Could someone please help with a no sequence

[ShadowUK]

I'm stuck with a number sequence.......

I'm trying to find a formula for the sequence 4, 10, 20, 35, 56.......
I've tried everything and haven't got anywhere.

I'm sure someone in here has to be cleverer than me.....

If anyone has any ideas it would be appretiated. Thanks a lot.

[wossname]

No problemo!

Where X is the number of the item in the sequence...

Item = .1666666666X^3 + X^2 + 1.83333333X + 1
so to get the 6th term:

.1666666666(6)^3 + (6)^2 + 1.83333333(6) + 1

More or less, thats a good approximation.

[transcendental]

2*2 +0 = 4
3*3 +1 = 10
4*4 +4 = 20
5*5 +10= 35
6*6 +20= 56

See the pattern above and below

4= k=2:k^2 + (k-2)^2
10= k=3:k^2 + (k-2)^2
20= k=4:k^2 + (k-2)^2 + (k-2*2)^2
35= k=5:k^2 + (k-2)^2 + (k-2*2)^2
56= k=6:k^2 + (k-2)^2 + (k-2*2)^2 + (k-2*3)^2

Formula for each term:

 B=k/2
 SUM   (k-2*B)^2
 B=0

Formula for the sequence

| B=k/2            |Inf
| SUM   (k-2*B)^2  |
| B=0              |k=2

Note: B=k/2 (rounded down)

[NotLKH]

The full formula ends up:

Y = 4 + [(x^3 + 9*x^2 + 26*x)/6]

and x = 0 thru 4 returns your sequence.

I don't know the name of the method, but given c0, c1, c2, c3, c4, ...
you can develop an equation by plugging them into :

Y = c0 + c1*X + c2*X*(X-1)/2! + c3*X*(X-1)*(X-2)/3! + ...

where c0, c1, c2, c3 are derived from known values of Y, such that: