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%I A160068 #34 Sep 08 2022 08:45:44
%S A160068 1,48,1054,-69408,-7284084,-2596032,45400915464,2198714182272,
%T A160068 -291719729560944,-35989688841645312,1554341893161645024,
%U A160068 524479521392325361152,3802815995858998255296,-7684657653083648501025792,-430659327280723849697798016
%N A160068 Numerator of Hermite(n, 24/25).
%H A160068 Robert Israel, <a href="/A160068/b160068.txt">Table of n, a(n) for n = 0..380</a>
%F A160068 From _Robert Israel_, Jan 02 2017 and Jan 24 2017: (Start)
%F A160068 Theorem: E.g.f.: exp(-x*(625*x-48)); a(n+2) = -1250*(n+1)*a(n)+48*a(n+1).
%F A160068 Proof:
%F A160068 The Hermite polynomials H(n,t) have e.g.f. g(x) = exp(-x^2 + 2 x t).
%F A160068 Thus b(n) := H(n, 24/25)*5^(2n) has e.g.f. exp(-x*(625*x -48)) and satisfies the recurrence b(n+2) = -1250*(n+1)*b(n)+48*b(n+1), n>=0.
%F A160068 To show that these are the numerators of H(n,24/25), we need to show that b(n) is never divisible by 5.  But taking the recurrence mod 5 we get b(n+2) == 3*b(n+1) mod 5.
%F A160068 Since b(0) and b(1) are not divisible by 5, induction finishes the proof.
%F A160068 (End)
%F A160068 From _G. C. Greubel_, Jun 02 2018: (Start)
%F A160068 a(n) = 25^n * Hermite(n, 24/25).
%F A160068 a(n) = numerator(Sum_{k=0..floor(n/2)} (-1)^k*n!*(48/25)^(n-2*k)/(k!*(n-2*k)!)). (End)
%e A160068 Numerators of 1, 48/25, 1054/625, -69408/15625, -7284084/390625, ...
%p A160068 A160068:=n->numer(orthopoly[H](n,24/25)): seq(A160068(n), n=0..20); # _Wesley Ivan Hurt_, Nov 20 2014
%t A160068 Table[Numerator@ HermiteH[n, 24/25], {n, 0, 30}] (* _Michael De Vlieger_, Jan 02 2017 *)
%t A160068 Table[25^n*HermiteH[n, 24/25], {n,0,30}] (* _G. C. Greubel_, Jul 11 2018 *)
%o A160068 (PARI) a(n)=numerator(polhermite(n,24/25)) \\ _Charles R Greathouse IV_, Jan 29 2016
%o A160068 (Magma) [Numerator((&+[(-1)^k*Factorial(n)*(48/25)^(n-2*k)/( Factorial(k) *Factorial(n-2*k)): k in [0..Floor(n/2)]])): n in [0..30]]; // _G. C. Greubel_, Jul 11 2018
%Y A160068 Cf. A009969 (denominators).
%K A160068 sign,frac
%O A160068 0,2
%A A160068 _N. J. A. Sloane_, Nov 12 2009

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