A343051 - OEIS

%I #37 Jan 20 2024 16:06:59

%S 1,16,3,256,800,125,4096,62720,115248,16807,65536,3096576,23514624,

%T 34012224,4782969,1048576,118947840,2518720512,13605588480,

%U 17148710480,2357947691,16777216,3898605568,185305595904,2609720475648,11485488551680,13234415217504,1792160394037

%N 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.

%C The examples will demonstrate how this works.

%H Andrew Howroyd, <a href="/A343051/b343051.txt">Table of n, a(n) for n = 0..1325</a> (rows 0..50)

%F 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.

%F T(n,k) = binomial(2*n+1, 2*k+1)*(2*n+1)^(2*k-1)*16^(n-k). - _Andrew Howroyd_, May 23 2021

%e 16*A^2 - 3 = 0, A = 0.433012... the area of an equilateral triangle with sides of length 1.

%e 256*A^4 - 800*A^2 + 125 = 0, A = 1.720477..., the area of a regular pentagon with sides of length 1.

%e 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.

%e 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.

%e This sequence can be expressed as a triangle:

%e       1;

%e      16,       3;

%e     256,     800,      125;

%e    4096,   62720,   115248,    16807;

%e   65536, 3096576, 23514624, 34012224, 4782969;

%e   ...

%t T[n_, k_] := Binomial[2n+1, 2k+1] (2n+1)^(2k-1) 16^(n-k);

%t Table[T[n, k], {n, 0, 6}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, May 29 2021, after _Andrew Howroyd_ *)

%Y Cf. A103327.

%K nonn,tabl

%O 0,2

%A _Peter Armstrong Maley_, May 16 2021