## Number spiral

(FB/Math/BM – Problem by Mauri Ericson Sombowadile)

Consider the number spiral:

37 36 35 34 33 32 31
38 17 16 15 14 13 30
39 18 ..5 ..4 .3 12 29
40 19 ..6 ..1 .2 11 28
41 20 ..7 ..8 .9 10 27
42 21 22 23 24 25 26
43 44 45 46 47 48 49

Taking numbers in a straight line, we generate a sequence such as
…, 41, 20, 7, 8, 9, 10, 27, … or 1, 3, 13, 31, …

Formulas for sequences of this type?

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I think I can find a formula for the sequence 1, 3, 13, 31, … , i e numbers on the 45 degree line from the starting point 1 upwards to the right.

First let a(n) denote the number at the nth corner (including the starting point 1), counting from 1:

a(1) = 1, a(2) = 2, a(3) = 3, a(4) =5, a(5) = 7, a(6) = 10, a(7) = 13, a(8) = 17,…
1st differences b(n) = a(n) – a(n-1), n ≥ 2: 1, 1, 2, 2, 3, 3, 4, …
2nd differences b(n) – b(n-1), n ≥ 3: 0,1,0,1,0,1,…

This is rather easily seen from the figure; main point: The lengths of the segments between corners increase by 1 at every other corner.

b(n) = (1 + (-1)n)

Now, b(n) = a(n) – 2a(n-1) + a(n-2), so we get a difference equation:

(*) a(n) – 2a(n-1) + a(n-2) = (1 + (-1)n)

The solution satisfying a(1) = 1 and a(2) = 2 is

(**) a(n) = (7 + 2n² + (-1)n) / 8

I will not go inte the details here since it would be too long; in case you are not aquainted with difference equations (they are usually not in the basic mathematics courses as far as I know), the subject can be found in text books on numerical analysis or discrete mathamatics. – In contrast to differential equations it easy to check the solution: Obviously a(n) + 2a(n-1) – a(n-2) + ( 1 + (-1)n ), gives unique values for a(n) for n > 4, given a(3) =3 and a(4) = 5. These two values agree with the formula (**). Further (**) satifies the difference equation (*). Thus (**) gives all a(n) for n ≥ 3. It is also easily seen that (**) in fact gives all a(n) for n ≥ 1.

Now the numbers in the sequence 1, 3, 13, 31, … lie on the 1st, 3 rd, 7nd and then on each fourth corner, so these numbers are given by t(k) = a(4k-5), for k ≥ 2 (except for the first one). In short:

The terms in the sequence 1, 3, 13, 31, … are given by t(1) = 1, and for k ≥ 2,
t(k) = (7 + 2(4k-5)² + (-1)4k-5) / 8 = (32k² – 80k + 56) / 8.

So, t(1) = 1, and t(k) = 4k² – 10k + 7 for k ≥ 2.

t(1) = 1
t(2) = 4·4 – 10·2 + 7 = 3
t(3) = 4·9 – 10·3 + 7 = 13
t(4) = 4·16 – 10·4 + 7 = 31
t(5) = 4·25 – 10·5 + 7 = 57, etc

For the other ”45 degree rays” you can construct formulas using a(n) and substitute something other than 4k-5 for n.
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Correction:
a(n) + 2a(n-1) – a(n-2) + (1 + (-1)n)
5-6 lines below the formula (**) should be
a(n) = 2a(n-1) – a(n-2) + (1 + (-1)n)

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