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A273894
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Table T(n,k), n >= 0, k = 1..2^n, read by rows, giving coefficients of iterations of polynomial x^2-x: see Comments for precise definition.
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1
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1, -1, 1, 1, 0, -2, 1, -1, 1, 2, -5, 2, 4, -4, 1, 1, 0, -4, 2, 12, -14, -20, 48, -14, -50, 60, -10, -28, 24, -8, 1, -1, 1, 4, -10, -8, 54, -24, -180, 270, 270, -960, 150, 2064, -2040, -2352, 5871, -1566, -7236, 8880, 120, -9120, 7980, 120, -5340, 4212, -756
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OFFSET
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0,6
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COMMENTS
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Let p(0) = t, p(n) = p(n-1)^2 - p(n-1) for i >= 1.
T(n,k) is coefficient of t^k in p(n).
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LINKS
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FORMULA
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T(n,k) = -T(n-1,k) + Sum_{j=1..k-1} T(n-1,j) T(n-1,k-j).
Column k is of the form
T(n,k) = b_k(n) + (-1)^n*c_k(n)
where b_k and c_k seem to be polynomials of degree floor(k/2) - 1 and floor((k-1)/2) respectively (except b_1 = 0).
Leading coefficient of b_k(n) + (-1)^n*c_k(n) seems to be
-(-2)^(k/2-2) - binomial(-3/2,k/2-1)*2^(k/2-2)*(-1)^n if k is even,
2^((k-1)/2)*binomial(-1/2,(k-1)/2)*(-1)^n if k is odd.
T(n,2) = 1/2 + (-1)^n/2 = A000035(n)
T(n,3) = -1/2 + (-n + 1/2)*(-1)^n = -A137501(n).
T(n,4) = -n + 5/4 + (3*n/2 - 5/4)*(-1)^n)
T(n,5) = 2*n - 11/4 + (3*n^2/2 - 5*n + 11/4)*(-1)^n
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EXAMPLE
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Table starts
1
-1, 1
1, 0, -2, 1
-1, 1, 2, -5, 2, 4, -4, 1
1, 0, -4, 2, 12, -14, -20, 48, -14, -50, 60, -10, -28, 24, -8, 1
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MAPLE
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P[0]:= t:
for n from 1 to 8 do
P[n]:= expand(P[n-1]^2 - P[n-1])
od:
seq(seq(coeff(P[n], t, j), j=1..2^n), n=0..8);
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MATHEMATICA
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CROSSREFS
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KEYWORD
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sign,tabf
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AUTHOR
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STATUS
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approved
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