# I cannot solve this pattern!

1,1,4,3,9,6,16,10,25,15

HINT: two different patterns are intertwined

BIGGER HINT: only look at every other number

This is a conjured up sequence not identifiable by any recursion formula. The first sequnece is
1 4 9 16 25 36 48... and the second the Fibonnaci sequence
1 3 6 10 15 21 28...

Their combination has no mathematical basis.

T chrwil, you're close, but the second one is actually the triangular numbers:
1, 1+2=3, 1+2+3=6, 1+2+3+4=10, etc,
The Fib. seq is 1,1,2,3,5,8,13...(the sum of the previous two terms.)
As for a mathematical basis, I wouldn't say there "can't" be one without suggesting some proof or conjecture.
Consider this for a definition of the seq.:
(1)If n is odd f(n) = ((n+1)/2)^2
(2)If n is even f(n) = ((n/2)((n/2)+1))/2
The seq. {f(n): n is a natural number} will generate this sequence with (1) and (2) defining it. Other definitions are possible too.
I'm not sure if this sequence can be written with a single rule of assignment though. I'm not entirely sure whether it can be written as a recursive seq. either. It is definitely contrived though.

So right you are, When I first looked at the sequence, the first things that came to mind were the Fibonacci sequence and the triangular numers. Clearly, I typed in the wrong one. As for a singe relationship covering the whole sequence, I doubt it. It appears to be one of the many concocted sequences devised over the years. But it is worth taking another ;ook at it.

After peering at your "hybrid" sequance a bit more I stand by my initial statement that no one formula will give you every nth term of the sequence. However, I did conjur up two obvious expressions for both the odd and even numbers of the sequ3nce. As the sequence proceeds:
n..1..2..3..4..5..6..7..8..9..10
N..1..2..4..3..9..6.16.10.25..15
The N's for the odd n's are simply n^2.
The N's for the even n's are
N = n(n+2)/8

That is as far as I have gotten so far. There is one other avenue I will pursue in my sleep tonight based on the following interesting nature of the numbers. Consider the difference between each successive pairs of numbers:

n.......1..2..3..4..5..6..7..8..9..10
N.......1..1..4..3..9..6.16.10.25..15
Odd Diff.0.....-1....-3...-6....-10
Ev Diff..,..3.....6....10....15.....21

Do the numbers look familiar? Yes! Our famous Triangular numbers again, defined by Tn = n(n + 1)/2.

Might the sequence of the same numbers in both differences, despite their opposite signs, be made to conform to one formula? Tune in later for further developments..

Out of the many expressions I was playing with, I typed in the wrong one for the odd nubers. My apologies.

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Posted by Tchrwill on Saturday, August 26, 2006 at 10:43pm.

After peering at your "hybrid" sequance a bit more I stand by my initial statement that no one formula will give you every nth term of the sequence. However, I did conjur up two obvious expressions for both the odd and even numbers of the sequ3nce. As the sequence proceeds:
n..1..2..3..4..5..6..7..8..9..10
N..1..2..4..3..9..6.16.10.25..15
The N's for the odd n's are
[(n+1)/2]^2. (sorry)
The N's for the even n's are
N = n(n+2)/8

That is as far as I have gotten so far. There is one other avenue I will pursue in my sleep tonight based on the following interesting nature of the numbers. Consider the difference between each successive pairs of numbers:

n.......1..2..3..4..5..6..7..8..9..10
N.......1..1..4..3..9..6.16.10.25..15
Odd Diff.0.....-1....-3...-6....-10
Ev Diff..,..3.....6....10....15.....21

Do the numbers look familiar? Yes! Our famous Triangular numbers again, defined by Tn = n(n + 1)/2.

Might the sequence of the same numbers in both differences, despite their opposite signs, be made to conform to one formula? Tune in later for further developments..

where is the appplication of fibonacci numbers.