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A340263
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T(n, k) = [x^k] ((1-x)^(2^n) + 2^(-n)*((2^n-1)*(x-1)^(2^n) + (x+1)^(2^n)))/2. Irregular triangle read by rows, for n >= 0 and 0 <= k <= 2^n.
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5
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1, 1, -1, 1, 1, -3, 6, -3, 1, 1, -7, 28, -49, 70, -49, 28, -7, 1, 1, -15, 120, -525, 1820, -4095, 8008, -10725, 12870, -10725, 8008, -4095, 1820, -525, 120, -15, 1
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OFFSET
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0,6
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COMMENTS
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Conjecture: for n >= 1 the polynomials are irreducible.
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LINKS
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FORMULA
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Let p_n(x) = b(n) - (2^n-1)*a(n-1), b(n) = Sum_{k=0..2^n} binomial(2^n, 2*k)* x^(2*k), and a(n) = x*Product_{k=0..n} b(k). Then T(n, k) = [x^k] p_n(x).
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EXAMPLE
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Polynomials begin:
[0] 1;
[1] x^2 - x + 1;
[2] x^4 - 3*x^3 + 6*x^2 - 3*x + 1;
[3] x^8 - 7*x^7 + 28*x^6 - 49*x^5 + 70*x^4 - 49*x^3 + 28*x^2 - 7*x + 1;
Triangle begins:
[0] [1]
[1] [1, -1, 1]
[2] [1, -3, 6, -3, 1]
[3] [1, -7, 28, -49, 70, -49, 28, -7, 1]
[4] [1, -15, 120, -525, 1820, -4095, 8008, -10725, 12870, -10725, 8008, -4095, 1820, -525, 120, -15, 1]
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MAPLE
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if n = 0 then return [1] fi;
b := n -> add(binomial(2^n, 2*k)*x^(2*k), k = 0..2^n);
a := n -> x*mul(b(k), k = 0..n);
expand(b(n) - (2^n-1)*a(n-1));
[seq(coeff(%, x, j), j = 0..2^n)] end:
for n from 0 to 5 do A340263_row(n) od;
# Alternatively:
CoeffList := p -> [op(PolynomialTools:-CoefficientList(p, x))]:
Tpoly := n -> ((1-x)^(2^n) + 2^(-n)*((2^n-1)*(x-1)^(2^n) + (x + 1)^(2^n)))/2:
seq(print(CoeffList(Tpoly(n))), n=0..5); # Peter Luschny, Feb 03 2021
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PROG
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(SageMath)
a, b, c = 1, 1, 1
yield [1]
while True:
c *= 2
a *= b
b = sum(binomial(c, 2 * k) * x ^ (2 * k) for k in range(c + 1))
yield ((b - (c - 1) * x * a)).list()
for _ in range(6):
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CROSSREFS
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Row sums are 2^(2^n - n - 1) = A016031(n-1).
Central terms of the rows are A037293(n) for n >= 2.
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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