My off the top of my head sequence. - Page 2

Uncwilly · 2004-08-13, 00:39

OK, post, spoilerised, you suggested key to the pattern.

jebeagles · 2004-08-13, 04:15

f(n)=f(n-1)*f(n-3) - f(n-2); for f(1)=3, f(2)=4, f(3)=5

axn · 2004-08-13, 04:54

f(n)=f(n-1)*f(n-2) - f(n-1) - f(n-2)

Uncwilly · 2004-08-13, 13:26

Ok, here is the way that I derived the sequence. axn1 has the correct sequence, but my formula is simpler.

n[sub]1[/sub]=3, n[sub]2[/sub]=4
n[sub]x[/sub] = (n[sub]x-1[/sub]*n[sub]x-2[/sub]) - (n[sub]x-1[/sub]+n[sub]x-2[/sub])

wblipp · 2004-08-13, 14:16

Quote:

Originally Posted by Uncwilly

Ok, here is the way that I derived the sequence. axn1 has the correct sequence, but my formula is simpler.

Do you really think so?
The only difference is that your expression adds two pairs of parentheses. I don't see how that counts as simpler.

tom11784 · 2004-08-13, 16:38

something else about this sequence:

Each term is the largest number that cannot be composed by a linear combination of a non-negative number
of each of the previous two terms (holds at least until 11 39 379)

I figured this part out back in middle school after a question at the MathCounts countdown
was what is the largest number that cannot be created by adding 3s and 4s.

also, could an admin please erase my post above this as the spoiler seems to have made my message on one line only

2004-08-13, 16:38	#17
tom11784 Aug 2003 Upstate NY, USA 326₁₀ Posts	something else about this sequence: Each term is the largest number that cannot be composed by a linear combination of a non-negative number of each of the previous two terms (holds at least until 11 39 379) I figured this part out back in middle school after a question at the MathCounts countdown was what is the largest number that cannot be created by adding 3s and 4s. also, could an admin please erase my post above this as the spoiler seems to have made my message on one line only Last fiddled with by tom11784 on 2004-08-13 at 16:39

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2004-08-13, 00:39	#12
Uncwilly 6809 > 6502 """"""""""""""""""" Aug 2003 101×103 Posts 10011000111010₂ Posts	OK, post, spoilerised, you suggested key to the pattern.

2004-08-13, 04:15	#13
jebeagles Jun 2004 Chicago 1C₁₆ Posts	f(n)=f(n-1)*f(n-3) - f(n-2); for f(1)=3, f(2)=4, f(3)=5

2004-08-13, 04:54	#14
axn Jun 2003 2²·3·421 Posts	f(n)=f(n-1)*f(n-2) - f(n-1) - f(n-2)

2004-08-13, 13:26	#15
Uncwilly 6809 > 6502 """"""""""""""""""" Aug 2003 101×103 Posts 9786₁₀ Posts	Ok, here is the way that I derived the sequence. axn1 has the correct sequence, but my formula is simpler. n[sub]1[/sub]=3, n[sub]2[/sub]=4 n[sub]x[/sub] = (n[sub]x-1[/sub]*n[sub]x-2[/sub]) - (n[sub]x-1[/sub]+n[sub]x-2[/sub])