A381423 Exponent of x of maximal coefficient in Hermite polynomial of order n.

0, 1, 2, 3, 4, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 4, 5, 6, 7, 4, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 9, 6, 7, 8, 9, 6, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 11, 8, 9, 10, 11, 8, 9, 10, 11, 12, 9, 10, 11, 12, 9, 10, 11, 12, 9, 10, 11, 12, 13, 10

Offset: 0

Author

Mike Sheppard, Feb 23 2025

Keywords

• nonn

Comments

The exponent is always unique. The coefficients, in absolute value, follow a unimodal pattern, and their signs alternate. If the maximum absolute coefficient appears twice due to symmetry (e.g., H_8(x)), the terms will have opposite signs, ensuring a unique exponent for the maximum signed coefficient.

Conjecture: Differences are either 1 or -3; more specifically the patterns (1,1,1,-3) or (1,1,1,1,-3), with position of those patterns appearing at linearly and quadratically spaced intervals, respectively. Seems to grow O(n^(1/2))

Examples

			For n = 5, H_5(x) = 32*x^5 - 160*x^3 + 120*x. The maximal coefficient is 120 (we take signs into account, so -160 < 120), occurring at x^1, hence a(5) = 1.
		

Crossrefs

Cf. A277280 (maximal coefficient).

Programs

Mathematica

Table[(PositionLargest@CoefficientList[HermiteH[n, x], x])[[1]] - 1, {n, 0, 100}]

PARI

a(n) = my(p=polhermite(n), m=vecmax(Vec(p))); for(i=0, poldegree(p), if (polcoef(p, i) == m, return(i))); \\ Michel Marcus, Feb 23 2025