What are the next two numbers in the following sequence:

0, 1, 4, 18, __, __.

Please explain your logic.


closed as too broad by MikeQ, NL628, athin, Peregrine Rook, ffao May 12 '18 at 5:24

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  • 9
    $\begingroup$ There are 6 pages of results on the OEIS for this sequence, including f(n) = n*n!, f(n) = n^2*fibonacci(n), f(n) = lcm(n^2, n!), and the recursive definition a(n) = 4*a(n-1) + 2*a(n-2), a(0) = 0, a(1) = 1. This is too broad. $\endgroup$ – Eric Tressler May 12 '18 at 0:07
  • $\begingroup$ Yes, and next time you come up with a similar number puzzle, please tell us if it is polynomial or not. I could find a quartic, quintic, sextic, septic, etc. polynomial that satisfies the above conditions. $\endgroup$ – NL628 May 12 '18 at 0:47
  • $\begingroup$ Agreed, @Eric Tressler and NL628. I could have applied no-computers tag, though. $\endgroup$ – Mea Culpa Nay May 12 '18 at 1:08

I think the next two numbers are:

96 and 600

I think the pattern is:

Starting with $n=0$, the $n$th term is $n*n!$. The factorial term suggests an interpretation of the title: the exclamation mark is also a mark of surprise.
$0*0! = 0*1 = 0$
$1*1! = 1*1 = 1$
$2*2! = 2*2 = 4$
$3*3! = 3*6 = 18$
$4*4! = 4*24 = 96$
$5*5! = 5*120 = 600$
EDIT: If you don't like zero-indexed sequences, this could also start with $n=1$ and with $n$th term being $(n-1)*(n-1)!=n!-(n-1)!$. In this way, it's clear that these are the differences between successive factorials $1,1,2,6,24,120,720,...$

  • $\begingroup$ That is correct. Too quick. Another interpretation of the terms of the sequence is n! - (n-1)!. $\endgroup$ – Mea Culpa Nay May 12 '18 at 0:15
  • $\begingroup$ I was just updating my answer to add another interpretation, actually. I think it's the one you mention, but the last part of your comment is cut off for me. EDIT: Ah, there's the rest of the comment now. I think we're playing the comment equivalent of Phone Tag at this point $\endgroup$ – ManyPinkHats May 12 '18 at 0:18

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