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

0,2

Comments

The examples will demonstrate how this works.

Links

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

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

Example

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;
  ...

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 *)

Crossrefs

Cf. A103327.

Sequence in context: A070569 A317316 A084473 * A040246 A075462 A082959

Adjacent sequences: A343048 A343049 A343050 * A343052 A343053 A343054

Keyword

nonn,tabl

Author

Peter Armstrong Maley, May 16 2021

Status

approved