|
|
A090805
|
|
A simple recurrence with one error.
|
|
3
|
|
|
1, 2, 6, 21, 85, 430, 2586, 18109, 144880, 1303929, 13039300, 143432311, 1721187744, 22375440685, 313256169604, 4698842544075, 75181480705216, 1278085171988689, 23005533095796420, 437105128820131999, 8742102576402640000, 183584154104455440021, 4038851390298019680484
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
I included this in the OEIS only because was published on a web page. The explanation is my own - perhaps the original proposer had a different explanation.
|
|
REFERENCES
|
Found on a puzzle page.
|
|
LINKS
|
|
|
FORMULA
|
a(0) = 1; a(n) = n*(a(n-1) + 1) but make an error if n = 4.
Hans Havermann points out that the first 7 terms could also be produced by the recurrence f[x] = f[x - 1]*(x - 1) + GCD[3*f[x - 1], (x - 1)] with f[1] = 1. (This gives the continuation 1, 2, 6, 21, 85, 430, 2586, 18103, 144825, 1303434, 13034342, ...) But given the nature of the other problems on this quiz, I think my explanation is more likely.
|
|
EXAMPLE
|
1..add.1..multiply.by 1 -> 2
2..add.1..multiply.by 2 -> 6
6......1............. 3 -> 21
21.....1............. 4 -> 88 but here you make a mistake and instead multiply by 4 and add 1, getting 85
85.....1............. 5 -> 430
430....1............. 6 -> 2586
etc
|
|
MAPLE
|
a:= proc(n) a(n):= n*a(n-1) + `if`(n=4, 1, n) end: a(0):= 1:
|
|
MATHEMATICA
|
a={1}; Do[n=Length[a]; a=Append[a, If[n==4, Last[a]n+1, (Last[a]+1)n]], 22]; a (* Jake L Lande, Jul 28 2024 *)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy,dumb
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|