lure law of small numbers

[benoit]

Indeed, sorry al and Neil. Some precisions are needed.

Let (u(n)) n>=0  having g.f.  2/(exp(x)+1)
Let v(n)=(2^( n+1)-1)*bernfrac(n+1)/(n+1)  (bernfrac=bernoulli)

Then this following equality must holds :

v(2*n+1)=(-1)*n*(2*n+1)*(2*n-1)!*u(2*n+1)

Note that a(n)=v(2n+1)/u(2n+1)=n*(2*n+1)*(2*n-1)! gives the nice  
enumerative sequence  A085990.

to avoid offset problems here my pari check :

? a(n)=polcoeff(2/(1+exp(x)),n)
? b(n)=(2^( n+1)-1)*bernfrac(n+1)/(n+1)
? for(n=2,100,print1(-b(2*n+1)/a(2*n+1)/(2*n-1)!/n/(2*n+1),","))
1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1, 
1,1,1,1,1,1,1,1,1,1,1,1,.....

BC
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