A283431 - OEIS

%I #25 May 20 2017 22:20:38

%S 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,

%T 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,

%U 8,7,8,7,8,7,8,7,8,7,8,9,8,9

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

%C The Hermite polynomials satisfy the following recurrence relation:

%C H(0,x) = 1,

%C H(1,x) = 2*x,

%C H(n,x) = 2*x*H(n-1,x) - 2*(n-1)*H(n-2,x).

%C The first few Hermite polynomials are:

%C H(0,x) = 1

%C H(1,x) = 2x

%C H(2,x) = 4x^2 - 2

%C H(3,x) = 8x^3 - 12x

%C H(4,x) = 16x^4 - 48x^2 + 12

%C H(5,x) = 32x^5 - 160x^3 + 120x

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HermitePolynomial.html">Hermite Polynomial.</a>

%H <a href="/index/He#Hermite">Index entries for sequences related to Hermite polynomials</a>

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

%e 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).

%p for n from 0 to 90 do:it:=0:

%p 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:

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

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

%K nonn

%O 0,3

%A _Michel Lagneau_, May 16 2017