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A075473 n^n + n! - 2*(n+1)^(n-1). 4

%I #3 Mar 30 2012 18:51:39

%S 0,0,0,1,30,653,13762,304295,7251598,187783369,5287733418,

%T 161516858963,5332258661782,189493508862461,7219703867130466,

%U 293780009979371503,12721918893479808030,584361555380576356625,28385640762100638931546

%N n^n + n! - 2*(n+1)^(n-1).

%C Possibly the number of patterns for forming an ordered sum of n values v1+v2+...+vn chosen (possibly with repetition) from {b1,b2,...,bn} with b1<b2<...<bn, where the question of whether v1+v2+...+vn is greater than, less than or equal to b1+b2+...+bn depends of the values of {b1,b2,...,bn}: for example with n=3 b2+b2+b2 is not immediately clear, while with n=4 there are 30 unclear possibilities namely b2+b2+b2+b2, b3+b3+b3+b3, 6 permutations of b1+b1+b4+b4, 6 permutations of b2+b2+b3+b3, 4 permutations of b2+b2+b2+b3, 4 permutations of b2+b3+b3+b3, 4 permutations of b2+b2+b2+b4 and 4 permutations of b1+b3+b3+b3.

%F a(n) =A000312(n)+A000142(n)-2*A000272(n+1)

%e a(2)=2^2+2!-2*3^1=4+2-6=0. a(3)=3^3+3!-2*4^2=27+6-32=1. a(4)=4^4+4!-2*5^3=256+24-250=30.

%K nonn

%O 0,5

%A _Henry Bottomley_, Oct 11 2002

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