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Recurrence (for n>5): (n-5)*(n-2)*a(n) = (n-6)*n*a(n-1) + (n-4)*(n-1)*n*a(n-2) . - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 1/4*sqrt(2)*exp(sqrt(n)-n/2-1/4)*n^(n/2+3/2) . - Vaclav Kotesovec, Oct 20 2012
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Vincenzo Librandi, <a href="/A135593/b135593.txt">Table of n, a(n) for n = 2..200</a>
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_Vladeta Jovovic (vladeta(AT)eunet.rs), _, Feb 25 2008
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Rest[Rest[CoefficientList[Series[x^2*(x+2)/2*E^(x*(x+2)/2), {x, 0, 20}], x]* Range[0, 20]!]] (* or _Vaclav Kotesovec_, Oct 20 2012 *)
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Recurrence (for n>5): (n-5)*(n-2)*a(n) = (n-6)*n*a(n-1) + (n-4)*(n-1)*n*a(n-2) . - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 1/4*sqrt(2)*exp(sqrt(n)-n/2-1/4)*n^(n/2+3/2) . - Vaclav Kotesovec, Oct 20 2012
Rest[Rest[CoefficientList[Series[x^2*(x+2)/2*E^(x*(x+2)/2), {x, 0, 20}], x]* Range[0, 20]!]] (* or *)
Flatten[{2, 9, RecurrenceTable[{(n-5)*(n-2)*a[n]==(n-6)*n*a[n-1]+(n-4)*(n-1)*n*a[n-2], a[4]==36, a[5]==140}, a, {n, 4, 20}]}] (* Vaclav Kotesovec, Oct 20 2012 *)
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