A376777 a(n) = Sum_{k=0..n} tan(k*Pi/(1+2*n))^(2*n).

1, 3, 90, 7077, 1070244, 264893255, 97371674686, 49810055605065, 33841518448166024, 29482777900878972939, 32045566134755984390370, 42511262856482596083333613, 67601184141908795841006166700, 126937986415384594402633688922447, 277898987279628989741077214849901894

Offset: 0

Author

N. J. A. Sloane, Nov 09 2024

Keywords

• nonn

Comments

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).

(End)

References

• Bill Gosper, Email to N. J. A. Sloane, Nov 02 2024.
• Shigeichi Moriguchi, Kanehisa Udagawa, Shin Hitotsumatsu, "Mathematics Formulas II", Iwanami Shoten, Publishers (1957), p. 14.

Links

• Peter Luschny, Table of n, a(n) for n = 0..100
• Bill Gosper, Extract from email message of Nov 02 2024.

Crossrefs

Main diagonal of A377657.

Cf. A014105, A377858, A376778.

Programs

Julia

using Nemo
RR = ArbField(1000)
function F(n)
    sum(RR(tanpi(QQBar(k) / (1 + 2 * n))^(2 * n)) for k in 0:n)
end
a(n) = unique_integer(F(n))[2]
println([a(n) for n in 0:14])  # Peter Luschny, Nov 10 2024

Mathematica

(* See Gosper link for his original Mathematica code. *)
a[0] = 1; a[n_] := ToNumberField@ Sum[Tan[k*Pi/(2*n + 1)]^(2*n), {k, 0, n}]; Array[a, 15, 0] (* Amiram Eldar, Nov 10 2024 *)

PARI

a(n) = { polsym(sum(m=0, n, x^m*binomial(2*n+1, 2*(n-m))*(-1)^(m+1)), n)[n+1]+(n==0) } \\ Thomas Scheuerle, Nov 11 2024

Formula

a(n) ~ exp(1) * 2^(4*n) * n^(2*n) / Pi^(2*n). - Vaclav Kotesovec, Nov 10 2024

Extensions

More terms from Amiram Eldar, Nov 10 2024

a(0) = 1 prepended by Peter Luschny, Nov 10 2024