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#23 by Vaclav Kotesovec at Wed Aug 06 08:15:56 EDT 2014
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#22 by Vaclav Kotesovec at Wed Aug 06 08:15:42 EDT 2014
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| FORMULA
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a(n) ~ n^n / (zr^n * exp((2*zr^2*n)/(1+2*zr^2)) * sqrt(3+2*zr^2 - 2/(1 + 2*zr^2))), where r is the root of the equation r*exp(r^2)*(1+2*r^2) = n.
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#21 by Vaclav Kotesovec at Wed Aug 06 08:14:53 EDT 2014
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a(n)=(n!*sum(m=floor((n+1)/2)..n, ((2*m-n)^(n-m))/((2*m-n)!*(n-m)!))). [Vladimir Kruchinin, Mar 09 2013]]]
From Vaclav Kotesovec, Aug 06 2014: (Start)
a(n) ~ n^n / (z^n * exp((2*z^2*n)/(1+2*z^2)) * sqrt(3+2*z^2 - 2/(1 + 2*z^2))), where r is the root of the equation r*exp(r^2)*(1+2*r^2) = n.
(a(n)/n!)^(1/n) ~ exp(1/(3*LambertW(2^(1/3)*n^(2/3)/3))) * sqrt(2/(3*LambertW(2^(1/3)*n^(2/3)/3))).
(End)
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| STATUS
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approved
editing
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#20 by Joerg Arndt at Mon Mar 18 08:59:54 EDT 2013
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#19 by Vincenzo Librandi at Mon Mar 18 03:13:08 EDT 2013
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#18 by Vincenzo Librandi at Mon Mar 18 03:13:01 EDT 2013
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| LINKS
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Vincenzo Librandi, <a href="/A216688/b216688.txt">Table of n, a(n) for n = 0..200</a>
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| STATUS
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approved
editing
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#17 by Joerg Arndt at Sat Mar 09 11:51:58 EST 2013
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#16 by Joerg Arndt at Sat Mar 09 11:51:53 EST 2013
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#15 by Joerg Arndt at Sat Mar 09 11:51:39 EST 2013
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| FORMULA
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a(n)=(n!*sum(m=floor((n+1)/2)..n, ((2*m-n)^(n-m))/((2*m-n)!*(n-m)!))), n>0, a(0)=1. [_)!))). [_Vladimir Kruchinin_, Mar 09 2013]]
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| STATUS
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proposed
editing
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#14 by Vladimir Kruchinin at Sat Mar 09 09:14:58 EST 2013
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Discussion
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| Sat Mar 09
| 11:07
| Joerg Arndt: The expression also gives a(0)=1.
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