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A281122
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Triangle T read by rows: n-th row (n>=0) gives the non-vanishing coefficients of the polynomial q(n,x) = ((x+1)^(2^n) - (x-1)^(2^n))/2.
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3
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1, 2, 4, 4, 8, 56, 56, 8, 16, 560, 4368, 11440, 11440, 4368, 560, 16, 32, 4960, 201376, 3365856, 28048800, 129024480, 347373600, 565722720, 565722720, 347373600, 129024480, 28048800, 3365856, 201376, 4960, 32
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OFFSET
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0,2
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COMMENTS
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The length of row n is 1 for n = 0 and 2^(n-1) = A000079(n-1) for n >= 1.
The polynomial q(n,x) = 2^n*Product_{k=0..n-1} p(k,x) with polynomial p(n,x) = ((x+1)^(2^n) + (x-1)^(2^n))/2, whose coefficients are tabulated in A201461.
The row polynomials are q(n, x) = Sum_{k=0..2^(n-1)-1} T(n, k)*x^(2^n-1-2*k) for n >= 1, and q(0,x) = 1. - Wolfdieter Lang, Jan 19 2017
A201461 and T(n,k) are a bisection of row 2^n of Pascal's triangle A007318.
The algorithm r(n) = 1/2*(r(n-1) + A/r(n-1)), starting with r(0) = A, used for approximating sqrt(A), which is known as the Babylonian method or Hero's method after the first-century Greek mathematician Hero of Alexandria and which can be derived from Newton's method, generates fractions beginning with (A+1)/2, (A^2 + 6*A + 1)/(4*A + 4), (A^4 + 28*A^3 + 70*A^2 + 28*A + 1)/(8*A^3 + 56*A^2 + 56*A + 8), ... This is sqrt(A)*p(n,sqrt(A))/q(n,sqrt(A)) with the given polynomials p(n,x) and q(n,x).
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LINKS
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FORMULA
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T(n, k) = 1 if n = 0 and k = 0, and T(n, k) = binomial(2^n,2*k+1) = A103328(2^(n-1),k) for k = 0..2^(n-1)-1 and n >= 1. - Wolfdieter Lang, Jan 19 2017
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EXAMPLE
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The triangle of non-vanishing coefficients starts with
1
2
4, 4
8, 56, 56, 8
16, 560, 4368, 11440, 11440, 4368, 560, 16
etc., since the first few polynomials are
q(0,x) = 1,
q(1,x) = 2*x,
q(2,x) = 4*x^3 + 4*x = 4*x*(x^2 + 1),
q(3,x) = 8*x^7 + 56*x^5 + 56*x^3 + 8*x = 8*x*(x^2 + 1)*(x^4 + 6*x^2 + 1),
q(4,x) = 16*x^15 + 560*x^13 + 4368*x^11 + 11440*x^9 + 11440*x^7 + 4368*x^5 + 560*x^3 + 16*x = 16*x*(x^2 + 1)*(x^4 + 6*x^2 + 1)*(x^8 + 28*x^6 + 70*x^4 + 28*x^2 + 1),
etc.
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MAPLE
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CoeffList := p -> remove(n->n=0, [op(PolynomialTools:-CoefficientList(p, x))]):
Tpoly := n -> (1/2)*((x+1)^(2^n) - (x-1)^(2^n));
seq(print(CoeffList(Tpoly(n))), n=0..5); # Peter Luschny, Feb 04 2021
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MATHEMATICA
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q[n_] := DeleteCases[ CoefficientList[ Expand[((x +1)^(2^n) - (x -1)^(2^n))/2], x], 0]; Array[q, 7, 0] // Flatten (* Robert G. Wilson v, Jan 16 2017 *)
t={1}; T[n_, k_]:=Table[Binomial[2^n, 2k+1], {n, 1, 6}, {k, 0, 2^(n-1)-1}]; Do[AppendTo[t, T[n, k]]]; Flatten[t] (* Indranil Ghosh, Feb 22 2017 *)
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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