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A122573
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Expansion of -x*(1+x)*(3*x^2-1) / ( 1-4*x^2+x^4 ).
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0
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1, 1, 1, 1, 3, 3, 11, 11, 41, 41, 153, 153, 571, 571, 2131, 2131, 7953, 7953, 29681, 29681, 110771, 110771, 413403, 413403, 1542841, 1542841, 5757961, 5757961, 21489003, 21489003, 80198051, 80198051, 299303201, 299303201, 1117014753
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OFFSET
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1,5
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COMMENTS
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4 X 4 vector matrix Markov for characteristic polynomial:(1 - 4 x^2 + x^4) Which Mathematica gives as a factor of the cubic polynomial: Factor[(1 - 14 x^4 + x^8)]=(1 - 4 x^2 + x^4)(1 + 4 x^2 + x^4)
A001835[n]=v[2*n][[1]]=4*a[n-1]-a[n-2]: The coefficient expansion gives an alternating even term zeros sequence: p[x_] := x^4 - 4x^2 + 1 q[x_] := ExpandAll[x^4*p[1/x]] Table[ SeriesCoefficient[Series[x/q[x], {x, 0, 30}], n], {n, 0, 30}] {0, 1, 0, 4, 0, 15, 0, 56, 0, 209, 0, 780, 0, 2911, 0, 10864, 0, 40545, 0, 151316, 0, 564719, 0, 2107560, 0, 7865521, 0, 29354524, 0, 109552575, 0}
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LINKS
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Table of n, a(n) for n=1..35.
Index to sequences with linear recurrences with constant coefficients, signature (0,4,0,-1).
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FORMULA
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M = {{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {-1, 0, 4, 0}}; v[1] = {1, 1, 1, 1}; v[n_] := v[n] = M.v[n - 1] a(n) = v[n][[1]]
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MATHEMATICA
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M = {{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {-1, 0, 4, 0}}; v[1] = {1, 1, 1, 1}; v[n_] := v[n] = M.v[n - 1] a1 = Table[v[n][[1]], {n, 1, 50}] (* alternative calculation method*) a[0] = 1; a[1] = 1; a[2] = 1; a[3] = 1; a[n_] := a[n] = 4*a[n - 2] - a[n - 4] Table[a[n], {n, 0, 50}]
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CROSSREFS
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Cf. A001835.
Sequence in context: A146828 A146583 A146458 * A136123 A045495 A045494
Adjacent sequences: A122570 A122571 A122572 * A122574 A122575 A122576
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KEYWORD
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nonn,uned
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AUTHOR
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Roger Bagula, Sep 17 2006
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STATUS
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approved
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