

A136239


Forced end points ( Infinity >1) integration of A137286: Triangle of coefficients of Integrated recursive orthogonal Hermite polynomials given in Hochstadt's book : P(x, n) = x*P(x, n  1)  n*P(x, n  2).


1



1, 0, 1, 1, 0, 1, 1, 3, 0, 1, 9, 0, 6, 0, 1, 1, 27, 0, 10, 0, 1, 19, 0, 65, 0, 15, 0, 1, 1, 165, 0, 135, 0, 21, 0, 1, 399, 0, 624, 0, 252, 0, 28, 0, 1, 1, 2145, 0, 1750, 0, 434, 0, 36, 0, 1
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OFFSET

1,8


COMMENTS

Because of error functions in the result where constants should be this is a difficult calculation.
Probably the wrong approach, but it is my best effort at getting Gaussian normal type functions to give integers. There has got to be a better way than this: maybe a conformal transform of the known Chebyshev Integration polynomials?
No recurrence formula was found for these polynomials, so they are probably wrong.
Row sums are:
{1, 1, 0, 3, 4, 17, 32, 51, 0, 793}


REFERENCES

page 8 and pages 42  43 : Harry Hochstadt, The Functions of Mathematical Physics, Dover, New York, 1986


LINKS

Table of n, a(n) for n=1..55.


FORMULA

P(x, n) = x*P(x, n  1)  n*P(x, n  2); L(x,n)=Integrate[Exp[y^2/4]*p(y,n1),{y,Infinity,x}]/(2*Exp[ x^2/4]) Here the weight function is taken as the square root of the Hermite weight function Exp[ x^2/2] and then divided out of the end result.


EXAMPLE

{1},
{0, 1},
{1, 0, 1},
{1, 3, 0, 1},
{9, 0, 6, 0, 1},
{1, 27, 0, 10, 0, 1},
{19, 0, 65, 0, 15, 0, 1},
{1, 165, 0, 135, 0, 21, 0,1},
{399, 0, 624, 0, 252, 0, 28, 0, 1},
{1, 2145, 0, 1750, 0, 434, 0, 36, 0, 1}


CROSSREFS

Cf. A137286.
Sequence in context: A157391 A099097 A152150 * A225443 A222060 A256549
Adjacent sequences: A136236 A136237 A136238 * A136240 A136241 A136242


KEYWORD

uned,tabl,sign


AUTHOR

Roger L. Bagula, Mar 16 2008


STATUS

approved



