# What is the formula for this sequence: 1,4,3,8,5,16?

Help!

Relevance

Don't know the formula, but

Second number (4) minus First number (1) = Third Number (3)

First, Second, and Third numbers added equal 8. (1+4+3 = 8)

Keep going, 2 numbers minus then 3 numbers plus. (example below)

8 - 3 = 5

3+8+5 = 16

So next one would be

16 - 5 = 11

5+16+11 = 32

....and on and on

hope that helps

THE formula?

There are infinitely many.

One of which is:

f(x) = (1/2)x^5 - 8.5x^4 + (54 1/6)x^3 - 159.5x^2 + (214 1/3)x - 100

mrs.mkt403 gives a valid recursive RULE, but its not a formula. you'd be hard pressed to find, say, the 10000th value of the sequence with it.

[(1 + (–1)ⁿ]/2

For odd numbers

[(1 + (–1)ⁿ]/2 = [(1 – 1)]/2 = 0

[(1 + (–1)ⁿˉ¹]/2 = [(1 + 1)]/2 = 1

For even numbers

[(1 + (–1)ⁿ]/2 = [(1 + 1)]/2 = 1

[(1 + (–1)ⁿˉ¹]/2 = [(1 – 1)]/2 = 0

General formula

t(n) = n[(1 + (–1)ⁿˉ¹]/2 + 2^(n/2 + 1)[(1 + (–1)ⁿ]/2

For odd

t(n) = n(1) + 2^(n/2 + 1)(0)

for even

t(n) = n(0) + 2^(n/2 + 1)(1)

t(1) = 1(1) + 2^(1/2 + 1)(0) = 1

t(2) = 2(0) + 2^(2/2 + 1)(1) = 4

t(3) = 3(1) + 2^(3/2 + 1)(0) = 3

t(4) = 4(0) + 2^(4/2 + 1)(1) = 8

t(5) = 5(1) + 2^(5/2 + 1)(0) = 5

t(6) = 6(0) + 2^(6/2 + 1)(1) = 16

t(7) = 7(1) + 2^(7/2 + 1)(0) = 7

t(8) = 8(0) + 2^(8/2 + 1)(1) = 32

t(9) = 9(1) + 2^(9/2 + 1)(0) = 9

t(10) = 10(0) + 2^(10/2 + 1)(1) = 64

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t(n) = n[(1 + (–1)ⁿˉ¹]/2 + 2^(n/2 + 1)[(1 + (–1)ⁿ]/2

t(2) = 2(0) + 2²(