%I #6 Jan 27 2019 18:03:21

%S 1,3,8,30,359,72385,4338080222,18448597098193762732,

%T 340282370354622283774333836315916425069,

%U 115792089237316207213755562747271079374483128445080168204415615259394085515423

%N a(n) = Sum_{k = 1...n} k^(2^(n - k)).

%C Number of ways to choose a constant integer partition of each part of a constant integer partition of 2^(n - 1).

%e The a(1) = 1 through a(4) = 30 twice-partitions:

%e (1) (2) (4) (8)

%e (11) (22) (44)

%e (1)(1) (1111) (2222)

%e (2)(2) (4)(4)

%e (11)(2) (22)(4)

%e (2)(11) (4)(22)

%e (11)(11) (22)(22)

%e (1)(1)(1)(1) (1111)(4)

%e (4)(1111)

%e (11111111)

%e (1111)(22)

%e (22)(1111)

%e (1111)(1111)

%e (2)(2)(2)(2)

%e (11)(2)(2)(2)

%e (2)(11)(2)(2)

%e (2)(2)(11)(2)

%e (2)(2)(2)(11)

%e (11)(11)(2)(2)

%e (11)(2)(11)(2)

%e (11)(2)(2)(11)

%e (2)(11)(11)(2)

%e (2)(11)(2)(11)

%e (2)(2)(11)(11)

%e (11)(11)(11)(2)

%e (11)(11)(2)(11)

%e (11)(2)(11)(11)

%e (2)(11)(11)(11)

%e (11)(11)(11)(11)

%e (1)(1)(1)(1)(1)(1)(1)(1)

%t Table[Sum[k^2^(n-k),{k,n}],{n,12}]

%Y Cf. A000123, A001970, A002577, A006171, A279787, A279789, A305551, A306017, A319056, A323766, A323774, A323776.

%K nonn

%O 1,2

%A _Gus Wiseman_, Jan 27 2019