4,4,5,7,9 - can anyone see the pattern, it’s driving me mad.

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Long story short, one answer amongst an infinity is: S(N) = -1/6 (N-1)^3 + 1/2 (N-1)^2 + 2/3 (N-1) + 4

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Well, you have a sequence which goes:

S(N) = 4, 4, 5, 7, 8 => s(n) = 0, 1, 2 ,2

(Here all caps means the original sequence, lower case means the difference sequence, s(n) = S(N) - S(N-1), n = N-1)

Solving the first condition leads to a sequence (one amongst an infinite number, anyways):

s(n) = (1 - a - b) n^3 - a n^2 + b n

Then we can solve the others via simple algebra:

(1 - a -b)*8 + a*4 + b*2 = 2 ==> -6 = -4*a - 6*b ==> a = 3/2( 1 - b )
(1 - a -b)*27 + a*9 + b*3 = 2 ==> -25 = -18*a -24*b ==> -25 = -27( 1 - b ) -24*b ==> b = 2/3

Therefore,

a = 1/2, b = 2/3

Ergo,

s(n) = -1/6 n^3 + 1/2 n^2 + 2/3 n

i.e.
s(0) = 0
s(1) = 1
s(2) = -8/6 + 2 + 8/6 = 2
s(3) = -(27)/6 + 9/2 + 2 = 2

Now then, we invert the relations to find:

S(N) = s(N-1) + A, where S(0) = 4. Solving the initial condition yields:

S(0) ==> 4 = A + s(-1) = A+ 1/6 + 3/6 -4/6 = A ==> A = 4


Ergo,

S(N) = -1/6 (N-1)^3 + 1/2 (N-1)^2 + 2/3 (N-1) + 4


I could use Newton's binomial theorem to get this into a proper polynomial for N, but frankly my interest in the problem wanes.