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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
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
KEYWORD
nonn,tabl
AUTHOR
STATUS
approved