Question

For n ≥ 3, the sequence of centred n-gon numbers is found by starting with a central dot, then adding layers consisting of n-gons of dots around this centre, where the number of dots on each side increases by 1 for each layer. For instance, the sequence of centred 7-gon numbers starts 1, 8, 22, 43,... What is the smallest n for which 2017 is in the sequence of centred n-gon numbers?

Answer 1

if n = 7 we get the sequence 1,8,22,43,....
which can be considered a recursive sequence where
t(1) = 1
t(k+1) = t(k) + 7k , where k > 1 and k ∊ N
in such a sequence it is easy to establish that the 2nd sequence is constant, so
we can express the sequence as a quadratic function.
let f(x) = ax^2 + bx + c
using the first 3 points we get:
a+b+c = 1
4a + 2b + c = 8
9a + 3b + c = 22
this is easy to solve and I found
f(x) = (7/2)x^2 - (7/2)x + 1 = (7/2)(x^2 - x) + 1
checking: f(4) = (7/2)(16-4) + 1 = 43

Notice we got f(x) = (7/2)(x^2 - x) + 1 , when n = 7
We can establish that if n = k, f(x) = (k/2)(x^2 - x) + 1
e.g. when n = 4, we would get f(x) = (4/2)(x^2 - x) + 1

We need to find n so that
(n/2)(x^2 - x) + 1 = 2017
(n/2)(x^2 - x) = 2016, for integer values of n and x

We will probably need to use "technology" to find such values, unless you are very
good at guessing.
I have this very old computer program from the early 80's and I set up the following:
10 FOR N = 1 TO 800
20 FOR X = 1 TO 50
30 Y=(N/2)*(X*X-X)
35 IF Y > 2016 THEN 60
40 IF Y = 2016 THEN PRINT N;X,
50 NEXT X
60 NEXT N

This gave me: 56,9 72,8 96,7 336,4 and 672,3

so it looks like <b>the smallest n value is 56</b> and the 9th term would be 2017
check:
term(1) = 1
term(2) = 1 + 1(56) = 57
term(3) = 57 + 2(56) = 169
term(4) = 169 + 3(56) = 337
term(5) = 337 + 4(56) = 561
term(6) = 561 + 5(56) = 841
term(7) = 841 + 6(56) = 1177
term(8) = 1177 + 7(56) = 1569
term(9) = 1569 + 8(56) = 2017 <----- YEAHHH

Answer 2

Interesting solution, Reiny.
I used a similar approach to acquire the equation, but then put it in standard form:
nx ² - nx - 4032 = 0
Then I applied the quadratic formula to get:
x = [n ± √(n ² + 16128n)]/2n
I divided out a factor of n:
x = [1 ± √(1 + 16128/n)]/2
Since x is the term position, then x ∈ N. And, since n is the number of sides of a polygon, then n ∈ N | n ≥ 3.
I noted that the discriminant must be a perfect square, with an odd square root, to satisfy x ∈ N.
I entered just the discriminant function into a graphing calculator:
Y = √(1 + 16128/X)
Then I viewed the resulting table of values, starting at X = 3, and looked for the first result for Y that was a an odd natural number. This occurred at X = 56 (i.e. n = 56).