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
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
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