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Let z(n,m) be the number of zeros in a period of T(n,k) modulo m, i.e., z(n,m) = pi(n,m)/E(n,m), then we have: for odd primes p that are not divisible by n^2 - 4, z(n,p) = 2 if ((n+2)/p) = -1; 1 or 2 if ((n+2)/p) = 1 (conjecturally 2/3 of the primes of this form are with z(n,p) = 1 and 1/3 are with z(n,p) = 2. The result is that among all odd primes that are not divisible by n^2 - 4, 2/3 of them are with z(n,p) = 1 and 1/3 are with z(n,p) = 2 if n + 2 is a square, and is the opposite otherwise). z(n,p) = 1 for odd primes p that are divisible by n - 2 and 2 for odd primes p that are divisible by n + 2. z(n,2) = 1. For all odd primes p, if E(n,p) is even then z(n,p) = 2 (the converse is not necessarily true)..

For all odd primes p, z(n,p) = 2 if and only if pi(n,p) is even, z(n,p) = 1 if and only if pi(n,p) is odd. For all odd primes p, if E(n,p) is even then z(n,p) = 2 (the converse is not necessarily true). [Comment revised by Jianing Song, Jul 06 2019]

By induction it is easy to show that T(n,k) = T(n,m+1)*T(n,k-m) - T(n,m)*T(k-m-1). Let k = 2m we have T(n,2m) = T(n,m)*(T(n,m+1)-T(m-1)); let k = 2m+1 we have T(n,2m+1) = T(n,m+1)^2 - T(n,m)^2 = (T(n,m+1)+T(n,m))*(T(n,m+1)-T(n,m)). So T(n,k) is composite if n >= 3, k >= 3. - Jianing Song, Jul 06 2019

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For n >= 2, Sum_{i=1..k} 1/T(n,2^i) = 2/n - ((u^(2^k-1) + v^(2^k-1))/(u + v)) * (1/T(n,2^k)), where u = (n + sqrt(n^2 - 4))/2, v = (n - sqrt(n^2 - 4))/2 are the two roots of the polynomial x^2 - n*x + 1. As a result, Sum_{i==>1..inf} 1/T(n,2^i) = (n - sqrt(n^2 - 4))/2. - Jianing Song, Apr 21 2019

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Jianing Song, <a href="/A316269/b316269.txt">Antidiagonals n = 12..100101, flattened</a>

Jianing Song: I changed the table to be read by upward antidiagonals like A172236.