[seqfan] Re: A conjecture for sums of binomials
Andrew N W Hone
A.N.W.Hone at kent.ac.uk
Mon Jun 19 23:42:31 CEST 2017
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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