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A283431
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a(n) is the number of zeros of the Hermite H(n, x) polynomial in the open interval (-1, +1).
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0
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0, 1, 2, 1, 2, 3, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 4, 3, 4, 3, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 4, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 5, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 6, 7, 8, 7, 8, 7, 8, 7, 8, 7, 8, 7, 8, 7, 8, 7, 8, 7, 8, 9, 8, 9
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OFFSET
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0,3
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COMMENTS
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The Hermite polynomials satisfy the following recurrence relation:
H(0,x) = 1,
H(1,x) = 2*x,
H(n,x) = 2*x*H(n-1,x) - 2*(n-1)*H(n-2,x).
The first few Hermite polynomials are:
H(0,x) = 1
H(1,x) = 2x
H(2,x) = 4x^2 - 2
H(3,x) = 8x^3 - 12x
H(4,x) = 16x^4 - 48x^2 + 12
H(5,x) = 32x^5 - 160x^3 + 120x
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LINKS
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FORMULA
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EXAMPLE
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a(5) = 3 because the zeros of H(5,x) = 32x^5 - 160x^3 + 120x are x1 = -2.0201828..., x2 = -.9585724..., x3 = 0., x4 = .9585724... and x5 = 2.020182... with three roots x2, x3 and x4 in the open interval (-1, +1).
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MAPLE
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for n from 0 to 90 do:it:=0:
y:=[fsolve(expand(HermiteH(n, x)), x, real)]:for m from 1 to nops(y) do:if abs(y[m])<1 then it:=it+1:else fi:od: printf(`%d, `, it):od:
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MATHEMATICA
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a[n_] := Length@ List@ ToRules@ Reduce[ HermiteH[n, x] == 0 && -1 < x < 1, x]; Array[a, 82, 0] (* Giovanni Resta, May 17 2017 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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