A343051 A triangle T(n,k) read by rows which can be used to calculate the area of a regular polygon with sides having length 1, provided that the polygon has an odd number of sides.

1, 16, 3, 256, 800, 125, 4096, 62720, 115248, 16807, 65536, 3096576, 23514624, 34012224, 4782969, 1048576, 118947840, 2518720512, 13605588480, 17148710480, 2357947691, 16777216, 3898605568, 185305595904, 2609720475648, 11485488551680, 13234415217504, 1792160394037

Offset: 0

Author

Peter Armstrong Maley, May 16 2021

Keywords

• nonn
• tabl

Comments

The examples will demonstrate how this works.

Examples

			16*A^2 - 3 = 0, A = 0.433012... the area of an equilateral triangle with sides of length 1.
256*A^4 - 800*A^2 + 125 = 0, A = 1.720477..., the area of a regular pentagon with sides of length 1.
4096*A^6 - 62720*A^4 + 115248*A^2 - 16807 = 0: A = 3.63391244..., the area of a regular heptagon with sides of length 1.
16777216*A^12 - 3898605568*A^10 + 185305595904*A^8 - 2609720475648*A^6 + 11485488551680*A^4 - 13234415217504*A^2 + 1792160394037 = 0: A = 13.185768328323878..., the area of a regular 13-gon with sides of length 1.
This sequence can be expressed as a triangle:
      1;
     16,       3;
    256,     800,      125;
   4096,   62720,   115248,    16807;
  65536, 3096576, 23514624, 34012224, 4782969;
  ...
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50)

Crossrefs

Cf. A103327.

Programs

Mathematica

T[n_, k_] := Binomial[2n+1, 2k+1] (2n+1)^(2k-1) 16^(n-k);
Table[T[n, k], {n, 0, 6}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2021, after Andrew Howroyd *)

Formula

This sequence can be generated from A103327. This example is for the pentagon: 5*L^4 - 10*L^2 + 1 = 0, L = (4/5)*A. Thus 256*A^4 - 800*A^2 + 125 = 0. In the case of a heptagon, L = (4/7)*A.

T(n,k) = binomial(2*n+1, 2*k+1)*(2*n+1)^(2*k-1)*16^(n-k). - Andrew Howroyd, May 23 2021