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%I #12 May 03 2015 08:53:50
%S 1,1,9,169,5625,292681,21930489,2236627849,297935847225,
%T 50229268482121,10454564139438969,2632936466960600329,
%U 789136169944454084025,277579719258755165321161,113238180214596650771616249,53030348046942317338336489609,28256184698070300360908567636025
%N a(n) = A000670(n)^2.
%C This is also the number of possible positions of n intervals on a line having a common non-punctual intersection. Proof: Let us denoted each interval Ai (1 <= i <= n) by the string AiAi. Then the set of all such relative positions is given by the S-language [A1 ⊗ A2 ... ⊗ An]^2. The cardinality of $A1 ⊗ A2 ... ⊗ An$ is given by A000670. - Sylviane R. Schwer (schwer(AT)lipn.univ-paris13.fr), Nov 26 2007
%F a(n) = sum(sum((k*l)^n/2^(k+l+2),k=0..infinity),l=0..infinity).
%F G.f.: sum(1/(2-exp(n*x))/2^(n+1),n=0..infinity).
%F Sum_{n>=0} a(n)*log(1+x)^n/n! = o.g.f. of A101370. [_Paul D. Hanna_, Nov 07 2009]
%F a(n) ~ (n!)^2 / (4 * (log(2))^(2*n+2)). - _Vaclav Kotesovec_, May 03 2015
%t Table[(PolyLog[ -z, 1/2]/2)^2, {z, 1, 25}] - Elizabeth A. Blickley (Elizabeth.Blickley(AT)gmail.com), Oct 10 2006
%o (PARI) {Stirling2(n, k)=if(k<0|k>n,0, sum(i=0,k,(-1)^i*binomial(k, i)/k!*(k-i)^n))}
%o {a(n)=sum(k=0,n,Stirling2(n, k)*k!)^2} \\ _Paul D. Hanna_, Nov 07 2009
%Y Cf. A101370, A055203.
%K easy,nonn
%O 0,3
%A _Vladeta Jovovic_, Sep 23 2006
%E More terms from Elizabeth A. Blickley (Elizabeth.Blickley(AT)gmail.com), Oct 10 2006
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