Location via proxy:   [ UP ]  
[Report a bug]   [Manage cookies]                
a(n) = a(n-1)*2^n+n, a(0)=1.
0

%I #8 Apr 05 2015 14:47:29

%S 1,3,14,115,1844,59013,3776838,483435271,123759429384,63364827844617,

%T 64885583712887818,132885675443994251275,544299726618600453222412,

%U 4458903360459574912797999117,73054672657769675371282417532942

%N a(n) = a(n-1)*2^n+n, a(0)=1.

%F a(n+1) = (2^(n + 1) + 1)*a(n) - 2^n*a(n - 1) + 1.

%F a(n+1) = ((a(n - 2) + 4*a(n - 1) + 4)*a(n) - 2*a(n - 1)^2 - 4*a(n)^2 + a(n - 2) - 4*a(n - 1))/(a(n - 2) - 2*a(n - 1)).

%F a(n) = 2^(n*(n+1)/2) + sum_{k=1..n} 2^( (n+k+1)*(n-k)/2 ) * k. - _Max Alekseyev_, Sep 05 2010

%t nxt[{n_,a_}]:={n+1,a*2^(n+1)+n+1}; Transpose[NestList[nxt,{0,1},20]] [[2]] (* _Harvey P. Dale_, Apr 05 2015 *)

%o (PARI) a(n)=if(n<=0,1, a(n-1)*2^n+n )

%Y Cf. A010842.

%K nonn

%O 0,2

%A _Georgi Guninski_, Sep 05 2010

%E Minor edits by _N. J. A. Sloane_, Sep 05 2010