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 *) -
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
A111883 Unsigned row sums of triangle A111595 (normalized rescaled squared Hermite polynomials).
1, 1, 4, 16, 100, 676, 5776, 53824, 583696, 6864400, 90174016, 1274204416, 19642583104, 323196798016, 5714394630400, 107112895415296, 2135062451773696, 44858948563673344, 994634863541502976, 23133227941938073600, 564474119626559497216, 14388648533002088866816
Offset: 0
Links
- Indranil Ghosh, Table of n, a(n) for n = 0..100
Programs
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Magma
m:=25; R
:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(Exp(x/(1-x))/Sqrt(1-x^2))); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, Jun 09 2018 -
Mathematica
Table[Abs[HermiteH[n, I/Sqrt[2]]]^2/2^n, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 11 2016 *) CoefficientList[Series[Exp[t/(1-t)]/Sqrt[1-t^2],{t,0,100}],t] Range[0, 12]! (* Emanuele Munarini, Aug 31 2017 *) -
PARI
a(n)=if(n<0, 0, n!*polcoeff(exp(x/(1-x)+x*O(x^n))/sqrt(1-x^2+x*O(x^n)),n)) /* Michael Somos, Aug 30 2005 */
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Python
from sympy import hermite, Poly, sqrt, I def a(n): return abs(Poly(hermite(n, I/sqrt(2)), x))**2/2**n # Indranil Ghosh, May 26 2017
Formula
E.g.f.: exp(x/(1-x))/sqrt(1-x^2).
a(n) = A000085(n)^2. - Michael Somos, Aug 30 2005
Conjecture: a(n) -n*a(n-1) -n*(n-1)*a(n-2) +(n-1)*(n-2)^2*a(n-3)=0. - R. J. Mathar, Oct 05 2014
Remark: the above conjectured recurrence is true and can be easily obtained by the e.g.f. - Emanuele Munarini, Aug 31 2017
a(n) = |H_n(i/sqrt(2))|^2 / 2^n = H_n(i/sqrt(2)) * H_n(-i/sqrt(2)) / 2^n, where H_n(x) is n-th Hermite polynomial, i = sqrt(-1). - Vladimir Reshetnikov, Oct 11 2016
a(n) ~ exp(2*sqrt(n) - n - 1/2) * n^n / 2. - Vaclav Kotesovec, Oct 01 2017
Comments