, N*t+r), r=1,...,N-1, has the
two main terms in asymptotic (2/N)*(2^(n-1)
+(2cos(Pi/N))^n*cos(Pi(n-2r)/N)). (For example,
cf. formulas for a(n) in A139398, A133476,A139714).

Best regards,
Vladimir

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