A283431 a(n) is the number of zeros of the Hermite H(n, x) polynomial in the open interval (-1, +1).

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

Offset

0,3

Comments

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

Links

Table of n, a(n) for n=0..81.

Eric Weisstein's World of Mathematics, Hermite Polynomial.

Index entries for sequences related to Hermite polynomials

Formula

Conjecture: a(n) = A257564(n+2).

Example

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).

Maple

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:

Mathematica

a[n_] := Length@ List@ ToRules@ Reduce[ HermiteH[n, x] == 0 && -1 < x < 1, x]; Array[a, 82, 0] (* Giovanni Resta, May 17 2017 *)

Crossrefs

Cf. A054373, A054374, A059343, A008611, A096713, A257564, A285872.

Sequence in context: A362257 A136625 A321861 * A258594 A364215 A086520

Adjacent sequences: A283428 A283429 A283430 * A283432 A283433 A283434

Keyword

nonn

Author

Michel Lagneau, May 16 2017

Status

approved