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The Ramanujan-type sequence number 6 for the argument 2Pi/7 (see also A214683, A215112, A006053, A006054, A215076, A215100, A120757 for the numbers: 1, 1a, 2, 2a, 3, 4 and 5 respectively).

The sequence a(n) is one of the three special sequences (the remaining two are A215569 and A215572) connected with the following recurrence relation: T(n):=49^(1/3)*T(n-2)+T(n-3), with T(0)=3, T(1)=0, and T(2)=2*49^(1/3) - see the comments to A214683.

It can be proved that

T(n) = (c(1)^4/c(2))^(n/3) + (c(2)^4/c(4))^(n/3) + (c(4)^4/c(1))^(n/3), where c(j):=2*cos(2*Pi*j/7), and the following decomposition hold true:

T(n) = at(n) + bt(n)*7^(1/3) + ct(n)*49^(1/3), where sequences at(n), bt(n), and ct(n) satisfy the following system of recurrence equations: at(n)=7*bt(n-2)+at(n-3),

bt(n)=7*ct(n-2)+bt(n-3), ct(n)=at(n-2)+ct(n-3), with at(0)=3, at(1)=at(2)=bt(0)=bt(1)=bt(2)=ct(0)=ct(1)=0, ct(2)=2 - for details see the first Witula reference.

It follows that a(n)=at(3*n), bt(3*n)=ct(3*n)=0.

Every difference of the form a(n)-a(n-2)-a(n-3) is divisible by 3. Because the difference a(n+1)-a(n) is congruent to the difference a(n-4)-a(n-2) modulo 3 we easily deduce that a(6)-a(5) and a(7)-a(6)-2 are both divisible by 3.

The Ramanujan-type sequence number 8 for the argument 2Pi/7 (see also A214683, A215112, A006053, A006054, A215076, A215100, A120757, A215560, A215569 for the numbers: 1, 1a, 2, 2a, 3-7 respectively). The sequence a(n) is one of the three special sequences (the remaining two are A215560 and A215569) connected with the following recurrence relation:

(c(1)^4/c(2))^(n/3) + (c(2)^4/c(4))^(n/3) + (c(4)^4/c(1))^(n/3) = at(n) + bt(n)*7^(1/3) + ct(n)*49^(1/3), where c(j):=2*cos(2*Pi*j/7), and the sequences at(n), bt(n), and ct(n) are defined in comments to A215560 (see also A215569). It follows that a(n)=ct(3*n+2), at(3*n+2)=bt(3*n+2)=0, which implies the first formula below.

We note that if a(n), a(n+1) and a(n+2) are all odd for some n in N then a(n+3) is even, a(n+4) is odd, a(n+5) and a(n+6) are both even, and the numbers a(n+7), a(n+8), a(n+9) are all odd again. In consequence, this situation hold for every n of the form 7*k+4, k=0,1,..., in the other words cyclical through all sequence a(n), n=4,5,... (from n=1 whenever we start from odd-even-even sequence).

The Ramanujan type sequence number 9 for the argument 2*Pi/9 defined by the following relation a(n)*3^(-n + 1/3) = ((-1)^n)*(c(1) - 1/3)^(n + 2/3) + ((-1)^n)*(c(2) - 1/3)^(n + 2/3) + (1/3 - c(4))^(n + 2/3), where c(j) := 2*cos(2*Pi*j/9). We note that c(4) = -cos(Pi/9). The conjugate with a(n) are the sequences A217052 and A217069.

In Witula's et al.'s paper the following sequence is discussed: S(n) := sum{j=0,1,2} (1/3 - c(2^j))^(n/3). It is proved that S(n+3) = (3^(1/3))*S(n+1) + S(n)/3, n=0,1,... Moreover the following decomposition holds true S(n) = x(n) + y(n)*3^(1/3) + z(n)*9^(1/3), which implies the system of recurrence relations: x(n+3) = 3*z(n+1) + x(n)/3, y(n+3) = x(n+1) + y(n)/3, z(n+3) = y(n+1) + z(n)/3, x(0)=3, y(0)=z(0)=x(1)=y(1)=z(1)=x(2)=z(2)=0, y(2)=2. Then it can be generated the relations X'(n+9) - 3*X'(n+6) - 24*X'(n+3) - X'(n) = 0, where X'(n) = X(n)*3^n for every X=x,y,z and n=0,1,..., from which we obtain that x(3*n+1)=x(3*n+2)=y(3*n)=y(3*n+1)=z(3*n)=z(3*n+2)=0 and a(n) = y(3*n+2)*3^n, A217052(n) = x(3*n)*3^(n-1), and A217069(n) = z(3*n+1)*3^(n-1).

Each a(n)+1 is divisible by 3 but which a(n)+1 are divisible by 9 - it is a question?

The Ramanujan sequence number 11 for the argument 2*Pi/9 defined by the relation a(n)*9^(1/3) = (3^(n-1))*(((-1)^(n-1))*(c(1) - 1/3)^(n + 1/3) + ((-1)^(n-1))*(c(2) - 1/3)^(n + 1/3) + (1/3 - c(4))^(n + 1/3)), where c(j) := 2*cos(2*Pi*j/9). The sequences A217052 and A217053 are conjugate with the sequence a(n). For more information on these connections - see Comments in A217053.

The 3-valuation of the sequence a(n) is equal to (0,2,1).

The Ramanujan-type sequence the number 9 for the argument 2*Pi/7. The sequence is connecting with the following decomposition: (s(4)/s(1))^(1/3)*s(1)^n + (s(1)/s(2))^(1/3)*s(2)^n + (s(2)/s(4))^(1/3)*s(4)^n = x(n)*(4-3*7^(1/3))^(1/3) + y(n)*(11-3*49^(1/3))^(1/3), where s(j) := sin(2*Pi*j/7), x(0)=1, x(1)=-7^(1/6)/2, x(2)=y(0)=y(1)=0, y(2)=7^(1/3)/4 and X(n)=sqrt(7)*(X(n-1)-X(n-3)) for every n=3,4,..., and X=x or X=y. It could be deduced the formula 4*y(n) = a(n)*7^(1/3 + (3+(-1)^n)/4), which implies a(0)=0, a(1)= 0, a(2)= 1/7, a(3)=1/7, a(4)=1, a(5)=6/7, i.e., A163260(n)=7*a(n) for every n=0,1,...,5. The sequence a(n) is discussed in third Witula paper.