A111595 Triangle of coefficients of square of Hermite polynomials divided by 2^n with argument sqrt(x/2).
1, 0, 1, 1, -2, 1, 0, 9, -6, 1, 9, -36, 42, -12, 1, 0, 225, -300, 130, -20, 1, 225, -1350, 2475, -1380, 315, -30, 1, 0, 11025, -22050, 15435, -4620, 651, -42, 1, 11025, -88200, 220500, -182280, 67830, -12600, 1204, -56, 1, 0, 893025, -2381400, 2302020, -1020600, 235494, -29736, 2052, -72
Offset: 0
Examples
The triangle a(n, m) begins: n\m 0 1 2 3 4 5 6 7 8 9 10 ... 0: 1 1: 0 1 2: 1 -2 1 3: 0 9 -6 1 4: 9 -36 42 -12 1 5: 0 225 -300 130 -20 1 6: 225 -1350 2475 -1380 315 -30 1 7: 0 11025 -22050 15435 -4620 651 -42 1 8: 11025 -88200 220500 -182280 67830 -12600 1204 -56 1 9: 0 893025 -2381400 2302020 -1020600 235494 -29736 2052 -72 1 10: 893025 -8930250 28279125 -30958200 15961050 -4396140 689850 -63000 3285 -90 1 -------------------------------------------------------------------------------------------------
References
- R. P. Boas and R. C. Buck, Polynomial Expansions of Analytic Functions, Springer, 1958, p. 41
- S. Roman, The Umbral Calculus, Academic Press, New York, 1984, p. 128.
Links
- G. C. Greubel, Rows n=0..100 of triangle, flattened
Programs
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Mathematica
row[n_] := CoefficientList[ 1/2^n*HermiteH[n, Sqrt[x/2]]^2, x]; Table[row[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Jul 17 2013 *)
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Python
from sympy import hermite, Poly, sqrt, symbols x = symbols('x') def a(n): return Poly(1/2**n*hermite(n, sqrt(x/2))**2, x).all_coeffs()[::-1] for n in range(11): print(a(n)) # Indranil Ghosh, May 26 2017
Formula
E.g.f. for column m>=0: (1/sqrt(1-x^2))*((x/(1+x))^m)/m!.
a(n, m)=((-1)^(n-m))*(n!/m!)*sum(binomial(2*k, k)*binomial(n-2*k-1, m-1)/(4^k), k=0..floor((n-m)/2)), n>=m>=1. a(2*k, 0)= ((2*k)!/(k!*2^k))^2 = A001818(k), a(2*k+1) = 0, k>=0. a(n, m)=0 if n
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