cp's OEIS Frontend

Bill Gosper's Mathematica program (see link) was able to simplify the sums, producing the integer values shown here. He asks, why are the a(n) integers? He also remarks that there are similar results with cot instead of tan, and with 2*n instead of n.

If we replace the sum with product, we get the well-known formula Product_{k=1..n} tan(k*Pi/(1+2*n))^(2*n) = (2n+1)^n. - Chai Wah Wu, Nov 10 2024

Comments from Fredrik Johansson, Nov 10 2024: (Start)

I don't think there is any particularly deep mystery about why these sums are integers.

Any symmetric polynomial function (with integer coefficients) of all the conjugates of an algebraic number x is a rational number, and in particular will be an integer when x is an algebraic integer.

I think x_m = tan(1/(2m + 1))^2 is an algebraic integer of degree <= m, and {tan(n/(2m + 1))^2, n = 1...m} is a set of all algebraic conjugates of x_m (when deg(x_m) = m) or a union of such sets (when deg(x_m) < m). So there is nothing special about the 2m power of the tangents: any fixed even power will give an integer sequence.

I'm guessing that

A_k(m) = sum(tan(n/(2m + 1))^(2k), n = 1...m) for fixed k

is a polynomial in m.

Indeed A_1(m) gives 0, 3, 10, 21, 36, 55, 78, 105, 136, 171, 210, 253, 300, 351, ... which is A014105, m*(2m+1).

A_2(m) gives 0, 9, 90, 371, 1044, 2365, 4654, 8295, 13736, 21489, 32130, 46299, ... which is A377858, (4*m^2 + 6*m - 1)*(2*m + 1)*m/3.

A_3(m) gives 0, 27, 850, 7077, 33300, 113311, 312390, 742665, 1581544, 3093219, 5653242, 9776173, ... which is A376778, (32*m^4 + 80*m^3 + 40*m^2 - 20*m + 3)*(2*m + 1)*m/15.

and so on...

I believe one can derive such a polynomial expression for any A_k(m) using Newton's identities, and this could maybe even result in some kind of recurrence relation (if not a closed form) for A_m(m), which is the present sequence (A376777).

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