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Formula for a pattern

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Formula for a pattern

Can you come up with a formula for the following sequence of numbers:

1 , 3 , 5 , 11 , 21, 43 , 85 , 171 , ...


Can you help me with my previous message on "Gibonacci" numbers ?

Putting the members on the number line, it's clear by http://planetmath.org/encyclopedia/NestedIntervalTheorem.html that the limit of the sequence exists. That it is 2/3, I cannot yet show. You may succeed...

I think in Sloane's register (http://www.research.att.com/~njas/sequences/) you will find several fitting formulas.

Yes, i can help.
the sequence is constant from after the 171 so all the terms
from then on are 171.

Hi Zee,
As pahio recommended you, the natural way is going to Sloan page. Unfortunately your sequence it is not reported there. Why? Because one may create infinite sequences. Now, your sequence is a bit tricky and also some depth. In finding out the law governing a sequence it is useful investigate its difference (a_{n+1}-a_n). Let me try to help you a little. (every below row there corresponds to the difference of the next above one)
1st row: 1, 3, 5, 11, 21, 43, 85, 171, (341), (683), ... (original sequence)
2nd row: 2, 2, 6, 10, 22, 42, 86, (170),(342), ...
3rd row: 0, 4, 4, 12, 20, 44,(84),(172), ...
4th row: 0,0, 8, 8, 24, (40), (88), ...
5th row: 0, 0, 0, 16, 16,(48), ... (this row is conjectured, see below)
-------------------------------------------------------------------
Well, now we may see the structure of your sequence. From the second row onwards you get the new sequence 2^1, 2^2, 2^3, ..., and these powers are repeated in every 'difference row', hence 2^4 must be appear repeated on the 5th row and so 16+24=(40) must be contained on the 4th row. But not just only that; from where does arise that term (48) on the last showed row? Notice that the next term following the powers of 2, at every difference row!, can be obtained by multiplying those repeated power of 2 by 3. Thus,
2^1.3=6, 2^2.3=12, 2^3.3=24,
so
2^4.3=(48),... (a new term on 5th row!)
Furthermore, now one can be returned up until reaching the original sequence getting new terms (and on each difference row too!). That is,
(48)+(40)=(88) (new term on the fourth row)
(40)+44=(84), and (88)+(84)=(172) (new terms on the third row)
(84)+86=(170), and (172)+(170)=(342) (new terms on the second row)
(170)+171=(341), and (342)+(341)=(683) (new terms on the original sequence)
So I hope with this brief explanation you may start working out to find a general formula to your sequence. A bit hard? Yeah, but not so much! Okay, I'll give you a hint: On (n+1)th row, that one having (n-1)zeros, you will have
0, ...(n-1)...0, 2^n, 2^n, (3.2^n), ...
But now you know how to descend from the top of this mountain.
perucho

Dear Zee,
I have not gone through previous replies.

Here: http://www.research.att.com/~njas/sequences/index.html?q=1+%2C+3+%2C+5+%...

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