Find the first 6 numbers (bis)?

Question

A sequence of integers is defined for n=1,2,3,....
by:

a_n= n + (n-1)(n-2)(n-3)(n-4)(n-5)(c-6)/120

where c is any given integer (positive or negative). What are
the 6 first terms of the sequence?

Answer 1

I didn't catch your 'c' the first time I read your question. I read it as n.

The sequence would be 1, 2, 3, 4, 5, c

a(1) = 1 + (0)(-1)(-2)(-3)(-4)(c-6) / 120 = 1 + 0/120 = 1
a(2) = 2 + (1)(0)(-1)(-2)(-3)(c-6) / 120 = 2 + 0/120 = 2
a(3) = 3 + (2)(1)(0)(-1)(-2)(c-6) / 120 = 3 + 0/120 = 3
a(4) = 4 + (3)(2)(1)(0)(-1)(c-6) / 120 = 4 + 0/120 = 4
a(5) = 5 + (4)(3)(2)(1)(0)(c-6) / 120 = 5 + 0/120 = 5
a(6) = 6 + (5)(4)(3)(2)(1)(c-6) / 120 = 6 + 120(c-6)/120 = 6 + c-6 = c

As you have shown, with any sequence of numbers there is a way to form a polynomial of the correct magnitude that will make the next number in the sequence be anything you want.

That's why you can always argue against someone's answer of "what's the next number in the sequence 1, 2, 3, 4, 5..." by supplying your generating function above or infinitely many variations on the formula to get any result you like.

However, if we apply Occam's Razor to the problem, I would think the most reasonable answer is 6, not some arbitrary answer that can be generated by a complicated 5th order polynomial. :-)

Answer 2

a(n) = n + (n-1)(n-2)(n-3)(n-4)(n-5)(c-6)/120

a(1) = 1 + (0)(....)/120 = 1

a(2) = 2 + (1)(0)(....)/120 = 2

a(3) = 3 + (2)(1)(0)(...)/120 = 3

a(4) = 4 + (3)(2)(1)(0)(...)/120 = 4

a(5) = 5 + (4)(3)(2)(1)(0)(c-6)/120 = 5

a(6) = 6 + (5)(4)(3)(2)(1)(c-6)/120 = 6+((c-6))/(120) = 6 + (c-6)/120 = 5.95 + c/120

Answer 3

0,1,2,3,4,5 and 6+(c-6)/120