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%I #40 Nov 24 2019 18:59:53
%S 1,0,1,4,18,96,600,4320,35280,322560,3265920,36288000,439084800,
%T 5748019200,80951270400,1220496076800,19615115520000,334764638208000,
%U 6046686277632000,115242726703104000,2311256907767808000,48658040163532800000,1072909785605898240000
%N Sum of all possible sums formed from all but one of the previous terms, starting 1.
%C Apart from initial 1, same sequence as A001563. Additive analog of A057438.
%C a(1) = 1, for n >= 2: a(n) = sum of previous terms * (n-2) = (Sum_(i=1...n-2) a(i)) * (n-2). a(n) = A001563(n-2) = A094258(n-1) for n >= 3. - _Jaroslav Krizek_, Oct 16 2009
%H Milan Janjic, <a href="http://www.pmfbl.org/janjic/">Enumerative Formulas for Some Functions on Finite Sets</a>
%F a(n) = (n-2)!(n-2) for n>=2. - _Emeric Deutsch_, May 01 2008
%F G.f.: x*T(0), where T(k) = 1 - x^2*(k+1)^2/(x^2*(k+1)^2 - (1 -x -2*x*k)*(1 -3*x -2*x*k)/T(k+1) ); (continued fraction). - _Sergei N. Gladkovskii_, Nov 10 2013
%F a(n) = S1(n,1) - S1(n-1,1), where S1 are the unsigned Stirling cycle numbers. - _Peter Luschny_, Apr 10 2016
%F a(n) = A122974(n-1,n-1). - _Alois P. Heinz_, Nov 24 2019
%e a(2) = 0 as there is only one previous term and empty sum is taken to be 0.
%e a(4) = (a(1) +a(2))+ (a(1) +a(3)) + (a(2) +a(3)) = (1+0) +(1+1) +(0+1) = 4.
%e a(5) = (a(1)+a(2)+a(3)) +(a(1)+a(2)+ a(4)) +(a(1)+a(3)+a(4)) +(a(2)+a(3)+a(4)) = (1+0+1) +(1+0+4) +(1+1+4) +(0+1+4) = 2 + 5 + 6 + 5 = 18.
%p a := n -> (n-2)*(n-2)!: 1,seq(a(n), n=2..23); # _Emeric Deutsch_, May 01 2008
%t In[2]:= l = {1}; Do[k = Length[l] - 1; p = Plus @@ Flatten[Select[Subsets[l], Length[ # ]==k& ]]; AppendTo[l, p], {n, 20}]; l (* _Ryan Propper_, May 28 2006 *)
%o (PARI) v=vector(30);v[1]=1;v[2]=0;for(n=3,#v,s=0;for(i=1,2^(n-1)-1, vb=binary(i); if(hammingweight(vb)==n-2,s=s+sum(j=1,#vb, if(vb[j], v[n-#vb+j-1]))));v[n]=s;print1(s,",")) /* _Ralf Stephan_, Sep 22 2013 */
%Y Cf. A001563, A057438, A122974.
%K nonn
%O 1,4
%A _Amarnath Murthy_, Apr 29 2004
%E Edited by _N. J. A. Sloane_, May 29 2006
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