Tough Sequence Equation
- Thread starter kaizar
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Must you find an "equation", or would a "formula" be enough?
Must you find an "equation", or would a "formula" be enough?
My fault, I'm seeking for a formula for this sequence.
D
Deleted member 4993
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There are several methods - what method/s have you been taight?
Which method/s did you apply?
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Plus two, plus three, plus three; plus two, plus three, plus three; ....
Plus two, plus three, plus three; plus two, plus three, plus three; ....
No kidding? Seriously sir... I could figure this out myself...
I'm not sure if it's a 'formula' or an 'expression' that I want, to represent this sequence containing 'n' in it where 'n' represents an integer starting from 1 and onwards.
Example:
2,5,10,17...
is (n squared + 1), a quick question, does this represent Geometric Sequence?
but I got no idea what is the 'formula' or 'expression' for this sequence:
4,6,9,12,14
where it increases by: 2,3,3 repeatedly
All I want is a formula/expression for this sequence containing 'n'.
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D
Deleted member 4993
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No the Joke is that you are not responding to our querries
what method/s have you been taught?
Which method/s did you apply?
You can always define polynomial such that a line will pass through those points within some tolerance.
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One way to express it is by recursion:
\(\displaystyle \displaystyle a_1 = 4\)
\(\displaystyle \displaystyle a_2 = 6\)
\(\displaystyle \displaystyle a_3 = 9\)
\(\displaystyle \displaystyle a_n = a_{n-3} + 8\), for \(\displaystyle n>3\)
Another way to get it would be to use a function that gives the integer part of a division. I have seen the functiion "FLOOR" used in that way, but i don't know what language that is from. In Fortran (my language of choice!), the function is called INT, for "integer part of":
\(\displaystyle \displaystyle a_n = \mathrm{INT}\left[\dfrac{n\times 8 + 4}{3}\right]\)
Great! thanks alot for your help. ^^
That's what I was looking for.
@Subhotosh Khan
Thanks for answering my quick question ^^
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Specifically, it is sequence A047408 at the bottom of the page at that link (if the pattern that stapel hinted at continues). <------ Edit. The grammar was fixed.
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\(\displaystyle Let\ \ F_n \ = \ the \ \ nth \ \ Fibonacci \ \ number, \ \ and \ \ let \ \ a_n \ = \ the \ \ nth \ \ term \ \ of \ \ the \ \ OP's \ \ sequence.\)
\(\displaystyle Then \ \ F_{n + 2} - 1 \ = \ \) one less than the (n + 2)th Fibonacci number.
(The brackets below are the usual brackets.)
\(\displaystyle a_n \ = \ \dfrac{1}{2}\bigg[ \ 1 \ + \ (-1)^{( \ F_{n + 2} \ - \ 1 \ )}\bigg] \ + \ 2\) \(\displaystyle \ \ \ \ \ \ \) <---- Edit: I left off the "+2."
\(\displaystyle Then \ \ F_{n + 2} - 1 \ = \ \) one less than the (n + 2)th Fibonacci number.
(The brackets below are the usual brackets.)
\(\displaystyle a_n \ = \ \dfrac{1}{2}\bigg[ \ 1 \ + \ (-1)^{( \ F_{n + 2} \ - \ 1 \ )}\bigg] \ + \ 2\) \(\displaystyle \ \ \ \ \ \ \) <---- Edit: I left off the "+2."
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HallsofIvy
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Do you understand that this is exactly the answer that Stapel gave you? You originally said that you wanted to know how the sequence continued and that was exactly what Stapel gave.