How would i describe this mathematically?

Question

Hi,

Here's what Im trying to describe, Ive written it out in steps to be as clear as possible:

i) I've taken the numbers n = 1,2,3,...,10.

ii) Then Ive split them into 2 sets, each with 5 numbers.i.e.,

A = {1,2,3,4,5}
B = {6,7,8,9,10}

iii) Then I've summed up each consecutive term in each set. i.e.,

Taking set A we get:

(1 + 2) = 3, (2+3) = 5, (3+4) = 7, (4+5) = 9

which gives another set C = {3,5,7,9}
[for set B the resulting set is D = {13,15,17,19}]

iv) Then I've summed each consecutive term in sets C and D, i.e.,

Taking set C only we get:

(3+5) = 8, (5+7) = 12, (7+9) = 16

which gives the set E = {8,12,16}
[for set D we get set F = {28,32,36}]

v) And again Ive summed the consecutive terms in sets E and F to get another 2 sets G and H:

G = {20, 28}
H = {60, 68}

vi) and once more ive summed the remaining terms to get 2 more sets:

I = {48}
J = {128}

This in sequence is just: 48,128

vii) Now if n = 1,2,3,...,20
and if we follow the above method for this set of values for n we end up with the sequence:

48, 128, 208, 288

This is equivalent to: 80 x − 32

What Im trying to do is describe the steps (i) - (vii) in mathematical terms?

Puzzling · Accepted Answer

Let's take your numbers to be a1, a2, a3, a4, a5

First you add them in pairs:
a1+a2, a2+a3, a3+a4, a4+a5

Then you add them in pairs again:
(a1 + 2a2 + a3). (a2 + 2a3 + a4), (a3 + 2a4 + a5)

Then again:
(a1 + 3a2 + 3a3 + a4), (a2 + 3a3 + 3a4 + a5)

Then again:
(a1 + 4a2 + 6a3 + 4a4 + 1a5)

Additionally, you have an arithmetic sequence where an = a(n-1) + 1

So we can equate this to:
a1 + 4(a1+1) + 6(a1+2) + 4(a1+3) + a1+4

Expanding this out further we have:
a1 + 4a1 + 4 + 6a1 + 12 + 4a1 + 12 + a1 + 4
= 16a1 + 32

You are doing this again for the next sequence of 5 numbers (6, 7, 8, 9, 10)
= 16a6 + 32

And again for (11, 12, 13, 14, 15)
= 16a11 + 32

And again for (16, 17, 18, 19, 20)
= 16a20 + 32

Let's figure out a way to compute the first digit in each group of 5.
Group 1 --> 1
Group 2 --> 6
Group 3 --> 11
Group 4 --> 16

If you are familiar with arithmetic sequences, this is:
= 1 + 5(n - 1)
And that can be simplified to:
= 5n - 4

Now plug that into the original formula:
= 16(5n - 4) + 32
= 80n - 64 + 32
= 80n - 32

That's exactly what you have done.

Grid · Answer

Just describe the general algorithm.

For instance, given a set of numbers n = 1 , ... , K partition into two subsets {1,...floor(K/2)} and {floor(K/2) + 1,...K} where it would be two equal subsets for even K and I arbitrarily decided to give the second subset the extra term for odd K.

Then proceed to explain the algorithm that at each step we add pairwise consecutive terms.

We terminate the algorithm when we reach sets consisting of one term.

If you actually run your algorithm for a general K (make it even for simplicity), you'll come up with a formula for a general n = 1 ,... K.

The description of the algorithm and the math are inseparable.

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How would i describe this mathematically?

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