|
Search: id:A075473
|
|
|
| A075473 |
|
n^n + n! - 2*(n+1)^(n-1). |
|
+0
8
|
|
| 0, 0, 0, 1, 30, 653, 13762, 304295, 7251598, 187783369, 5287733418, 161516858963, 5332258661782, 189493508862461, 7219703867130466, 293780009979371503, 12721918893479808030, 584361555380576356625, 28385640762100638931546
(list; graph; listen)
|
|
|
OFFSET
|
0,5
|
|
|
COMMENT
|
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.
|
|
FORMULA
|
a(n) =A000312(n)+A000142(n)-2*A000272(n+1)
|
|
EXAMPLE
|
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.
|
|
CROSSREFS
|
Adjacent sequences: A075470 A075471 A075472 this_sequence A075474 A075475 A075476
Sequence in context: A001777 A136661 A111779 this_sequence A051563 A027475 A035520
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Henry Bottomley (se16(AT)btinternet.com), Oct 11 2002
|
|
|
Search completed in 0.002 seconds
|
